# 5.1 Module 1: Rates, Equilibrium and ph

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 5.1 Module 1: Rates, Equilibrium and ph How Fast? The rate of reaction is defined as the change in concentration of a substance in unit time Its usual unit is mol dm 3 s 1 When a graph of concentration of reactant is plotted vs time, the gradient of the curve is the rate of reaction. The initial rate is the rate at the start of the reaction where it is fastest Reaction rates can be calculated from graphs of concentration of reactants or products concentration Initial rate = gradient of tangent time Rate Equations The rate equation relates mathematically the rate of reaction to the concentration of the reactants. For the following reaction, aa + bb products, the generalised rate equation is: r = k[a] m [B] n r is used as symbol for rate The unit of r is usually mol dm 3 s 1 m, n are called reaction orders Orders are usually integers 0,1,2 0 means the reaction is zero order with respect to that reactant 1 means first order 2 means second order NOTE: the orders have nothing to do with the stoichiometric coefficients in the balanced equation. They are worked out experimentally The square brackets [A] means the concentration of A (unit mol dm 3 ) k is called the rate constant The total order for a reaction is worked out by adding all the individual orders together (m+n) Calculating orders from initial rate data initial For zero order: the concentration of A has no effect on the rate of reaction r = k[a] 0 = k For first order: the rate of reaction is directly proportional to the concentration of A r = k[a] 1 For second order: the rate of reaction is proportional to the concentration of A squared r = k[a] 2 Graphs of initial rate against concentration show the different orders. The initial rate may have been calculated from taking gradients from concentration /time graphs For a rate concentration graph to show the order of a particular reactant the concentration of that reactant must be varied whilst the concentrations of the other reactants should be kept constant. 1

2 Continuous rate experiments [A] The rate constant (k) If halflives are constant then the order is 1 st order t ½ t ½ t ½ Time (min) Continuous rate data This is data from one experiment where the concentration of a substance is followed throughout the experiment. This data is processed by plotting the data and calculating successive halflives. The halflife of a firstorder reaction is independent of the concentration and is constant If halflives rapidly increase then the order is 2nd order 1. The units of k depend on the overall order of reaction. It must be worked out from the rate equation 2. The value of k is independent of concentration and time. It is constant at a fixed temperature. 3. The value of k refers to a specific temperature and it increases if we increase temperature For a 1 st order overall reaction the unit of k is s 1 For a 2 nd order overall reaction the unit of k is mol 1 dm 3 s 1 For a 3 rd order overall reaction the unit of k is mol 2 dm 6 s 1 Example (first order overall) Rate = k[a][b] 0 m = 1 and n = 0 reaction is first order in A and zero order in B overall order = = 1 usually written: Rate = k[a] Calculating units of k Remember: the values of the reaction orders must be determined from experiment; they cannot be found by looking at the balanced reaction equation 1. Rearrange rate equation to give k as subject k = Rate [A] 2. Insert units and cancel k = mol dm 3 s 1 mol dm 3 Unit of k = s 1 Example: Write rate equation for reaction between A and B where A is 1 st order and B is 2 nd order. r = k[a][b] 2 overall order is 3 Calculate the unit of k 1. Rearrange rate equation to give k as subject 2. Insert units and cancel 3. Simplify fraction k = Rate [A][B] 2 k = mol dm 3 s 1 mol dm 3.(moldm 3 ) 2 k = s 1 mol 2 dm 6 Unit of k = mol 2 dm 6 s 1 2

3 Working out orders from experimental initial rate data Normally to work out the rate equation we do a series of experiments where the initial concentrations of reactants are changed (one at a time) and measure the initial rate each time. This data is normally presented in a table. Example: work out the rate equation for the following reaction, A+ B+ 2C D + 2E, using the initial rate data in the table Experiment [A] mol dm 3 [B] mol dm 3 [C] mol dm Rate mol dm 3 s 1 In order to calculate the order for a particular reactant it is easiest to compare two experiments where only that reactant is being changed If conc is doubled and rate stays the same: order= 0 If conc is doubled and rate doubles: order= 1 If conc is doubled and rate quadruples : order= 2 For reactant A compare between experiments 1 and 2 For reactant A as the concentration doubles (B and C staying constant) so does the rate. Therefore the order with respect to reactant A is first order For reactant B compare between experiments 1 and 3 : As the concentration of B doubles (A and C staying constant) the rate quadruples. Therefore the order with respect to B is 2 nd order For reactant C compare between experiments 1 and 4 : As the concentration of C doubles (A and B staying constant) the rate stays the same. Therefore the order with respect to C is zero order The overall rate equation is r = k [A] [B] 2 The reaction is 3 rd order overall and the unit of the rate constant =mol 2 dm 6 s 1 Working out orders when two reactant concentrations are changed simultaneously In most questions it is possible to compare between two experiments where only one reactant has its initial concentration changed. If, however, both reactants are changed then the effect of both individual changes on concentration are multiplied together to give the effect on rate. In a reaction where the rate equation is r = k [A] [B] 2 If the [A] is x2 that rate would x2 If the [B] is x3 that rate would x3 2 = x9 If these changes happened at the same time then the rate would x2x9= x 18 3

4 Example work out the rate equation for the reaction, between X and Y, using the initial rate data in the table Experiment Initial concentration of Initial concentration X/ mol dm 3 of Y/ mol dm 3 Initial rate/ mol dm 3 s x x x 10 6 For reactant X compare between experiments 1 and 2 For reactant X as the concentration doubles (Y staying constant) so does the rate. Therefore the order with respect to reactant X is first order Comparing between experiments 2 and 3 : Both X and Y double and the rate goes up by 8 We know X is first order so that will have doubled rate The effect of Y, therefore, on rate is to have quadrupled it. Y must be second order The overall rate equation is r = k [X] [Y] 2 The reaction is 3 rd order overall and the unit of the rate constant =mol 2 dm 6 s 1 Calculating a value for k using initial rate data Using the above example, choose any one of the experiments and put the values into the rate equation that has been rearranged to give k. Using experiment 3: r = k [X] [Y] 2 k = r [X] [Y] 2 k = 2.40 x x k = 3.0 x 10 4 mol 2 dm 6 s 1 Remember k is the same for all experiments done at the same temperature. Increasing the temperature increases the value of the rate constant k Increasing temperature increases the rate constant k. The relationship is given by the Arrhenius equation k = Ae Ea/RT where A is a constant R is gas constant and Ea is activation energy. k You do not need to learn or use the Arrhenius equation but be aware that the relationship between k and Temperature is not directly proportional but is shown on the graph to the right temperature 4

5 Rate Equations and mechanisms A mechanism is a series of steps through which the reaction progresses, often forming intermediate compounds. If all the steps are added together they will add up to the overall equation for the reaction Each step can have a different rate of reaction. The slowest step will control the overall rate of reaction. The slowest step is called the ratedetermining step. The molecularity (number of moles of each substance) of the molecules in the slowest step will be the same as the order of reaction for each substance. e.g. 0 moles of A in slow step would mean A is zero order. 1 mole of A in the slow step would mean A is first order Example 1 overall reaction A + 2B + C D + E Mechanism Step 1 A + B X + D slow Step 2 X + C Y fast Step 3 Y + B E fast r = k [A] 1 [B] 1 [C] o C is zero order as it appears in the mechanism in a fast step after the slow step Example 3 Overall Reaction NO 2 (g) + CO(g) NO(g) + CO 2 (g) Mechanism: Step 1 NO 2 + NO 2 NO + NO 3 Step 2 NO 3 + CO NO 2 + CO 2 NO 3 is a reaction intermediate r = k [NO 2 ] 2 slow fast NO 2 appears twice in the slow steps so it is second order. CO does not appear in the slow step so is zero order. Example 2 r = k [X]1 [C]1 Example 5: S N 1 or S N 2? You don t need to remember the details here. overall reaction A + 2B + C D + E Mechanism Step 1 A + B X + D fast Step 2 X + C Y slow Step 3 Y + B E fast The intermediate X is not one of the reactants so must be replaced with the substances that make up the intermediate in a previous step A + B X + D r = k[a] 1 [B] 1 [C] 1 Example 4 Using the rate equation rate = k[no] 2 [H 2 ] and the overall equation 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g), the following threestep mechanism for the reaction was suggested. X and Y are intermediate species. Step 1 NO + NO X Step 2 X + H 2 Y Step 3 Y + H 2 N 2 + 2H 2 O Which one of the three steps is the ratedetermining step? Step 2 as H 2 appears in rate equation and combination of step 1 and 2 is the ratio that appears in the rate equation. Remember the nucleophilic substitution reaction of haloalkanes and hydroxide ions. This is a one step mechanism H 3 C HO: H δ + δ C Br H CH 3 CH 2 Br + OH CH 3 CH 2 OH + Br slow step The rate equation is r = k [CH 3 CH 2 Br] [OH ] C H 3 H C H OH + :Br This is called S N 2. Substitution, Nucleophilic, 2 molecules in rate determining step The same reaction can also occur via a different mechanism Overall Reaction (CH 3 ) 3 CBr + OH (CH 3 ) 3 COH + Br Mechanism: (CH 3 ) 3 CBr (CH 3 ) 3 C + + Br (CH 3 ) 3 C + + OH (CH 3 ) 3 COH The rate equation is r = k [(CH 3 ) 3 CBr] slow fast This is called S N 1. Substitution, Nucleophilic, 1 molecule in rate determining step 5

6 5.1.2 How Far? Equilibrium Equilibrium constant Kc Kc = equilibrium constant For a generalised reaction ma + nb pc + qd [ C] p [D] Kc= q m,n,p,q are the stoichiometric balancing [ A] m [B] n numbers A,B,C,D stand for the chemical formula [ ] means the equilibrium concentration Example 1 N 2 (g) + 3H 2 (g) 2 NH 3 (g) [NH 3 (g) ] Kc= 2 [N 2 (g) ] [H 2 (g) ] 3 The unit of Kc changes and depends on the equation. Working out the unit of Kc Put the unit of concentration (mol dm 3 ) into the Kc equation [NH 3 (g) ] 2 Cancel out Kc= [mol dm3 ] 2 units 1 Unit = Unit = [N 2 (g) ] [H 2 (g)] 3 [mol dm 3 ] [mol dm 3 ] 3 [mol dm 3 ] 2 Example 2: writing Kc expression Working out the unit H 2 (g) +Cl 2 (g) 2HCl (g) [HCl Kc= (g) ] 2 [mol dm 3 ] Unit Kc= 2 [H 2 (g) ] [Cl 2 (g)] [mol dm 3 ] [mol dm 3 ] Unit = [mol dm 3 ] 2 Unit = mol 2 dm +6 = no unit Calculating Kc Most questions first involve having to work out the equilibrium moles and then concentrations of the reactants and products. Usually the question will give the initial amounts (moles) of the reactants, and some data that will help you work out the equilibrium amounts. Calculating the moles at equilibrium moles of reactant at equilibrium = initial moles moles reacted moles of product at equilibrium = initial moles + moles formed Example 3 For the following equilibrium H 2 (g) +Cl 2 (g) 2HCl (g) In a container of volume 600cm 3 there were initially 0.5mol of H 2 and 0.6 mol of Cl 2. At equilibrium there were 0.2 moles of HCl. Calculate Kc H 2 Cl 2 HCl Initial moles Equilibrium moles 0.2 It is often useful to put the mole data in a table. Using the balanced equation if 0.2 moles of HCl has been formed it must have used up 0.1 of Cl 2 and 0.1 moles of H 2 (as 1:2 ratio) Work out the moles at equilibrium for the reactants moles of hydrogen at equilibrium = = 0.4 H 2 Cl 2 HCl Initial moles Equilibrium moles moles of reactant at equilibrium = initial moles moles reacted moles of chlorine at equilibrium = = 0.5 If the Kc has no unit then there are equal numbers of reactants and products. In this case you do not have to divide by volume to work out concentration and equilibrium moles could be put straight into the kc expression Equilibrium concentration (M) Work out the equilibrium concentrations 0.4/0.6 = /0.6 =0.83 conc = moles/ vol (in dm 3 ) 0.2/0.6 =0.33 Finally put concentrations into Kc expression Kc= Kc [HCl (g) ] 2 [H 2 (g) ] [Cl 2 (g)] = x0.83 = no unit 6

7 Example4 For the following equilibrium N 2 (g) + 3H 2 (g ) 2 NH 3 (g) Initially there were 1.5 moles of N 2 and 4 mole of H 2, in a 1.5 dm 3 container. At equilibrium 30% of the Nitrogen had reacted. Calculate Kc N 2 H 2 NH 3 Initial moles Equilibrium moles 30% of the nitrogen had reacted = 0.3 x1.5 = 0.45 moles reacted. Using the balanced equation 3 x 0.45 moles of H 2 must have reacted and 2x 0.45 moles of NH 3 must have formed Work out the moles at equilibrium for the reactants and products moles of reactant at equilibrium = initial moles moles reacted moles of nitrogen at equilibrium = = 1.05 moles of hydrogen at equilibrium = x3 = 2.65 moles of product at equilibrium = initial moles + moles formed moles of ammonia at equilibrium = 0 + (0.45 x 2) = 0.9 N 2 H 2 NH 3 Initial moles Finally put concentrations into Kc expression Equilibrium moles Equilibrium concentration (M) 1.05/1.5 = /1.5 = /1.5 =0.6 Kc= [NH 3 (g) ] 2 [N 2 (g) ] [H 2 (g)] 3 Work out the equilibrium concentrations conc = moles/ vol (in dm 3 ) Kc = x = mol 2 dm +6 Effect of changing conditions on value of Kc The larger the Kc the greater the amount of products. If Kc is small we say the equilibrium favours the reactants Effect of Temperature on position of equilibrium and Kc Kc only changes with temperature. It does not change if pressure or concentration is altered. A catalyst also has no effect on Kc Both the position of equilibrium and the value of Kc will change it temperature is altered In this equilibrium which is exothermic in the forward direction N 2 (g) + 3H 2 (g ) 2 NH 3 (g) If temperature is increased the reaction will shift to oppose the change and move in the backwards endothermic direction. The position of equilibrium shifts left. The value of Kc gets smaller as there are fewer products. Effect of Concentration on position of equilibrium and Kc Changing concentration would shift the position of equilibrium but the value of Kc would not change. H 2 (g) +Cl 2 (g) 2HCl (g) Increasing concentration of H 2 would move equilibrium to the right lowering concentration of H 2 and Cl 2 and increasing concentration of HCl. The new concentrations would restore the equilibrium to the same value of Kc 7

8 Effect of Pressure on position of equilibrium and Kc The position of equilibrium will change it pressure is altered but the value of Kc stays constant as Kc only varies with temperature In this equilibrium which has fewer moles of gas on the product side N 2 (g) + 3H 2 (g ) 2 NH 3 (g) If pressure is increased the reaction will shift to oppose the change and move in the forward direction to the side with fewer moles of gas. The position of equilibrium shifts right. The value of Kc stays the same though as only temperature changes the value of Kc. Increasing pressure does not change Kc. The increased pressure increases concentration terms on bottom of Kc expression more than the top. The system is now no longer in equilibrium so the equilibrium shifts to the right increasing concentrations of products and decreases the concentrations of reactants. The top of Kc expression therefore increases and the bottom decreases until the original value of Kc is restored Kc= [NH 3 (g) ] 2 [N 2 (g) ] [H 2 (g)] 3 Effect of catalysts on position of equilibrium and Kc Catalysts have no effect on the value of Kc or the position of equilibrium as they speed up both forward and backward rates by the same amount Acids, Bases and Buffers BronstedLowry Definition of Acid Base behaviour A BronstedLowry acid is defined as a substance that can donate a proton. A BronstedLowry base is defined as a substance that can accept a proton. HCl (g) + H 2 O (l) H 3 O + (aq) + Cl (aq) Acid 1 base 2 Acid 2 Base 1 Each acid is linked to a conjugate base on the other side of the equation. HNO 3 + HNO 2 NO 3 + H 2 NO 2 + Acid 1 Base 2 Base 1 Acid 2 HCOOH + CH 3 (CH 2 ) 2 COOH HCOO + CH 3 (CH 2 ) 2 COOH 2 + Acid 1 Base 2 Base 1 Acid 2 In these reactions the substance with bigger Ka will act as the acid The acidic role of H + in the reactions of acids with metals, carbonates, bases and alkalis acid + metal salt + hydrogen 2CH 3 CH 2 CO 2 H + Mg (CH 3 CH 2 COO) 2 Mg + H 2 acid + alkali (NaOH) salt + water 2HNO 2 + Ca(OH) 2 Ca(NO 2 ) 2 + 2H 2 O Ionic Equations 2H + + Mg Mg 2+ + H 2 H + + OH H 2 O acid + carbonate (Na 2 CO 3 ) salt + water + CO 2 2CH 3 CO 2 H + Na 2 CO 3 2CH 3 CO 2 Na + + H 2 O + CO 2 2H + + CO 3 2 H 2 O + CO 2 8

9 Calculating ph ph = log [H + ] Where [H + ] is the concentration of hydrogen ions in the solution Calculating ph of strong acids Strong acids completely dissociate The concentration of hydrogen ions in a monoprotic strong acid will be the same as the concentration of the acid. For HCl and HNO 3 the [H + (aq)] will be the same as the original concentration of the acid. For 0.1M HCl the ph will be log[0.1] =1.00 Finding [H + ] from ph On most calculators this is [H + ] = 1 x 10 ph done by pressing Inv (or 2 nd function) log number(ph) Always give ph values to 2d.p. In the exam Example 1 What is the concentration of HCl with a ph of 1.35? [H + ] = 1 x = 0.045M Ionic Product for water In all aqueous solutions and pure water the following equilibrium occurs: H 2 O (l) H + (aq) + OH (aq) This equilibrium has the following equilibrium expression Kc= [H + (aq) ][OH (aq)] [H 2 O(l)] Rearrange to Kc x [H 2 O (l)] = [H + (aq) ][OH (aq)] Because [H 2 O (l)] is much bigger than the concentrations of the ions, we assume its value is constant and make a new constant Kw Kw = [H + (aq) ][OH (aq) ] Learn this expression At 25 o C the value of Kw for all aqueous solutions is 1x10 14 mol 2 dm 6 The Kw expression can be used to calculate [H + (aq) ] ions if we know the [OH (aq) ] ions and vice versa Finding ph of pure water Pure water/ neutral solutions are neutral because the [H + (aq) ] = [OH (aq)] Using Kw = [H + (aq) ][OH (aq) ] then when neutral Kw = [H + (aq) ] 2 and [H + (aq) ] = Kw At 25 o C [H + (aq) ] = 1x10 14 = 1x10 7 so ph = 7 Example 2 : Calculate the ph of water at 50ºC given that K w = x mol 2 dm 6 at 50ºC [H + (aq) ] = Kw = x =2.34 x 10 7 mol dm 3 ph = log 2.34 x 10 7 = 6.6 It is still neutral though as [H + (aq) ] = [OH (aq)] At different temperatures to 25 o C the ph of pure water changes. Le Chatelier s principle can predict the change. The dissociation of water is endothermic (because bonds are broken) so increasing the temperature would push the equilibrium to the right giving a bigger concentration of H + ions and a lower ph 9

10 Calculating ph of Strong Base For bases we are normally given the concentration of the hydroxide ion. To work out the ph we need to work out [H + (aq)] using the kw expression. Strong bases completely dissociate into their ions NaOH Na + + OH Example 3: What is the ph of the strong base 0.1M NaOH Assume complete dissociation. Kw = [H + (aq) ][OH (aq) ] = 1x10 14 [H + (aq)] = kw/ [OH (aq) ] = 1x10 14 / 0.1 = 1x10 13 M ph = log[1x10 13 ] =13.00 Weak acids Weak acids only slightly dissociate when dissolved in water, giving an equilibrium mixture HA + H 2 O (l) H 3 O + (aq) + A We can simplify this to (aq) HA (aq) H + (aq) + A (aq) Weak acids dissociation expression [H + (aq) ][A (aq)] [HA (aq)] The K a for ethanoic acid is 1.7 x 10 5 mol dm 3. The larger ka the stronger the acid Calculating ph of a weak acid Example 4 Write equation for dissociation of propanoic acid and its ka expression CH 3 CH 2 CO 2 H (aq) [H + (aq) ][CH 3CH 2 CO 2 (aq) ] [CH 3 CH 2 CO 2 H (aq) ] H + (aq) + CH 3 CH 2 CO 2 (aq) pka Sometimes Ka values are quoted as pka values pka = log Ka so Ka = 10 pka To make the calculation easier two assumptions are made to simplify the Ka expression: 1) [H + (aq)] eqm = [A (aq)] eqm because they have dissociated according to a 1:1 ratio 2) As the amount of dissociation is small we assume that the initial concentration of the undissociated acid has remained constant. So [HA (aq) ] eqm = [HA (aq) ] initial [H + (aq) ][A (aq)] [HA (aq)] Simplifies to [H + (aq) ]2 [HA (aq) ] initial Example 5 What is the ph of a solution of 0.01M ethanoic acid (ka is 1.7 x 10 5 mol dm 3 )? CH 3 CO 2 H (aq) H + (aq) + CH 3 CO 2 (aq) [H + (aq) ][CH 3CO 2 (aq) ] [CH 3 CO 2 H (aq) ] [H + (aq) ]2 [CH 3 CO 2 H (aq) ] initial 1.7x 10 5 = [H + (aq) ] [H + (aq) ]2 = 1.7 x 10 5 x 0.01 [H + (aq) ] = 1.7 x 107 = 4.12 x 10 4 ph = log [H + ] = log (4.12 x10 4 ) ph =3.38 Example 6 What is the concentration of propanoic acid with a ph of 3.52 (ka is 1.35 x 10 5 mol dm 3 )? CH 3 CH 2 CO 2 H (aq) H + (aq) + CH 3 CH 2 CO 2 (aq) [H + ] = 1 x = M [H + (aq) ][CH 3CH 2 CO 2 (aq) ] [CH 3 CH 2 CO 2 H (aq) ] [H + (aq) ]2 [CH 3 CH 2 CO 2 H (aq) ] initial [ ] x 10 5 = [CH 3 CH 2 CO 2 H (aq) ] initial [CH 3 CH 2 CO 2 H (aq) ] = 9.12 x 10 8 /1.35 x 10 5 [CH 3 CH 2 CO 2 H (aq) ] = 6.75 x 10 3 M 10

13 Titration curves Strong acid Strong base ph 13 7 e.g. HCl and NaOH Long vertical part from around 3 to 9 ph at equivalence point = 7 There are 4 main types of curve 1. Strong acid and strong base 2. Weak acid and strong base 3. Strong acid and weak base 4. Weak acid and weak base 1 25 cm 3 of base You may also have to work out the neutralisation volume from titration data given in the question. These are done by standard titration calculations from module 1. The equivalence point lies at the mid point of the extrapolated vertical portion of the curve. The Key points to sketching a curve: Initial and final ph Volume at neutralisation General Shape (ph at neutralisation) Weak acid Strong base ph ph starts near e.g. CH 3 CO 2 H and NaOH Equivalence point >7 Vertical part of curve >7 (around 7 to 9) 25 cm 3 of base Half neutralisation volume For weak acids [H + (aq) ][A (aq)] [HA (aq)] At ½ the neutralisation volume the [HA] = [A ] So [H + ] and pka = ph If we know the Ka we can then work out the ph at ½ V or vice versa At the start the ph rises quickly and then levels off. The flattened part is called the buffer region and is formed because a buffer solution is made 13

14 Strong acid Weak base e.g. HCl and NH 3 Weak acid Weak base e.g. CH 3 CO 2 H and NH 3 ph 13 ph 13 No vertical part of the curve 7 Vertical part of curve <7 (around 4 to 7) 7 Equivalence point < cm 3 of base 1 25 cm 3 of base Choosing an Indicator Indicators can be considered as weak acids. The acid must have a different colour to its conjugate base An indicator changes colour from HIn to In over a narrow range. Different indicators change colours over a different ranges The endpoint of a titration is reached when [HIn] = [In ]. To choose a correct indicator for a titration one should pick an indicator whose endpoint coincides with the equivalence point for the titration HIn (aq) colour A In (aq) + H + (aq) colour B We can apply Le Chatelier to give us the colour. In an acid solution the H + ions present will push this equilibrium towards the reactants. Therefore colour A is the acidic colour. In an alkaline solution the OH ions will react and remove H + ions causing the equilibrium to shift to the products. Colour B is the alkaline colour. An indicator will work if the ph range of the indicator lies on the vertical part of the titration curve. In this case the indicator will change colour rapidly and the colour change will correspond to the neutralisation point. Only use phenolphthalein in titrations with strong bases but not weak bases ph 13 strong base Colour change: colourless acid pink alkali Use methyl orange with titrations with strong acids but not weak acids Colour change: red acid yellow alkali (orange end point) 7 1 weak acid strong acid weak base 25 cm 3 of base ph range for phenolphthalein ph range for methyl orange 14

15 Enthalpy change of Neutralisation The standard enthalpy change of neutralisation is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water. Enthalpy changes of neutralisation are always exothermic. For reactions involving strong acids and alkalis, the values are similar, with values between 56 and 58 kj mol 1 Example. 25cm 3 of 2M HCl was neutralised by 25cm 3 of 2M NaOH. The Temperature increased 13.5 o C What was the energy change per mole of HCl? Step 1: Calculate the energy change for the amount of reactants in the test tube. Q = m x c p x T Q = 50 x 4.18 x13.5 Q = J Step 2 : calculate the number of moles of the HCl. Note the mass equals the mass of acid + the mass of alkali, as they are both solutions. moles of HCl = conc x vol = 2 x 25/1000 = mol Step 3 : calculate H, the enthalpy change per mole which might be called the enthalpy change of neutralisation Hr = Q/ no of moles = /0.05 = J mol 1 = 56.4 kj mol 1 to 3 sf Exothermic and so is given a minus sign Remember in these questions: sign, unit, 3 sig figs. 15

### 4. Acid Base Equilibria

4. Acid Base Equilibria BronstedLowry Definition of acid Base behaviour A BronstedLowry acid is defined as a substance that can donate a proton. A BronstedLowry base is defined as a substance that can

### 1.12 Acid Base Equilibria

.2 Acid Base Equilibria BronstedLowry Definition of acid Base behaviour A BronstedLowry acid is defined as a substance that can donate a proton. A BronstedLowry base is defined as a substance that can

### The rate equation relates mathematically the rate of reaction to the concentration of the reactants.

1.9 Rate Equations Rate Equations The rate equation relates mathematically the rate of reaction to the concentration of the reactants. For the following reaction, aa + bb products, the generalised rate

### The rate of reaction is defined as the change in concentration of a substance in unit time Its usual unit is mol dm -3 s -1

5.1.1 ow Fast? The rate of reaction is defined as the change in concentration of a substance in unit time Its usual unit is mol dm -3 s -1 When a graph of concentration of reactant is plotted vs time,

### The rate equation relates mathematically the rate of reaction to the concentration of the reactants.

1.9 Rate Equations Rate Equations The rate equation relates mathematically the rate of reaction to the concentration of the reactants. For the following reaction, aa + bb products, the generalised rate

### 4.3 ANSWERS TO EXAM QUESTIONS

4. ANSWERS TO EXAM QUESTIONS. (a) (i) A proton donor () (ii) Fully ionised or fully dissociated () (iii) 0 0 4 () mol dm 6 () 4 (b) (i) 50 0 /5 000 () = 0 06 mol dm () () (ii) Mol OH added = 50 0 50/000

### chemrevise.org 20/08/2013 Titration curves N Goalby Chemrevise.org 25 cm 3 of base

chemrevise.org 20/08/203 Titration curves N Goalby Chemrevise.org Titration curves 3 Titration curves are made by measuring the of the solution in the conical flask each time a small amount of acid or

### Lesmahagow High School AHChemistry Inorganic and Physical Chemistry Lesmahagow High School CfE Advanced Higher Chemistry

Lesmahagow High School CfE Advanced Higher Chemistry Unit 1 Inorganic and Physical Chemistry Chemical Equilibrium 1 Dynamic Equilibrium Revision Dynamic equilibrium happens in a closed system when the

### 14-Jul-12 Chemsheets A

www.chemsheets.co.uk 14-Jul-12 Chemsheets A2 009 1 BRONSTED-LOWRY ACIDS & BASES Bronsted-Lowry acid = proton donor (H + = proton) Bronsted-Lowry base = proton acceptor (H + = proton) Bronsted-Lowry acid-base

### 5.1.2 How Far? Equilibrium

5.1.2 How Far? Equilibrium Equilibrium constant Kc Kc = equilibrium constant For a generalised reaction ma + nb pc + qd [ C] p [D] q m,n,p,q are the stoichiometric balancing [ A] m [B] n numbers A,B,C,D

### Chemical Equilibria Part 2

Unit 1 - Inorganic & Physical Chemistry 1.4 Chemical Equilibria Part 2 Acid / Base Equilibria Indicators ph Curves Buffer Solutions Pupil Notes Learning Outcomes Questions & Answers KHS ChemistrySept 2015

### Acid Base Equilibria

Acid Base Equilibria N Goalby Chemrevise.org ACIDBASE : Arrhenius The most basic of the acidbase concepts is the Arrhenius theory Acids are substances that dissociate in water to produce hydronium ions,

### Chemical Equilibrium

Chemical Equilibrium Many reactions are reversible, i.e. they can occur in either direction. A + B AB or AB A + B The point reached in a reversible reaction where the rate of the forward reaction (product

### All reversible reactions reach an dynamic equilibrium state.

11. Equilibrium II Many reactions are reversible + 3 2 All reversible reactions reach an dynamic equilibrium state. Dynamic equilibrium occurs when forward and backward reactions are occurring at equal

### ANSWERS Unit 14: Review Acids and Bases

ANSWERS Unit 14: Review Acids and Bases 1) CH 3 COOH(aq) + H 2 0(l) H 3 0 + (aq) + CH 3 COO - (aq) In the equilibrium above, what are the two conjugate bases? A. CH 3 COOH and H 2 0 B. CH 3 COO - and H

### Chemical Equilibrium. Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B

Chemical Equilibrium Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B The point reached in a reversible reaction where the rate of the forward reaction (product formation,

### Enthalpy changes

2.3.1. Enthalpy changes In an exothermic change energy is transferred from the system (chemicals) to the surroundings. The have less energy than the If an enthalpy change occurs then energy is transferred

### 1 Chapter 19 Acids, Bases, and Salts

1 Chapter 19 Acids, Bases, and Salts ACID-BASE THEORIES Acids and bases are all around us and part of our everyday life (ex. bodily functions, vinegar, carbonated drinks, citrus fruits, car batteries,

### cm mol l -1 NaOH added to 50.0 cm 3 of 0.10 mol l -1 HCl

cm 3 0.10 mol l -1 NaOH added to 50.0 cm 3 of 0.10 mol l -1 HCl Acids have been described as substances that dissolve in water to form H + (aq) ions, whilst bases are substances that react with acids.

### Titration curves, labelled E, F, G and H, for combinations of different aqueous solutions of acids and bases are shown below.

Titration curves, labelled E, F, G and H, for combinations of different aqueous solutions of acids and bases are shown below. All solutions have concentrations of 0. mol dm 3. (a) In this part of the question,

### Review: Acid-Base Chemistry. Title

Review: Acid-Base Chemistry Title Basics General properties of acids & bases Balance neutralization equations SA + SB water + salt Arrhenius vs. Bronsted-Lowry BL plays doubles tennis match with H+) Identify

### AS Paper 1 and 2 Kc and Equilibria

AS Paper 1 and 2 Kc and Equilibria Q1.When one mole of ammonia is heated to a given temperature, 50 per cent of the compound dissociates and the following equilibrium is established. NH 3(g) ½ N 2 (g)

### IB Chemistry ABS Introduction An acid was initially considered a substance that would produce H + ions in water.

IB Chemistry ABS Introduction An acid was initially considered a substance that would produce H + ions in water. The Brønsted-Lowry definition of an acid is a species that can donate an H + ion to any

### 10.1 Acids and Bases in Aqueous Solution

10.1 Acids and Bases in Aqueous Solution Arrhenius Definition of Acids and Bases An acid is a substance that gives hydrogen ions, H +, when dissolved in water. In fact, H + reacts with water and produces

### + H 2 O HPO 4. (a) In this system, there are two acid-base conjugate pairs. These are (1) HPO4

1 The dihydrogenphosphate-hydrogenphosphate ion system is an important buffer in the human body. H 2 PO 4 H 2 O HPO 4 2 H 3 O (a) In this system, there are two acid-base conjugate pairs. These are acid

### PHYSICAL SCIENCES/ P2 1 SEPTEMBER 2015 CAPS CAPE WINELANDS EDUCATION DISTRICT

PHYSICAL SCIENCES/ P2 1 SEPTEMBER 2015 CAPE WINELANDS EDUCATION DISTRICT MARKS 150 TIME 3 hours This question paper consists of 15 pages and 4 data sheets. PHYSICAL SCIENCES/ P2 2 SEPTEMBER 2015 INSTRUCTIONS

### (iii) Use the diagram to determine the value (mol 2 dm 6 ) of K w at 50 C. [1]

2. (a) The diagram shows the variation of the ionic product of water, K w, with temperature. 6 K w /10 14 mol 2 dm 6 Temperature/ C (i) Give the expression for the ionic product of water, K w. [1] (ii)

### Arrhenius base is one that dissociates in water to form hydroxide ions.

Chemistry Notes: Acids and Bases Arrhenius base is one that dissociates in water to form hydroxide ions. Arrhenius acid is on that dissociates in water to form hydrogen ions (protons). A Bronsted-Lowry

### Collision energy. E a. The mean energy of the particles is not at the peak of the curve

8 Reaction Kinetics Collision theory Reactions can only occur when collisions take place between particles having sufficient energy. The energy is usually needed to break the relevant bonds in one or either

### Ch. 17 Applications of Aqueous Equilibria: Buffers and Titrations

Ch. 17 Applications of Aqueous Equilibria: Buffers and Titrations Sec 1 The Common-Ion Effect: The dissociation of a weak electrolyte decreases when a strong electrolyte that has an ion in common with

### Chemical Equilibria. OCR Chemistry A H432

Chemical Equilibria Chemical equilibrium is a dynamic equilibrium. Features of a dynamic equilibrium, which can only be established in a closed system (nothing added or removed): - rates of forward and

### Acids and Bases Written Response

Acids and Bases Written Response January 1999 4. Consider the salt sodium oxalate, Na2C2O4. a) Write the dissociation equation for sodium oxalate. (1 mark) b) A 1.0M solution of sodium oxalate turns pink

### Acids And Bases. H + (aq) + Cl (aq) ARRHENIUS THEORY

Acids And Bases A. Characteristics of Acids and Bases 1. Acids and bases are both ionic compounds that are dissolved in water. Since acids and bases both form ionic solutions, their solutions conduct electricity

### ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor

ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) Acid Base Conjugate acid Conjugate

### CHAPTER 7.0: IONIC EQUILIBRIA

Acids and Bases 1 CHAPTER 7.0: IONIC EQUILIBRIA 7.1: Acids and bases Learning outcomes: At the end of this lesson, students should be able to: Define acid and base according to Arrhenius, Bronsted- Lowry

### Experiment [RCH 2 Cl] [OH ] Initial rate/mol dm 3 s

1. The kinetics of the hydrolysis of the halogenoalkane RCH 2 Cl with aqueous sodium hydroxide (where R is an alkyl group) was studied at 50 ºC. The following results were obtained: Experiment [RCH 2 Cl]

### Acid Base Equilibria

Acid Base Equilibria Acid Ionization, also known as acid dissociation, is the process in where an acid reacts with water to produce a hydrogen ion and the conjugate base ion. HC 2 H 3 O 2(aq) H + (aq)

### I. Multiple Choice Questions (Type-I) is K p

Unit 7 EQUILIBRIUM I. Multiple Choice Questions (Type-I) 1. We know that the relationship between K c and K p is K p K c (RT) n What would be the value of n for the reaction NH 4 Cl (s) NH 3 (g) + HCl

### ACIDS, BASES, PH, BUFFERS & TITRATION WEBINAR. Dr Chris Clay

ACIDS, BASES, PH, BUFFERS & TITRATION WEBINAR Dr Chris Clay http://drclays-alevelchemistry.com/ Q1.Titration curves, labelled E, F, G and H, for combinations of different aqueous solutions of acids and

### Acids, Bases, and Salts Review for Sections

1. Consider the following: Review for Sections 4.1 4.9 I H 2 CO 3 + F HCO 3 + HF 2 II HCO 3 + HC 2 O 4 H 2 CO 3 + C 2 O 4 2 III HCO 3 + H 2 C 6 H 6 O 7 H 2 CO 3 + HC 6 H 5 O 7 The HCO 3 is a base in A.

### capable of neutralizing both acids and bases

Buffers Buffer n any substance or mixture of compounds that, added to a solution, is capable of neutralizing both acids and bases without appreciably changing the original acidity or alkalinity of the

### NH 3(g) N 2(g) + H 2(g) What is the total number of moles of gas present in this equilibrium mixture? D 3.0 (Total 1 mark)

Q1.When one mole of ammonia is heated to a given temperature, 50% of the compound dissociates and the following equilibrium is established. NH 3(g) N 2(g) + H 2(g) What is the total number of moles of

### SCH4U: EXAM REVIEW. 2. Which of the following has a standard enthalpy of formation of 0 kj mol -1 at 25ºC and 1.00 atm?

SCH4U_08-09 SCH4U: EXAM REVIEW 1. The heat of a reaction is equal to: a. enthalpy (products) + enthalpy (reactants) b. enthalpy (reactants) enthalpy (products) c. enthalpy (products) enthalpy (reactants)

### , for C 2. COOH is mol dm [1] COOH by adding water until the total volume is cm 3. for C 2 COOH.

1 A student is supplied with 0.500 mol dm 3 potassium hydroxide, KOH, and 0.480 mol dm 3 propanoic acid, C 2 COOH. The acid dissociation constant, K a, for C 2 COOH is 1.35 10 5 mol dm 3. (a) C 2 COOH

### 1. Which statement is correct for a crystal of iron(ii) sulfate in a state of equilibrium with a saturated solution of iron(ii) sulfate?

1. Which statement is correct for a crystal of iron(ii) sulfate in a state of equilibrium with a saturated solution of iron(ii) sulfate? A. The colour of the solution darkens as the crystal continues to

### Atoms, Elements, Atoms, Elements, Compounds and Mixtures. Compounds and Mixtures. Atoms and the Periodic Table. Atoms and the.

Atoms, Elements, Compounds and Mixtures Explain how fractional distillation can be used to separate a mixture. 1 Atoms, Elements, Compounds and Mixtures Fractional distillation is used to separate components

### Chapter 8 Acid-Base Equilibria

Chapter 8 Acid-Base Equilibria 8-1 Brønsted-Lowry Acids and Bases 8-2 Water and the ph Scale 8-3 The Strengths of Acids and Bases 8-4 Equilibria Involving Weak Acids and Bases 8-5 Buffer Solutions 8-6

### CHEMICAL EQUILIBRIA. Dynamic Equilibrium Equilibrium involves reversible reactions which do not go to completion.

CHEMICAL EQUILIBRIA Dynamic Equilibrium Equilibrium involves reversible reactions which do not go to completion. If we consider a reaction between A and B to form C and D which is reversible. When A and

### Grace King High School Chemistry Test Review

CHAPTER 19 Acids, Bases & Salts 1. ACIDS Grace King High School Chemistry Test Review UNITS 7 SOLUTIONS &ACIDS & BASES Arrhenius definition of Acid: Contain Hydrogen and produce Hydrogen ion (aka proton),

### General Chemistry II CHM 1046 E Exam 2

General Chemistry II CHM 1046 E Exam 2 Dr. Shanbhag Name: 1. The formation of ammonia from elemental nitrogen and hydrogen is an exothermic process. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H= -92.2 kj Which of

### The Common Ion Effect

Chapter 17 ACID BASE EQUILIBRIA (Part I) Dr. Al Saadi 1 17.1 The Common Ion Effect A phenomenon known as the common ion effect states that: When a compound containing an ion in common with an already dissolved

### 5.111 Lecture Summary #22 Wednesday, October 31, 2014

5.111 Lecture Summary #22 Wednesday, October 31, 2014 Reading for Today: Sections 11.13, 11.18-11.19, 12.1-12.3 in 5 th ed. (10.13, 10.18-10.19, 11.1-11.3 in 4 th ed.) Reading for Lecture #23: Sections

### 19.3 Strengths of Acids and Bases > Chapter 19 Acids, Bases, and Salts Strengths of Acids and Bases

Chapter 19 Acids, Bases, and Salts 19.1 Acid-Base Theories 19.2 Hydrogen Ions and Acidity 19.3 Strengths of Acids and Bases 19.4 Neutralization Reactions 19.5 Salts in Solution 1 Copyright Pearson Education,

### Unit 4: Acid/Base I. abinotes. I) Introduction to Acids and Bases What is an acid?

Unit 4: Acid/Base I I) Introduction to Acids and Bases What is an acid? http://www.kidsknowit.com/flash/animations/acidsbases.swf What are properties of acids? 1) Acids react with. 2) Acids create when

### Page 2. M1.(a) [H + ] = or = Allow ( ) M1 1 = If [HX] / [X - ] or M2. M2 1

M.(a) [H + ] = or =.74 0 5 Allow ( ) M = 3.08 0 5 If [HX] / [X - ] or. upside down, or any addition or subtraction lose M & ph = 4.5 (correct answer scores 3) Can score for correct ph conseq to their [H

### Edexcel GCSE Chemistry. Topic 3: Chemical changes. Acids. Notes.

Edexcel GCSE Chemistry Topic 3: Chemical changes Acids Notes 3.1 Rec that acids in solution are sources of hydrogen ions and alkalis in solution are sources of hydroxide ions Acids produce H + ions in

### Grade A buffer: is a solution that resists changes in its ph upon small additions of acid or base.sq1

Chapter 15 Lesson Plan Grade 12 402. The presence of a common ion decreases the dissociation. BQ1 Calculate the ph of 0.10M CH 3 COOH. Ka = 1.8 10-5. [H + ] = = ( )( ) = 1.34 10-3 M ph = 2.87 Calculate

### Atoms What subatomic particles make up the atom?

Atoms What subatomic particles make up the atom? What are the masses of the subatomic particles? What do atomic and mass number represent? What does 7 3 Li represent? How are elements arranged in the periodic

### Acid-Base Equilibria and Solubility Equilibria Chapter 17

PowerPoint Lecture Presentation by J. David Robertson University of Missouri Acid-Base Equilibria and Solubility Equilibria Chapter 17 The common ion effect is the shift in equilibrium caused by the addition

### AP Chapter 15 & 16: Acid-Base Equilibria Name

AP Chapter 15 & 16: Acid-Base Equilibria Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. AP Chapter 15 & 16: Acid-Base Equilibria 2 Warm-Ups (Show

### Q.1 Write out equations for the reactions between...

1 CHEMICAL EQUILIBRIUM Dynamic Equilibrium not all reactions proceed to completion some end up with a mixture of reactants and products this is because some reactions are reversible; products revert to

### CHAPTER 8: ACID/BASE EQUILIBRIUM

CHAPTER 8: ACID/BASE EQUILIBRIUM Already mentioned acid-base reactions in Chapter 6 when discussing reaction types. One way to define acids and bases is using the Brønsted-Lowry definitions. A Brønsted-Lowry

### Acid-Base Titration Solution Key

Key CH3NH2(aq) H2O(l) CH3NH3 (aq) OH - (aq) Kb = 4.38 x 10-4 In aqueous solution of methylamine at 25 C, the hydroxide ion concentration is 1.50 x 10-3 M. In answering the following, assume that temperature

### Acids and Bases. A strong base is a substance that completely ionizes in aqueous solutions to give a cation and a hydroxide ion.

Acid-Base Theories Arrhenius Acids and Bases (1884) Acids and Bases An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. A base is a substance that, when

### ACIDS AND BASES. HCl(g) = hydrogen chloride HCl(aq) = hydrochloric acid HCl(g) H + (aq) + Cl (aq) ARRHENIUS THEORY

ACIDS AND BASES A. CHARACTERISTICS OF ACIDS AND BASES 1. Acids and bases are both ionic compounds that are dissolved in water. Since acids and bases both form ionic solutions, their solutions conduct electricity

### -log [H+][OH-] = - log [1 x ] Left hand side ( log H + ) + ( log OH - ) = ph + poh Right hand side = ( log 1) + ( log ) = 14 ph + poh = 14

Autoionization of Water H 2 O H + + OH - K = [H + ][OH - ]/[H 2 O] = 1.802 x 10-16 Concentration of [H 2 O] is so HIGH autoionization is just a drop in the bucket, so [H 2 O] stays constant at 55.5 M,

### IB Chemistry 2 Summer Assignment

Name IB Chemistry 2 Summer Assignment 1. What is the sum of all coefficients when the following equation is balanced using the smallest possible whole numbers? C 2 H 2 + O 2 CO 2 + H 2 O A. 5 B. 7 C. 11

### CHEM 1412 Zumdahl & Zumdahl Practice Exam II (Ch. 14, 15 & 16) Multiple Choices: Please select one best answer. Answer shown in bold.

CHEM 1412 Zumdahl & Zumdahl Practice Exam II (Ch. 14, 15 & 16) Multiple Choices: Please select one best answer. Answer shown in bold. 1. Consider the equilibrium: PO -3 4 (aq) + H 2 O (l) HPO 2-4 (aq)

### Chem 105 Tuesday March 8, Chapter 17. Acids and Bases

Chem 105 Tuesday March 8, 2011 Chapter 17. Acids and Bases 1) Define Brønsted Acid and Brønsted Base 2) Proton (H + ) transfer reactions: conjugate acid-base pairs 3) Water and other amphiprotic substances

### Chem12 Acids : Exam Questions M.C.-100

Chem12 Acids : Exam Questions M.C.-100 1) Given : HPO 4 2- (aq) + NH 4 + (aq) H 2 PO 4 - (aq) + NH 3 (aq), the strongest acid in the above equation is : a) NH 4 + b) HPO 4 2- c) NH 3 d) H 2 PO 4-2)

### NCERT. [H O] Since water is in large excess, its concentration can be assumed to be constant and combining it with K provides a new constant K w

UNIT-5 PH AND PH CHANGE IN AQUEOUS SOLUTIONS YOU have already performed experiments on dynamic equilibrium between unionised salt and the ions produced by it on dissolving in a solvent. In this unit we

### Chemistry 3202 Pre-Public Examination May 2012 Name:

Chemistry 3202 Pre-Public Examination May 2012 Name: Section A: Multiple Choice This section contains 40 multiple choice covering concepts from the entire course. Please answer all multiple choice items

### Chemical Equilibrium

Chemical Equilibrium THE NATURE OF CHEMICAL EQUILIBRIUM Reversible Reactions In theory, every reaction can continue in two directions, forward and reverse Reversible reaction! chemical reaction in which

### HALFWAY to EQUIVALENCE POINT: ph = pk a of the acid being titrated.

CHEMISTRY 109 Help Sheet #33 Titrations Chapter 15 (Part II); Section 15.2 ** Cover topics appropriate for your lecture** Prepared by Dr. Tony Jacob http://www.chem.wisc.edu/areas/clc (Resource page) Nuggets:

### ACID-BASE TITRATION AND PH

ACID-BASE TITRATION AND PH Section 1 Aqueous Solutions and the Concept of ph Hydronium and Hydroxide Ions Acids and bases form hydroxide and hydronium ions These ions are not the only ones in an aqueous

### Chapter 4 Reactions in Aqueous Solution

Chapter 4 Reactions in Aqueous Solution Homework Chapter 4 11, 15, 21, 23, 27, 29, 35, 41, 45, 47, 51, 55, 57, 61, 63, 73, 75, 81, 85 1 2 Chapter Objectives Solution To understand the nature of ionic substances

### Chem 5 PAL Worksheet Acids and Bases Smith text Chapter 8

D.CHO3HE.KOHB.NHC.CHC3OHHCl3F.H.CHHCH3COG.H2HCHEM 5 PAL Worksheet Acids and Bases Fall 2017 Chem 5 PAL Worksheet Acids and Bases Smith text Chapter 8 Many substances in the body are acids and bases. Many

### Worksheet 4.1 Conjugate Acid-Base Pairs

Worksheet 4.1 Conjugate AcidBase Pairs 1. List five properties of acids that are in your textbook. Acids conduct electricity, taste sour, neutralize bases, change the color of indicators, and react with

### A is capable of donating one or more H+

Slide 1 / 48 1 According to the Arrhenius concept, an acid is a substance that. A is capable of donating one or more H+ B C D E causes an increase in the concentration of H+ in aqueous solutions can accept

### COOH, and thioglycolic acid, HSCH 2. (a) Glycolic acid reacts with bases, such as aqueous sodium hydroxide, NaOH(aq), to form salts.

1 Glycolic acid, HOCH 2 COOH, and thioglycolic acid, HSCH 2 COOH, are weak acids. (a) Glycolic acid reacts with bases, such as aqueous sodium hydroxide, NaOH(aq), to form salts. A student pipetted 25.0

### No Brain Too Small CHEMISTRY AS90700 Describe properties of aqueous systems. ph of weak acids, weak bases and salt solutions

ph of weak acids, weak bases and salt solutions 2010: 1 An aqueous solution of ammonium chloride (NH 4 Cl) has a ph of 4.66. (a) (i) Write the equation for solid ammonium chloride dissolving in water.

### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Which one of the following is the weakest acid? 1) A) HF (Ka = 6.8 10-4) B) HNO2 (Ka

### Chemistry 12 Provincial Exam Workbook Unit 04: Acid Base Equilibria. Multiple Choice Questions

R. Janssen, MSEC Chemistry 1 Provincial Workbook (Unit 0), P. 1 / 69 Chemistry 1 Provincial Exam Workbook Unit 0: Acid Base Equilibria Multiple Choice Questions 1. Calculate the volume of 0.00 M HNO needed

### Unit 8: Equilibrium Unit Review

1. Predict the effect of increasing pressure on the position of equilibrium in the following systems: a. CH 4 (g) + 2H 2 O(g) CO 2 (g) + 4H 2 (g) b. N 2 O 5 (g) + NO(g) 3NO 2 (g) c. NO(g) + NO 2 (g) N

### Acids, Bases, & Neutralization Chapter 20 & 21 Assignment & Problem Set

Acids, Bases, & Neutralization Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. Acids, Bases, & Neutralization 2 Study Guide: Things You Must Know

### Unit 4a Acids, Bases, and Salts Theory

Unit 4a Acids, Bases, and Salts Theory Chemistry 12 Arrhenius Theory of Acids and Bases The first theory that was proposed to explain the actions of acids and bases was by Svante Arrhenius. It is still

### SOLUBILITY PRODUCT (K sp ) Slightly Soluble Salts & ph AND BUFFERS (Part Two)

SOLUBILITY PRODUCT (K sp ) Slightly Soluble Salts & ph AND BUFFERS (Part Two) ADEng. PRGORAMME Chemistry for Engineers Prepared by M. J. McNeil, MPhil. Department of Pure and Applied Sciences Portmore

OCR Advanced Subsidiary GCE in Chemistry 2811 June 2002 Unifying concepts Subject: Unifying Concepts Code: 2816/1 Session: June Year: 2002 Final Mark Scheme 22/6/2002 MAXIMUM MARK 60 Page 1 of 6 2816 June

### 1B Equilibrium. 3) Equilibrium is a dynamic state At equilibrium the rate in both directions must be the same.

1B Equilibrium The equilibrium constant, K c Characteristics of the equilibrium state 1) Equilibrium can only be established in a closed system. Matter cannot be exchanged with the surroundings (this will

### School of Chemistry, University of KwaZulu-Natal, Howard College Campus, Durban. CHEM191 Tutorial 1: Buffers

School of Chemistry, University of KwaZulu-Natal, Howard College Campus, Durban CHEM191 Tutorial 1: Buffers Preparing a Buffer 1. How many moles of NH 4 Cl must be added to 1.0 L of 0.05 M NH 3 to form

### (Label the Conjugate Pairs) Water in the last example acted as a Bronsted-Lowry base, and here it is acting as an acid. or

Chapter 16 - Acid-Base Equilibria Arrhenius Definition produce hydrogen ions in aqueous solution. produce hydroxide ions when dissolved in water. Limits to aqueous solutions. Only one kind of base. NH

### Lecture 12. Acid/base reactions. Equilibria in aqueous solutions.

Lecture 12 Acid/base reactions. Equilibria in aqueous solutions. Titrations Kotz 7 th ed. Section 18.3, pp.821-832. In a titration a solution of accurately known concentration is added gradually added

### CHEM Dr. Babb s Sections Exam #3 Review Sheet

CHEM 116 Dr. Babb s Sections Exam #3 Review Sheet Acid/Base Theories and Conjugate AcidBase Pairs 111. Define the following terms: Arrhenius acid, Arrhenius base, Lewis acid, Lewis base, BronstedLowry

### A 95 g/mol B 102 /mol C 117 g/mol D 126 g/mol E 152 g/mol

Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

### Bond C=O C H C O O H. Use the enthalpy change for the reaction and data from the table to calculate a value for the H H bond enthalpy.

Many chemical processes release waste products into the atmosphere. Scientists are developing new solid catalysts to convert more efficiently these emissions into useful products, such as fuels. One example

### Acceptable Answers Reject Mark + HS. (2) Initially 0.100

(a) (K a =) [H 3 O + (aq)] [HS (aq)] [H 2 S(aq)] [H 3 O + ] 2 numerator (2) (K a2 =) [H 3 O + (aq)][ S 2 (aq)] [HS (aq)] [H 3 O + ] 2 numerator Allow H + (aq) for H 3 O + (aq) Ignore missing / incorrect

### ACIDS AND BASES CONTINUED

ACIDS AND BASES CONTINUED WHAT HAPPENS WHEN AN ACID DISSOLVED IN WATER? Water acts as a Brønsted Lowry base and abstracts a proton (H+) from the acid. As a result, the conjugate base of the acid and a

### K w. Acids and bases 8/24/2009. Acids and Bases 9 / 03 / Ionization of water. Proton Jumping Large proton and hydroxide mobility

Chapter 2 Water Acids and Bases 9 / 03 / 2009 1. How is the molecular structure of water related to physical and chemical behavior? 2. What is a Hydrogen Bond? 3Wh 3. What are Acids Aid and db Bases? 4.