# 1.12 Acid Base Equilibria

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1 .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 accept a proton. HCl (g) + H 2 O (l) H 3 O + (aq) + Cl (aq) acid base acid base Each acid is linked to a conjugate base on the other side of the equation. Calculating = log [H + ] Where [H + ] is the concentration of hydrogen ions in the solution. Calculating 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.M HCl the will be log[0.] =.00 Finding [H + ] from On most calculators this is [H + ] = x 0 done by pressing Ionic Product for water Inv (or 2 nd function) log number() Always give values to 2d.p. In the exam Example What is the concentration of HCl with a of.35? [H + ] = x 0.35 = 0.045M 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 x0 4 mol 2 dm 6 Finding of pure water The Kw expression can be used to calculate [H + (aq) ] ions if we know the [OH (aq) ] ions and vice versa. 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) ] = x0 4 = x0 so = Example 2 : Calculate the of water at 50ºC given that K w = 5.46 x 0 4 mol 2 dm 6 at 50ºC [H + (aq) ] = Kw = 5.46 x 0 4 =2.34 x 0 mol dm 3 = log 2.34 x 0 = 6.6 It is still neutral though as [H + (aq) ] = [OH (aq)] At different temperatures to 25 o C the of pure water changes. Le Chatelier s principle can predict the change. The dissociation of water is endothermic so increasing the temperature would push the equilibrium to the right giving a bigger concentration of H + ions and a lower.

2 Calculating of a Strong Base For bases we are normally given the concentration of the hydroxide ion. To work out the we need to work out [H + (aq)] using the Kw expression. Strong bases completely dissociate into their ions. NaOH Na + + OH Weak Acids Example 3: What is the of the strong base 0.M NaOH Assume complete dissociation. Kw = [H + (aq) ][OH (aq) ] = x0 4 [H + (aq)] = Kw/ [OH (aq) ] = x0 4 / 0. = x0 3 M = log[x0 3 ] =3.00 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. x 0 5 mol dm 3. The larger ka the stronger the acid. Calculating of a Weak Acid Example 4 Write an 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 = 0 pka To make the calculation easier two assumptions are made to simplify the Ka expression: ) [H + (aq)] eqm = [A (aq)] eqm because they have dissociated according to a : 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 of a solution of 0.0M ethanoic acid (ka is. x 0 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.x 0 5 = [H + (aq) ]2 0.0 [H + (aq) ]2 =. x 0 5 x 0.0 [H + (aq) ] =. x 0 = 4.2 x 0 4 = log [H + ] = log (4.2 x0 4 ) =3.38 Example 6 What is the concentration of propanoic acid with a of 3.52 (ka is.35 x 0 5 mol dm 3 )? CH 3 CH 2 CO 2 H (aq) H + (aq) + CH 3 CH 2 CO 2 (aq) [H + ] = 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 [ ] 2.35 x 0 5 = [CH 3 CH 2 CO 2 H (aq) ] initial [CH 3 CH 2 CO 2 H (aq) ] = 9.2 x 0 8 /.35 x 0 5 [CH 3 CH 2 CO 2 H (aq) ] = 6.5 x 0 3 M 2

3 Calculations involving Neutralisation Reactions These can be quite complex calculations working out the of a partially neutralised acid or the of the solution if too much alkali has been added and has gone past neutralisation. The method differs if the acid is strong or weak for the partially neutralised case. Strong Acid and Strong Base Neutralisations Work out moles of original acid Work out moles of base added Work out which one is in excess If excess acid If excess alkali Example 5cm 3 of 0.5M HCl is reacted with 35cm 3 of 0.55M NaOH. What will be the of the resulting mixture? Moles HCl = conc x vol = 0.5 x 0.05 = 0.005mol Moles NaOH = conc x vol = 0.55 x = HCl + NaOH NaCl + H 2 O Moles of NaOH in excess = = 0.05 (as : ratio) [OH ] = moles excess OH = 0.05/ 0.05 = 0.235M [H + ] = K w /[OH ] = x0 4 / = 4.25x0 4 = log 4.25x0 4 = 3.3 Work out new concentration of excess H + ions [H + ] = moles excess H + Total volume = vol of acid + vol of base added Work out new concentration of excess OH ions [OH ] = moles excess OH [H + ] = K w /[OH ] Total volume = vol of acid + vol of base added Example 8 45cm 3 of M HCl is reacted with 30cm 3 of 0.65M NaOH. What will be the of the resulting mixture? Moles HCl = conc x vol = x = 0.045mol Moles NaOH = conc x vol = 0.65 x = HCl + NaOH NaCl + H 2 O Moles of HCl in excess = = (as : ratio) [H + ] = moles excess H + = / 0.05 = 0.34M = log 0.34 = 0.4 Weak Acid and Strong Base Neutralisations Work out moles of original acid Work out moles of base added Work out which one is in excess If excess acid If excess alkali use the same method with excess alkali and strong acid above Work out new concentration of excess HA [HA] = initial moles HA moles OH Work out concentration of salt formed [A ] [A ] = moles OH added Rearrange ka = [H + ] [A ] to get [H + ] [HA] Example 9 55cm 3 of 0.5M CH 3 CO 2 H is reacted with 25cm 3 of 0.35M NaOH. What will be the of the resulting mixture? Moles CH 3 CO 2 H = conc x vol =0.5x = 0.025mol Moles NaOH = conc x vol = 0.35 x = CH 3 CO 2 H+ NaOH CH 3 CO 2 Na + H 2 O ka is. x 0 5 mol dm 3 Moles of CH 3 CO 2 H in excess = = (as : ratio) [CH 3 CO 2 H ] = moles excess CH 3 CO 2 H = 0.085/ 0.08 = 0.234M [CH 3 CO 2 ] = moles OH added = / 0.08 = 0.09M ka = [H + ] [CH 3 CO 2 ] [ CH 3 CO 2 H ] [H + ] = ka x[ CH 3 CO 2 H ] / [CH 3 CO 2 ] =. x 0 5 x / 0.09 = 3.64 x 0 5 = log 3.64 x 0 5 =

6 Calculating change in of buffer on addition of small amount of acid or alkali If a small amount of alkali is added to a buffer then the moles of the buffer acid would reduce by the number of moles of alkali added and the moles of salt would increase by the same amount so a new calculation of can be done with the new values. CH 3 CO 2 H (aq) +OH CH 3 CO 2 (aq) + H 2 O (l) If a small amount of acid is added to a buffer then the moles of the buffer salt would reduce by the number of moles of acid added and the moles of buffer acid would increase by the same amount so a new calculation of can be done with the new values. CH 3 CO 2 (aq) + H + CH 3 CO 2 H (aq) Example 5: mol of NaOH is added to 500cm 3 of a buffer where the concentration of ethanoic acid is mol dm 3 and the concentration of sodium ethanoate is mol dm 3. (Ka =. x 0 5 ) Calculate the of the buffer solution after the NaOH has been added. Work out the moles of acid and salt in the initial buffer solution Moles ethanoic acid= conc x vol = x = 0.00mol Moles sodium ethanoate = conc x vol = 0.25 x = 0.25mol Work out the moles of acid and salt in buffer after the addition of 0.005mol NaOH Moles ethanoic acid = = mol Moles sodium ethanoate = = 0.30 mol [CH 3 COOH (aq) ] [CH 3 COO (aq) ] We can enter moles of acid and salt straight into the equation as they both have the same new final volume [H + (aq) ] =. x 05 x [H + (aq) ] =.24x = log.24x 0 5 = 4.9 6

7 Titration curves Constructing a PH curve Measure initial of the acid Add alkali in small amounts noting the volume added Stir mixture to equalise the Measure and record the to dp When approaching endpoint add in smaller volumes of alkali Add until alkali in excess Calibrate meter first by measuring known of a buffer solution. This is necessary because meters can lose accuracy on storage Can improve accuracy by maintaining constant temperature Practical Strong acid Strong base 3 e.g. HCl and NaOH Long steep part from around 3 to 9 at equivalence point = There are 4 main types of curve. Strong acid and strong base 2. Weak acid and strong base 3. Strong acid and weak base 4. Weak acid and weak base 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. 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 Volume at neutralisation General Shape ( at neutralisation) Weak acid Strong base e.g. CH 3CO 2 H and NaOH At the start the rises quickly and then levels off. The flattened part is called the buffer region and is formed because a buffer solution is made 3 Equivalence point > Half neutralisation volume For weak acids [H + (aq) ][A (aq)] [HA (aq)] At ½ the neutralisation volume the [HA] = [A ] So [H + ] and pka = starts near 3 ½ V V Steep part of curve > (around to 9) cm 3 of base If we know the Ka we can then work out the at ½ V or vice versa. If a curve is plotted then the of a weak acid at half neutralisation (½ V) will equal the pka

8 Strong acid Weak base e.g. HCl and NH 3 Weak acid Weak base e.g. CH 3 CO 2 H and NH No Steep part of the curve Steep part of curve < (around 4 to ) Equivalence point < 25 cm 3 of base 25 cm 3 of base Choosing an Indicator How indicators work Indicators can be considered as weak acids. The acid must have a different colour to its conjugate base HIn (aq) colour A In (aq) + H + (aq) colour B An indicator changes colour from HIn to In over a narrow range. Different indicators change colours over different ranges. The endpoint of a titration is defined as the point when the colour of the indicator changes colour 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. 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 range of the indicator lies on the steep 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 3 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) weak acid strong acid weak base 25 cm 3 of base range for phenolphthalein range for methyl orange 8

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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,

### Acids, Bases and the Common Ion Effect

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### H 2 SO 4. HCl. HNO 3 nitric acid. TOPIC: Acids and Bases. e.g. HCl! H + + Cl - sulphuric acid. hydrochloric acid

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

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### Chemistry 12 UNIT 4 ACIDS AND BASES

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### chemrevise.org 22/08/2013 Titrations N Goalby Chemrevise.org Titrations

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

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### 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

### 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:

### Buffers. How can a solution neutralize both acids and bases? Beaker B: 100 ml of 1.00 M HCl. HCl (aq) + H 2 O H 3 O 1+ (aq) + Cl 1 (aq)

Buffers How can a solution neutralize both acids and bases? Why? Buffer solutions are a mixture of substances that have a fairly constant ph regardless of addition of acid or base. They are used in medicine,

### 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

### AP Chemistry. CHAPTER 17- Buffers and Ksp 17.1 The Common Ion Effect Buffered Solutions. Composition and Action of Buffered Solutions

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### 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

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

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### 2/4/2016. Chapter 15. Chemistry: Atoms First Julia Burdge & Jason Overby. Acid-Base Equilibria and Solubility Equilibria The Common Ion Effect

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### 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?

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

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### 1. Know and be capable of applying the Bronsted-Lowery model of acids and bases (inculdig the concepts related to conjugate acid-base pairs.

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

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### M1.C [1] M2.B [1] M3.D [1] M4.B [1] M5.D [1] M6.D [1] M7.A [1] M8.A [1] M9.C [1] M10.D [1] M11.C [1] M12. B [1] M13. C [1]

M.C [] M.B [] M.D [] M4.B [] M5.D [] M6.D [] M7.A [] M8.A [] M9.C [] M0.D [] M.C [] M. B [] M. C [] Page of M4. (a) Burette Because it can deliver variable volumes (b) The change in ph is gradual / not

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

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### Principles of Reactivity: The Chemistry of Acids and Bases. Acids, Bases and Arrhenius

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### Operational Skills. Operational Skills. The Common Ion Effect. A Problem To Consider. A Problem To Consider APPLICATIONS OF AQUEOUS EQUILIBRIA

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### Unit 4-1 Provincial Practice Questions Page 1

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