# EQUILIBRIUM GENERAL CONCEPTS

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1 WHEN THE REACTION IS IN EQUILIBRIUM EQUILIBRIUM GENERAL CONCEPTS The concentrations of all species remain constant over time, but both the forward and reverse reaction never cease When a system is at equilibrium, the forward and reverse reactions are proceeding at the same rate. Equilibrium is signified with double arrows or the equal sign in reaction equation. 1 The same equilibrium composition is reached from either the forward or reverse direction, provided the overall system composition is the same. Pure NO is brown and pure N O 4 is colorless. The amber color of the equilibrium mixture indicates that both species are present at the equilibrium. The decomposition of N O 4 (g) into NO (g). The concentrations of N O 4 and NO change relatively quickly at first, but eventually stop changing with time when equilibrium is reached. 4 1

2 There is a simple relationship between the concentrations of the reactants and products for any chemical system at equilibrium. It is called the mass action expression, and it is derived from thermodynamics. Consider the equilibrium: H ( g) I ( g) HI( g) Four experiments to study the equilibrium among H, I, and HI gases. Different amounts of the reactants and products are placed in a 10.0 L reaction vessel at 440 o C where the gases establish equilibrium. When equilibrium is reached, different amounts of reactants and products remain. 5 6 The numerical value of the mass action expression is called the reaction quotient, Q. [HI] Expt. [H] [I] [HI] [H ][I I II III IV Average 49.5 ] 7 The reaction quotient can be evaluated at any concentrations. [HI] Q [H ][I ] At equilibrium (and 440 o C) for the this reaction the reaction quotient has the value 49.5 (an unit less number). 8

3 The value 49.5 is called the equilibrium constant, K c, and characterizes the system [HI] o K c 49.5 (at 440 C) [H ][I ] This relationship is called the equilibrium law for the system. For chemical equilibrium to exist, the reaction quotient Q must be equal to the equilibrium constant K c. dd ee At equilibrium f [ F] [ G] d [ D] [ E] ff gg K The form is always products over reactants raised to the appropriate powers The exponents in the mass action expression are the same as the stoichiometric coefficients. g e c 9 10 QUIZ For the following reaction find the formula of K C : H + N NH a) c) b) QUIZ Given the equilibrium concentrations for the reaction below, what is the value of the equilibrium constant in terms of concentrations? A (aq) + B (aq) C (aq) + D (aq) At equlibrium: [A]= 1 M; [C]= M [B]= M; [D] = 4 M 11 1

4 SOLUTION QUIZ K C = 6 1 For the reaction below, the equilibrium constant is equal to one. Which of the these sets of equilibrium concentrations is possible? A (aq) + B (aq) 4C (aq) A) [A]= M, [B]= 1 M, [C]= 1 M B) [A]= M, [B]= M, [C]= M C) [A]= 1 M, [B]= 1 M, [C]= M D) [A]= 1 M, [B]= 1 M, [C]= 1 M 14 VARIOUS OPERATIONS CAN BE PERFORMED ON EQUILIBRIUM EXPRESSIONS Changing the direction of equilibrium Multiplying coefficients Adding chemical equilibria Changing the direction of equilibrium When the direction of an equilibrium is reversed, the new equilibrium constant is the reciprocal of the original: PCl Cl PCl 5 PCl PCl Cl 5 [ PCl5 ] Kc [ PCl ][ Cl ] K ' c [ PCl ][ Cl] 1 [ PCl ] K 5 c

5 Multiplying the coefficients by a factor When the coefficients in an equation are multiplied by a factor, the equilibrium constant is raised to a power equal to that factor; PCl Cl PCl 5 PCl PCl 5 Cl [ PCl5 ] Kc [ PCl ][ Cl ] K " c [ PCl ] [ Cl ] [ PCl 5] K c Adding chemical equilibriums When chemical equilibria are added, their equilibrium constants are multiplied N N O N O O 4O N O Adding gives : 4NO 4NO K K K c c1 c [ N O] [ N [ O ] ] [ NO ] [ N [ NO ] [ N O] 4 4 ] [ O ] 4 [ O ] K c1 K c EXAMPLE Using the equilibrium constants from Table below calculate the value of K for the reaction CaCO (s) + H + (aq) Ca + (aq) + HCO (aq) CaCO (s) Ca + (aq) + CO (aq) K 1 = 5.01 x 10 7 HCO (aq) H + (aq) + CO (aq) K = 5.01 x Solution: The net reaction is the sum of reaction 1 and the reverse of reaction : CaCO (s) Ca + (aq) + CO (aq) H + (aq) + CO (aq) HCO (aq) CaCO (s) + H + (aq) Ca + (aq) + HCO (aq) K 1 = 5.01 x

6 The magnitude of K and the position of equilibrium. LARGE K C VALUES A large amount of product and very little reactant at equilibrium gives K>>1 (large K). When K 1, approximately equal amounts of reactant and product are present at equilibrium. When K<<1, mostly reactant and very little product are present at equilibrium. Large K C value means that reaction proceeds almost completely to the rights (products are favoured, products predominate over reactants) e.g. H + O H O 1 SMALL K C Small K C value means that reaction proceeds almost completely to the lefts (reactants are favoured, reactants predominate over products) e.g. N + O NO INTERMEDIATE K C When K C has a value between 10 - and 10 it tells that at equilibrium all components of reaction are present in significant amounts, and that the equilibrium reaction is fairly evenly balanced (neither lying to the left nor to the right). e.g. H + I HI at 400 o C K C = 50.5 (reactants and products are both present in approximately equal amounts). 4 6

7 QUIZ Look at the following equilibrium constants. Do these equilibrium reactions lie to the left, to the right, or in the middle? A) H CO +H O HCO - + H O + K C = B) NO + O NO K c = C) CH 4 + H O CO + H K c = D) CH COOH + H O CH COO - + H O + K c = PREDICTING THE DIRECTION OF A REACTION aa bb cc dd Q< K C aa bb cc dd. Q> K C aa bb cc dd Q (reaction quotient) K c (equilibrium constant) Q (reaction quotient) K c (equilibrium constant) [A] and [B] decrease (must be consumed) Greater than at equilibrium [C] and [D] increase (must be formed) [A] and [B] increase must be formed Less than at equilibrium [C] and [D] decrease; must be consumed Reaction proceeds from the left to the right 7 Reaction proceeds from the right to the left 8 7

8 Q=K C The reaction quotient is the same as the equilibrium constant. The reaction is already at equilibrium, no changes in the balance between the forward and back reactions. PROBLEM TO SOLVE The concentration of the three substances in the following reaction: H + I HI are as follows: [H ] = 1.5 mol L -1 [I ] = 0.5 mol L -1 [HI] = 1 mol L -1 K C = In which direction would the reaction need to proceed for equilibrium to be reached? 9 0 SOLVING Q = < 50.5 The reaction must proceed from the left to the right (forward reaction) for equilibrium to be reached 1 TO SELF-CHECK The concentration of the three substances in the following reactions: H + I HI are as follows: [H ] = 0.05 mol L -1 [I ] = 1.05 mol L -1 [HI] = 1.9 mol L -1 K C = In which direction would the reaction need to proceed for equilibrium to be reached? 8

9 The gas law can be used to write the equilibrium constant in terms of partial pressures molarity PV nrt P n V RT MRT Equilibrium constants written in terms of partial pressures are given the symbol K p N ( g) H K c ( g) [ NH ] [ N ][ H ] NH and ( g) NH N P H The size of the equilibrium constant gives a measure of how the reaction proceeds. General statements can be made about the equilibrium constant (either K c or K P ). K P P P 4 The two different forms of the equilibrium constants can be related PV nrt P K P n K ( RT ) g n RT MRT V c n g so that where (moles of gaseous products) -(moles of gaseous reactants) QUIZ Nitrogen gas and hydrogen gas combine to form ammonia in the Haber Bosch process. What equation can be used to convert K C to K p for the Haber Bosch process N (g) + H (g) NH (g) K p = K c (RT) Δn 5 6 9

10 QUIZ CONTINUATION For the same process at 5 o C, K C =.5 x10 8. What is K p? K p = K C (RT) - K p =( ). ( ) - = In a homogeneous reactions, all the reactants and products are in the same phase. Heterogeneous reactions involve more than one phase e.g. Thermal decomposition of sodium bicarbonate (baking soda) heat NaHCO ( s) NaCO ( s) HO( g) CO ( g) Heterogeneous reactions can come to equilibrium just like homogeneous systems 9 HETEROGENOUS EQUILIBRIUM In general, equilibrium constant expressions are presented in terms of molar concentration or gas phase pressures. For pure liquids or solids, experiments have shown that: The position of the equilibrium state of a system does not depend on the amount of liquid or solid in the reaction, provided that some exists. Pure liquids and solids do not appear in the equilibrium expression

11 If NaHCO is placed in a sealed container, heterogeneous equilibrium is established NaHCO ( s) Na CO ( s) CO ( g) H O( g) The equilibrium constant is : [ NaCO ][ CO ][ H O] K [ NaHCO ] The equilibrium law involving pure liquids and pure solids can be simplified For a pure liquid or solid, the ratio of amount of substance to volume of substance is constant The concentration of a substance in a solid is constant. Doubling the number of moles doubles the volume, but the ratio of moles to volume remains the same The equilibrium law for a heterogeneous reaction is written without concentration terms for pure solids or pure liquids. K c [ CO ][ H O ] QUIZ What is the equilibrium constant expression K p for the following reaction? CO (g) + H (g) < - > CO(g) + H O(l) The equilibrium constants found in tables represent all the constants combined

12 Solution Write the equilibrium constant expression the normal way for gases, in terms of the partial pressures of each gas. The liquid water on the right hand side of the equation will not have a term in the expression, so the answer is K = P CO /P CO P H FREE GIBBS ENERGY AND EQUILIBRIUM ΔG is related to equilibrium constant K c through van t Hoff isotherm ΔG = - RT ln K c ΔG = 0 at equilibrium because both reactant and product concentrations remain constant APPLICATION OF VAN T HOFF ISOTERM 1. CO + Cl COCl ΔG = -RT ln K K = 4.57 x10 9 at 100 o C ΔG = - (8.1 J K -1 mol -1 ) (( )K) (ln 4.57 x10 9 ) = kj mol -1 ΔG is negative, so reaction occurs spontaneously, reaction lies to the right (the products predominate over the reactants) and it is energetically favourable APPLICATION OF VAN T HOFF ISOTERM. N + O NO ΔG = -RT ln K K = 4.7 x10-1 at 5 o C ΔG = - (8.1 J K -1 mol -1 ) (( )K) (ln 4.7 x10-1 ) = kj mol -1 ΔG is positive, so reaction does not occur spontaneously, reaction lies to the left (very few products form) and it is energetically unfavourable

13 According to Le Châtelier s principle: If an outside influence upsets an equilibrium, the system undergoes a change in the direction that counteracts the disturbing influence and, if possible, returns the system to equilibrium common stresses Adding or removing a product or reactant Temperature Pressure Catalyst Inert gas Adding or removing a product or reactant The equilibrium shifts to remove reactants or products that have been added. The equilibrium shifts to replace reactants or products that have been removed REACTANTS OR PRODUCTS ADD/ REMOVE A + B C + D increase the concentration of A. It means that the position of equilibrium will move so that the concentration of A decreases again - by reacting it with B and turning it into C and D. The position of equilibrium moves to the right. Decreasing A A + B C + D The position of equilibrium will move so that the concentration of A increases again. It means that more C and D will react to replace the A that has been removed. The position of equilibrium moves to the left

14 TEMPERATURE A + B C + D ΔH= -50 KJ MOL -1 Increasing the temperature shifts a reaction in a direction that produces an endothermic (heat-absorbing) change. Decreasing the temperature shifts a reaction in a direction that produces an exothermic (heat-releasing) change. This shows that 50 kj is evolved (hence the negative sign) when 1 mole of A reacts completely with moles of B. For reversible reactions, the value is always given as if the reaction was one-way in the forward direction Suppose the system is at equilibrium at 00 C, and the temperature is increasing to 500 C. How can the reaction counteracts the change it has been made? How can it cool itself down again?

15 PRESSURE/ VOLUME PRESSURE A (G) + B (G) C (G) + D (G) This only applies to reactions involving gases: A (g) + B (g) C (g) + D (g) Increasing pressure The more molecules are present in the container, the higher the pressure will be. The system can reduce the pressure by reacting in such a way as to produce fewer molecules. Decreasing pressure Producing more molecules. In this case, the position of equilibrium will move towards the left-hand side of the reaction. QG6oc Cz8Y Catalysts have no effect on the position of equilibrium Catalysts change how fast a system achieves equilibrium, not the relative distribution of reactants and products Adding an inert gas at constant volume If the added gas cannot react with any reactants or products it is inert towards the substances in the equilibrium No concentration changes occur, so Q still equals K and no shift in equilibrium occurs 59 TO SUM UP Change Concentration Pressure Temperature A (G) + B (G) C (G) + D (G) + H Effect Increasing (reactants)favours the forward reaction Decreasing (reactants)favours the back reaction Increasing (products)favours the back reaction Decreasing (products) favours the forward reaction Increasing the pressure favours the reactions which yields the smaller number of molecules Decreasing the pressure favours the reactions which yields the largest number of molecules Increasing the temperature favours the endothermic reaction Decreasing the temperature favours the exothermic reactions 60 15

16 QUIZ What properties can be changes to shift the reaction equilibrium? A) volume B) temperature C) changing the reactant or product concentration D) catalyst E) inert gas F) the true answers are. 61 QUIZ Le Chatelier's principle states that a) if a chemical system at equilibrium is stressed, the system will adjust to increase the stress b) if a chemical system at equilibrium is stressed, the system will adjust to reduce the stress c) if a chemical system at equilibrium is stressed, the system will not adjust 6 QUIZ EXAMPLE Changes in pressure will only affect substances that are in the state. a. gaseous b. liquid c. solid Predicting the Effect of Changing Volume on Gas- Phase Reactions: Predict whether a decrease in the volume of the container will drive an equilibrium system for each reaction toward more products, toward more reactants, or neither. a. NH (g) + O (g) HNO (l) + H O (l) b. CO (g) + CF 4(g) COF (g) c. C (s) + H O (g) CO (g) + H (g)

17 NH (G) + O (G) HNO (L) + H O (L) CO (G) + CF 4(G) COF (G) Decreased volume shifts the system to the side of the reaction that has fewer moles of gas. This reaction has the same number of moles of gaseous reactants and products, EQULIBRIUM CALCULATIONS C (S) + H O (G) CO (G) + H (G) Be careful with this one. Although there are the same number of moles of reactants and products, one of the reactants is a solid. Equilibrium calculations can be divided into two main categories: 1) Calculating equilibrium constants from known equilibrium concentrations or partial pressures. ) Calculating one or more equilibrium concentrations or partial pressures using the known value of K c or K P

18 If mol N O 4(g) is placed in a 1 L flask at 5 o C, at equilibrium the following concentration are present: [ N O ] 0.09 mol/l 4 [ NO ] mol/l, and [ NO ] (0.0116) Kc [ N O ] (0.09) 4 N O 4(g) NO (g) Calculating the equilibrium constant this way is easy. 69 More commonly, a set of initial conditions and an equilibrium constant are used for calculation of reagents and products concentration at equilibrium. The Initial, Change, Equilibrium or ICE table is a useful way to summarize the problem. 70 Example: Ethyl acetate, CH CO C H 5, is produced from acetic acid and ethanol by the reaction CH CO H l) C H OH( l) CH CO C H ( l) H O( ( 5 5 l At 5 o C, K c = 4.10 for this reaction. Suppose mol of ethyl acetate and mol of water are placed in a 1.00 L reaction vessel. What are the concentrations of all species at equilibrium? ANALYSIS: Use an ICE table and the equilibrium constant to find the concentrations. ) All species are in the liquid phase. Let x concentration that reacts, then I ( M ) C ( M ) CH CO H C H OH 5 CH CO C H H O x x - x - x E ( M ) x x x x ( x)( x) Kc 4.10 x x x x

19 This can be solved by putting it in quadratic form: For an equation in the form ax bx c 0, For thissystem: b x b 4ac a x x Kc 4.10 or x.10x 0.50x and x 0.11 and x Negative concentrations are not allowed, so x and at equilibriu m [ CH CO H ] x M [ C H OH] x M [ CH CO C H ] x M [ H O] x 0.011M 5 5 A similar procedure can be used to calculate partial pressures using K P 7 74 Sometime simplifications can be made Example: Nitrogen and oxygen react to form nitrogen monoxide N ( g) O ( g) NO( ) g with K c = 4.8x10-1. In air at 5 o C and 1 atm, the N concentrations and O are initially 0.0 M and M. What are the equilibrium concentrations? ANALYSIS: The equilibrium constant is very small, very little of the reactants will be converted into products 75 I ( M ) C( M ) E( M ) c N x 0.0 K x x 1 O x (x), (0.0- x)( x) NO x x 4x or x (0.0)( )

20 Substituting: [N ]=0.0-x=0.0 M [O ]= x= M [NO]=x=1.60x10-17 M THE SAME CALCULATION EQUILIBRIA IN SOLUTIONS OF WEAK ACIDS AND BASES All weak acids behave the same way in aqueous solution: they partially ionize. In terms of the general weak acid HA, this process can be written as: HA HO HO A Following the procedures K a [ HO ][ A ] [ H ][ A ] [ HA] [ HA] 79 K a is called the acid ionization constant. This is often reported as the pk a pk log a K a list the K a and pk a for a number of acids A large pk a, means a small value of K a and only a small fraction of the acid molecules ionizes A small pk a, means a large value of K a and a large fraction of the acid molecules ionizes 80 0

21 QUIZ What is the H O + concentration and ph of a 0.10 M solution of hypochlorous acid (HOCl)? Hint: K a =.5 x 10-8 HOCl + H O = H O + + OCl ANSWER x = [H O + ] = 5.9 x 10-5 = [OCl - ] (also) ph = -log[h O + ] = -log(5.9 x 10-5 ) = 4. [HOCl] = x 10-5 = 0.10 M Weak bases behave in a similar manner like acids in water For the general base B: B H O HB OH the base ionization constant is [ HB ][ OH ] Kb [ B] Values of K b and pk b for a number of weak bases are listed in special Tables. Where, like for acids: pk log b K b

22 QUIZ What is the OH - concentration and ph of a 0.05 M solution of NH? Hint: K b = 1.8 x 10-5 There is an interesting relationship between the acid and base ionization constants for a conjugate acid-base pair Using the general weak acid HA: For theweak acid : HA H O A H O the product is H O A for theconjugate base : HA OH [ HO ][ A ] [ HA][ OH ] Ka Kb [ HA] [ A ] K [ H O ][ OH ] K a [ HA][ OH ] Kb [ A ] w [ HO ][ A ] [ HA] Thus, for any conjugate acid-base pair: K K K and a pk a b pk b w pk w o (at 5 C) Most tables of ionization constants only give values for the molecular member of the conjugate acid-base pair. The ionization constant of the ion member of the conjugate acid-base pair is then calculated as needed

23 Relative strengths of conjugate acidbase pairs. The stronger the acid is, the weaker the conjugate base. The weaker the acid, the stronger the conjugate base. Very strong acids ionize 100% and their conjugate bases do not react to any measurable extent. 89 COMMON CALCULATION The primary goal is usually to determine the equilibrium concentration for all species in the mass action expression The percentage ionization of the acid or base is defined as: percentage moles ionized per liter ionization 100% moles available per liter This, and the ph, are often used or requested in equilibrium calculations 90 Example: Morphine is very effective at relieving intense pain and it is a weak base. What is the K b, pk b, and percentage ionization of morphine if a M solution has a ph of 10.10? ANALYSIS: The reaction can be represented as: I C E B( aq) H O x x BH ( aq) OH x At equilibrium, [OH - ] = x = 10 -poh 0 x 0 x x [ BH ][ OH ] Kb [ B] x x SOLUTION: Use poh = ph, substituting: [ OH ] ( ) 6 4 M, then 4 x (1. 10 ) Kb x ( so pk 5.80, and x % ionization 100% % b 4 ) 91 9

24 SIMPLIFYING ASSUMPTIONS The quadratic equation is usually used to solve equilibrium problems time consuming Simplifying assumptions : the initial concentration of a weak acid or base in pure water is more that 400 times the ionization constant C O >400* K 9 94 I C Example: Calculate the ph of a M solution of dimethylamine for which K b =9.6x10-4. ANALYSIS: 400* K b > M, so use of the quadratic equation is indicated SOLUTION: Set the problem up B( aq) H O x E x BH ( aq) OH ( aq) x 0 x 0 x x K b 4 - [ BH ][ OH ] [ B] x ( x) 95 Put in standard form x Kb ( x) ( x) K x 0 x K x K 4 7 x x Solve for x and the equilibrium concentrations Only positivesolutions are allowed, so x M 4 [ BH ] [ OH ] x M [ B] ( x) M.910 b (9.610 b 4 (1) and 4 M ) 4(1)( b 7 ) 96 4

25 BUFFERS HOW IT'S WORKING? Minimizes ph changes when a small amount of strong acid or base is added to certain solutions Buffer solution usually contains two solutes, one providing a weak acid and the other a weak base If the weak acid is molecular, then the conjugate base can be supplied as a soluble salt of the acid Consider the general buffer made so that both acid HA and salt A - are present in solution 1. When base (OH - ) is added, the weak acid react with added base: HA( aq) OH ( aq) A ( aq) HO Net result: small changes in ph HOW IT'S WORKING? Consider the general buffer made so that both acid HA and salt A - are present in solution. When acid (H + ) is added, the salt A - react with added acid A ( aq) H ( aq) HA( aq) Net result: small changes in ph CALCULATIONS INVOLVING BUFFER SOLUTIONS It is worth remembering that for buffers: the initial concentration of both the weak acid and its conjugate base can be used as though they were equilibrium values molar concentrations or moles can be used in the K a (or K b ) (the same units must be used for both members of the pair)

26 Example: What is the ph of a buffer made by adding 0.10 mol NH and 0.11 mol NH 4 Cl to.0 L of solution? The K b for ammonia is 1.8x10-5 ANALYSIS: This is a buffer, initial concentrations can be used as equilibrium values: NH ( aq) H O NH ( aq) OH ( aq) mol [ NH ] 0.050M.0 L 0.11mol [ NH 4 ] 0.055M (from NH 4Cl).0 L [ NH 4 ][ OH ] Kb [ NH ] SOLUTION: Solve for [OH - ] and use this to calculate the ph [ NH 4 ][ OH ] 5 (0.055)[ OH ] K b [ NH ] (0.050) 5 (0.050) [ OH ] poh 4.79 and ph WHAT DETERMINES THE PH OF THE BUFFER? For the general weak acid HA: HA( aq) [ H Thus both : rearrangin g gives ] K the value of K a H ( aq) A ( aq) a [ HA] or [ H [ A ] ] K mol HA mol A the ratio of the molarities (or the ratio of moles) affect the ph. [ H ][ A ] Ka [ HA] a 10 WHAT DETERMINES THE PH OF THE BUFFER? ph pk ph pk a a [ A ] log [ HA] initial [ salt] log [ acid ] initial the Henderson-Hasselbalch equation it's worth remembering!!! the ph is mostly determined by the pk a of the acid;(in buffer, the concentration ratio is usually near 1) it's worth remembering!!! or the Henderson-Hasselbalch equation it's worth remembering!!! 104 6

27 Example: A solution of 0.0 M in acetic acid (HAc) and 0.10 M in sodium acetate (NaAc) were mixed. What is the ph of new solution? ICE TABLE Compare x to the K a value. In our sample problem, the concentrations of the weak acid and its conjugate base are 10 - and the K a is The difference is greater than 100 so both of the x quantities may be ignored. ph = -log[h O + ] = -log[.6 x 10-5 ] =

28 BUFFER CAPACITY BUFFER CAPACITY The goal of a buffer is to keep the ph of a solution within a narrow range Generally, the ph change in an experiment must be limited to about 0.1pHunit Buffer capacity =the effectiveness of a buffer A buffer s capacity is determined by the magnitudes of the molarities of its components. For a useful buffer : ph pk a Example: 0.0 mol of HCl was added to a buffer made from 0.10 mol HA (pk a =7.0) and 0.15 mol NaA in.0 L with no volume change. What is the ph change? ANALYSIS: This buffer problem is best solved in terms of moles. HCl is a strong acid. The H + it contributes to solution increases the amount of HA present at the expense of A -. SOLUTION: The ph before addition of HCl was: [ A ] [ HA] final final ( ) mol 0.1 mol ( ) mol 0.1 mol the phof the new solution is : 0.1 mol ph 7.0 log 7., and the phchange is 0.1 mol ph ph ph finial initial ph p K a [ A ] 0.15 mol log 7.0 log 7.8 [ HA] 0.10 mol

29 AND IF THERE WAS NO BUFFER If the HCl had been added to pure water, the ph change would have been much larger: 0.0 ph log THANK YOU FOR YOUR KIND ATTENTION!!! 115 9

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