# 1 Weak Acids and Bases A weak acid or base is an acid or base that partially (less than 100%) dissociated in water. The equilibrium constant is usually.

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1 Weak Acids and Bases A weak acid or base is an acid or base that partially (less than 100%) dissociated in water. The equilibrium constant is usually small and, in most cases, one can use the concepts mentioned in the equilibrium calculations section discussed previously with application of the assumption that the amount dissociated is negligible as compared to original concentration. This assumption is valid if the equilibrium constant is very small and the concentration of the acid or base is high enough.

2 Example Calculate the pH and pOH for a 0.10 M acetic acid solution. K a = 1.75x10 -5 Solution [H + ] solution = [H + ] HOAc + [H + ] water However, in absence of an acid the dissociation of water is extremely small and in presence of an acid dissociation of water becomes negligible due to the common ion effect. Therefore, we can neglect the [H + ] water (which is equal to [OH - ]) in presence of an acid since the hydroxide ion concentration is insignificant in an acid solution, therefore we can write

3 [H + ] solution = [H + ] HOAc The first point is to write the equilibrium where HOAc  H + + OAc -

4 K a = [H + ][OAc - ]/[HOAc] K a = x * x / (0.10 – x) K a is very small. Assume 0.10 >> x 1.75*10 -5 = x 2 /0.10 x = 1.3x10 -3 Relative error = (1.3x10 -3 /0.10) x 100 = 1.3% The assumption is valid and the [H + ] = 1.3x10 -3 M pH = 2.88 pOH = 14 – 2.88 = 11.12 Now look at the value of [OH - ] = 10 -11.12 = 7.6x10 -12 M = [H + ] from water. Therefore, the amount of H + from water is negligible

5 Example Calculate the pH and pOH for a 1.00x10 -3 M acetic acid solution. K a = 1.75x10 -5 Solution The first point is to write the equilibrium where HOAc  H + + OAc -

6 K a = [H + ][OAc - ]/[HOAc] K a = x * x / (1.00*10 -3 – x) K a is very small. Assume 1.00*10 -3 >> x 1.75*10 -5 = x 2 /1.00*10 -3 x = 1.32x10 -4 Relative error = (1.32x10 -4 /1.00x10 -3 ) x 100 = 13.2% The relative error is more than 5% therefore, the assumption is invalid and we have to use the quadratic equation to solve the problem.

7 Example Find the pH and pOH of a 0.20 M ammonia solution. K b = 1.8x10 -5. Solution The same treatment above can be used to solve this problem where: [OH - ] solution = [OH - ] ammonia + [OH - ] water [H + ] = [OH - ] water However, in absence of a base the dissociation of water is extremely small and in presence of the base the dissociation of water becomes negligible due to the common ion effect.

8 Therefore, we can neglect the [H + ] in presence of ammonia since the hydrogen ion concentration is insignificant in basic solution, therefore we can write OH - ] solution = [OH - ] ammonia NH 3 + H 2 O  NH 4 + + OH -

9 K b = [NH 4 + ][OH - ]/[NH 3 ] 1.8*10 -5 = x * x / (0.20 – x) k b is very small that we can assume that 0.20>>x. We then have: 1.8*10 -5 = x 2 / 0.2 x = 1.9x10 -3 M Relative error = (1.9x10 -3 /0.2) x 100 = 0.95% The assumption is valid, therefore: [OH - ] = 1.9x10 -3 M, [H + ] = 5.3x10 -12 M = [OH - ] water which is very small. pOH = 2.72 pH = 11.28

10 4. Salts of Weak Acids and Bases Imagine that an acid is formed from two species a hydrogen ion and a conjugate base attached to it. The acid is said to be strong if its conjugate base is weak while a weak acid has a strong conjugate base. Therefore, we can fairly recognize conjugate bases like Cl -, NO 3 -, and ClO 4 - as weak bases that do not react with water and thus will not change the pH of water (pH = 7). On the other hand, conjugate bases derived from weak acids are strong bases which react with water and alter its pH.

11 Examples of strong conjugate bases include OAc -, NO 2 -, CN -, etc.. The same applies for bases where weak bases are weak because their conjugate acids are strong which means they react with water and thus alter its pH. One important piece of information with regards to salts of weak acids and bases is that we have to find their k a or k b as the equilibrium constants given in problems are for the parent acid or base. Let us look at the following argument for acetic acid: HOAc  H + + OAc - K a = [H + ][OAc - ]/[HOAc]

12 For the conjugate base of acetic acid (acetate) we have OAc - + H 2 O  HOAc + OH - K b = [HOAc][OH - ]/[OAc - ] Let us multiply k a times k b we get K a k b = [H + ][OH - ] = k w, or K a k b = k w Therefore, if we know the k a for the acid we can get the equilibrium constant for its conjugate base since we know k w. We can find k a for the conjugate acid by the knowledge of the equilibrium constant of the parent base.

13 Example Find the pH of a 0.10 M solution of sodium acetate. K a = 1.75x10 -5 Solution [OH - ] solution = [OH - ] acetate + [OH - ] water [OH - ] water = [H + ] Since the hydrogen ion concentration is very small in a solution of a base, we can neglect [OH - ] water and we then have [OH - ] solution = [OH - ] acetate

14 OAc - + H 2 O  HOAc + OH - K b = k w /k a K b = 10 -14 /1.75x10 -5 = 5.7x10 -10

15 K b = [HOAc][OH - ]/[OAc - ] K b = x * x/(0.10 – x) K b is very small and we can fairly assume that 0.10>>x 5.7x10 -10 = x 2 /0.1 x = 7.6 x 10 -6 Relative error = (7.6x10 -6 /0.10) x100 = 7.6x10 -3 % The assumption is valid. [OH - ] = 7.6x10 -6 M [H + ] = 1.3x10 -9 M = [OH - ] water

16 The relative error in neglecting H + from water = (1.3x10 -9 /7.6x10 -6 ) x 100 = 0.017% This validate our assumption at the beginning of the solution that [OH - ] acetate >> [OH - ] water pOH = 5.12 pH = 14 – 5.12 = 8.88

17 Example Calculate the pH of a 0.25 M ammonium chloride solution. K b = 1.75x10 -5 Solution [H + ] solution = [H + ] ammonium + [H + ] water However, in absence of an acid the dissociation of water is extremely small and in presence of an acid dissociation of water becomes negligible due to the common ion effect. Therefore, we can neglect the [H + ] water in presence of an acid since the hydroxide ion concentration is insignificant in an acid solution

18 therefore we can write [H + ] solution = [H + ] ammonium The first point is to write the equilibrium where NH 4 +  H + + NH 3

19 K a = 10 -14 /1.75x10 -5 = 5.7x10 -10 K a = [H + ][NH 3 ]/[NH 4 + ] K a = x * x / (0.25 – x) K a is very small. Assume 0.25 >> x 5.7*10 -10 = x 2 /0.25 x = 1.2x10 -5 Relative error = (1.2x10 -5 /0.25) x 100 = 4.8x10 -3 % The assumption is valid and the [H + ] = 1.2x10 -5 M

20 Now look at the value of [OH - ] = 10 -14 /1.2x10 -5 = 8.3x10 -10 M = [H + ] from water. Therefore, the amount of H + from water is negligible when compared to that from the acid. The relative error for neglecting the H + from water = (8.3x10 -10 /1.2x10 -5 ) x 100 = 6.9x10 -3 % pH = 4.92 pOH = 14 – 4.92 = 9.08 We should remember that dissociation of water is negligible in presence of an acid or base.

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