# K_{a} and K_{b} calculations

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###### Topic Notes

In this lesson, we will learn:

- How to apply the K
_{a}expression to find the pH of weak acid solutions. - The assumptions made in K
_{a}calculations at equilibrium and how to justify them. - How to solve for K
_{b}using K_{w}and find the pH of weak base solutions.

__Notes:__- We know that weak acids and bases are any acid/base species that does not completely dissociate in water. Dissolving a weak acid in water then has two chemical effects:
- Some of the weak acid, HX, will interact with water and dissociate into H
_{3}O^{+}and X^{-}ions. - The rest of the HX will stay un-dissociated.

HX + H _{2}O $\rightleftharpoons$ H_{3}O^{+}+ Cl^{-}

Start concentration (M)

Equilibrium conc. (M)

HX

A

A - B

H

_{3}O^{+}$\approx$ 0

B

X

^{-}0

B

The acid dissociation constant, K_{a}can be expressed in these terms:*K*= $\frac{[X^{-}][H_3O^+]}{[HX]}$ = $\frac{[X^2]}{[A] -B}$_{a} - Some of the weak acid, HX, will interact with water and dissociate into H
- Taking an example with 0.1M methanoic acid HCOOH, we can write the following:
Start concentration (M)

Equilibrium conc. (M)

HCOOH

0.1

0.1 - B

H

_{3}O^{+}0

B

X

^{-}0

B

We can apply the acid dissociation constant, K_{a}to this equilibrium. Methanoic acid^{1}has a K_{a}value of 1.8$*$10^{-4}so the K_{a}equation can be written fully:*K*= $\frac{[HCOO^-][H_3O^+]}{[HCOOH]}$ = $\frac{[H_3O^+]^2}{[0.1] -[H_3O^+]}$ = 1.8 $*$ 10_{a}^{-4}__Some assumptions are made to complete this calculation__:- The starting concentration of H
_{3}O^{+}ions, in neutral water is only 1*10^{-7}M (see Autoionization of water) and an equally tiny amount of hydroxide ions are also present. This is an incredibly small amount, so this H_{3}O^{+}is not taken into the calculation;__only H___{3}O^{+}__due to the weak acid is used in the calculation__. - With weak acids, we assume that
__the acid is weak enough that the amount of dissociation doesn’t affect acid concentration__. Using the table above, 0.1 M HCOOH added to neutral water will still have concentration of approximately 0.1M at equilibrium, and we can ignore the ‘– B’ in ‘A-B’. Where [HX]_{eq}= equilibrium concentration, [HX]_{i}= start concentration of HX:[HX] _{eq}- B $\cong$ [HX]_{i}__YOU MUST STATE THIS ASSUMPTION IN CALCULATIONS__.

With the assumptions, we have a final expression:*K*= $\frac{[HCOO^-][H_3O^+]}{[HCOOH]}$ = $\frac{[H_3O^+]^2}{[0.1] }$ = 1.8 $*$ 10_{a}^{-4}0.1 $*$ 1.8 $*$ 10 ^{-4}= [H_{3}O^{+}]^{2}$\small\sqrt{1.8 * 10^{-5}}$ = [H _{3}O^{+}] = 4.24 $*$ 10^{-3}*pH*= -log [H_{3}O^{+}] = 2.37

Note that the assumption we made was that [HX] – B is approximately equal to [HX]. We now know that B = 4.24*10^{-3}, so our assumption in this example was to say that 0.1 – 4.24*10^{-3}= 0.0958.__The assumption can be justified if percentage dissociation is less than 5%__which we can work out:

% dissociation = $\frac{[H_3O^+]_{eq}}{[HX]_i}$ $*$ 100

Therefore:

% dissociation = $\frac{[4.24 *10^{-3}]}{[0.1]}$ $*$ 100 = 4.24%

As the calculation shows, the assumption was justified as only 4.24% dissociation occurs. - The starting concentration of H
__K__:_{b}calculations are similar to K_{a}calculations with some changes- Because acidity strength tables give only K
_{a}, K_{b}of a weak base will need to be found by the calculation in the autoionization of water expression. You will need to__find the K__._{a}of the conjugate acid in the acidity strength table to do this - The equilibrium concentrations you obtain
__using K___{b}will give you [OH^{-}], so pH will need to be found by__solving: pH = 14 – pOH__.

Taking an example with 0.5M of the weak base ammonia, NH_{3}, we can write the following:

NH _{3}+ H_{2}O $\rightleftharpoons$ NH_{4}^{+}+ OH^{-}Start concentration (M)

Equilibrium conc. (M)

NH

_{3}0.5

0.5 - B

NH

_{3}^{+}0

B

OH

^{-}0

B

The conjugate acid of ammonia is the ammonium ion, NH_{4}^{+}which has a K_{a}value^{1}of 5.6 $*$ 10^{-10}. Solving the autoionization expression for K_{b}(NH_{3}) gives:*K*= $\frac{K_w}{K_a}$ = $\frac{10^{-14}}{5.6 * 10^{-10}}$ = 1.79 $*$ 10_{b}^{-5}

Using our value for K_{b}(now rounding to 1.8 $*$ 10^{-5}or 2 significant figures) we can solve for the equilibrium concentration of hydroxide ions. Again, we make the assumption that the concentration of NH_{3}isn’t significantly affected by the dissociation into NH_{4}^{+}:[NH _{3}]_{eq}- B $\cong$ [NH_{3}]_{i}

Now we can find the hydroxide ion concentration:*K*= $\frac{[NH_{4}^{+}][OH^-]{}} {[NH_3]}$ = $\frac{[OH^-]^2}{[0.5]}$ = 1.8 $*$ 10_{b}^{-5}[OH ^{-}] = $\sqrt{((1.8 * 10^{-5}) * 0.5)}$ = 3 $*$ 10^{-3}pOH = -log[OH ^{-}] = -log (3 $*$ 10^{-3}) = 2.52pH = 14 - pOH = 14 - 2.52 = 11.48

Testing the assumption can now be done:% dissociation = $\frac{[OH^-]_{eq}}{[NH_3]_i}$ $*$ 100 = $\frac{3 * 10^{-3}}{0.5}$ $*$ 100 = 0.6 %

With only 0.6% dissociation, the assumption is justified and the pH has been found.- Because acidity strength tables give only K
- Another calculation that shows the difference between strong and weak acids and bases is
__the effect of dilution on pH__.__The effect on strong acids is straightforward__, because we assume 100% dissociation:- If a solution of strong acid, e.g. HCl is 1M, then [H
_{3}O^{+}_{ (aq)}] = 1M. Taking the negative log of this:

pH = -log[1] = 0

Diluting this by a factor of 10 will give a concentration of 0.1M

pH = -log[0.1] = 1

A further 10, or 100 fold from the original:

pH = -log[0.01] = 2

In short, diluting a strong acid or base has a direct logarithmic effect on pH.

__The effect of dilution on weak acids and bases is different__:- If a solution of weak acid, e.g. CH
_{3}COOH is 1M, then pH and [H_{3}O^{+}_{ (aq)}] is worked out using the K_{a}expression:

K_{a}(CH_{3}COOH) = 1.4*10^{-5}*K*= $\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}$_{a}

The calculation to find pH using this expression has been explained above, so moving forward to an answer (using the assumptions needed)1.4 * 10 ^{-5}= $\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[1]}$ where [H_{3}O^{+}] = [CH_{3}COOH^{-}]$\sqrt{1.4 * 10^{-5}}$ = [H _{3}O^{+}] = 3.74 * 10^{-3}__pH = -log[ 3.74 * 10__^{-3}__] = 2.42__

A dilution of this weak acid solution to make it 0.1M would have the following effect on the calculation:1.4 * 10 ^{-5}= $\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[0.1]}$ where [H_{3}O^{+}] = [CH_{3}COOH^{-}]$\sqrt{1.4 * 10^{-6}}$ = [H _{3}O^{+}] = 1.18 * 10^{-3}__pH = -log[ 1.18 * 10__^{-3}__] = 2.92__

Another dilution by a factor of ten:1.4 * 10 ^{-5}= $\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[0.01]}$ where [H_{3}O^{+}] = [CH_{3}COOH^{-}]$\sqrt{1.4 * 10^{-7}}$ = [H _{3}O^{+}] = 3.74 * 10^{-4}__pH = -log[ 3.74 * 10__^{-4}__] = 3.42__

In short, diluting a weak acid has a lesser effect on pH than in strong acids.

- If a solution of strong acid, e.g. HCl is 1M, then [H

remaining today

remaining today