# Calculating the pH of a 'buffer' solution?

Question: Determine the $$\mathrm{pH}$$ of the solution resulting when $$\pu{100 cm^3}$$ of $$\pu{0.50 mol dm-3}$$ $$\ce{CH2ClCOOH}$$ is mixed with $$\pu{200 cm^3}$$ of $$\pu{0.10 mol dm-3}$$ $$\ce{NaOH}$$.

The given solution:

Concentration of acid: $$\frac{0.030}{0.300} = \pu{0.10 mol dm-3}$$

Concentration of base/salt: $$\frac{0.020}{0.300} = \pu{0.067 mol dm-3}$$

$$\ce{[H+]} = \frac{K_\mathrm{a} \times \ce{[HA]}}{\ce{[A-]}} = \frac{1.3 \times 10^{-3} \times 0.10}{0.067} = \pu{1.9 \times 10^{-3} mol dm-3}$$

$$\mathrm{pH} = 2.72$$

Now, here is why I am confused: this calculation is for a buffer solution, where $$\ce{HA}$$ and $$\ce{A-}$$ are a conjugate acid/base pair. However, chloroethanoic acid and sodium hydroxide are not conjugate acid/base (are they?) - so why does this calculation work? What is the intuitive way to think about this?

• $\ce{NaOH}$ is reacting with the acid to produce it's conjugate base and thus the calculation. Please consider typing the question out instead of using images so that it is searchable. Question in the form of images are quickly closed due to sites policies. Apr 5 '21 at 9:09
• @NisargBhavsar - Thank you, but I am not sure I understand. What is the conjugate base of NaOH!? And thank you for the feedback - I will edit the question now :) Apr 5 '21 at 9:26
• I mean that the hydroxide will react with chloroethaonic acid to form sodium chloroethanoate which is the conjugate base of chloroethanoic acid. Apr 5 '21 at 11:30
• Please do not attach wordy excerpts as an image. Just type it in. Apr 5 '21 at 14:52

Question: Determine the $$\mathrm{pH}$$ of the solution resulting when $$\pu{100 cm^3}$$ of $$\pu{0.50 mol dm-3}$$ $$\ce{CH2ClCOOH}$$ is mixed with $$\pu{200 cm^3}$$ of $$\pu{0.10 mol dm-3}$$ $$\ce{NaOH}$$.

I'm not sure what level of chemistry is OP's in, but the given solution for the question is for the chemistry students with appreciable knowledge of stoichiometry of chemical reactions and buffers. That's probably why OP seems so confused of the solution. The answer elsewhere explains to certain level but has not explore the total picture. Hence, I thought it would be beneficial to OP if I explain it a little further:

$$\ce{CH2ClCOOH}$$ is a weak acid (about 100 times acidic than acetic acid) with $$K_\mathrm{a}$$ of $$1.3 \times 10^{-3}$$ (which is $$1.8 \times 10^{-5}$$ for acetic acid). On the other hand, $$\ce{NaOH}$$ is a strong base, which reacts with strong or weak acids irreversibly. When add $$\ce{CH2ClCOOH}$$ and $$\ce{NaOH}$$ together, following reaction would take place:

$$\ce{CH2ClCOOH (aq) + NaOH (aq) -> CH2ClCOONa (aq) + H2O (l)} \tag1$$

The net reaction is:

$$\ce{CH2ClCOOH (aq) + OH- (aq) -> CH2ClCOO- (aq) + H2O (l)} \tag2$$

The acid to base mole ratio of the reaction is 1:1. However, added amounts of acid and base are:
Weak acid, $$\ce{CH2ClCOOH}$$: $$\pu{0.50 mol dm-3} \times \pu{0.100 dm3} = \pu{0.050 mol}$$
Strong base, $$\ce{NaOH}$$: $$\pu{0.10 mol dm-3} \times \pu{0.200 dm3} = \pu{0.020 mol}$$

That makes $$\ce{NaOH}$$ the limiting reagent (recall your general chemistry knowledge). Thus, $$\pu{0.020 mol}$$ of $$\ce{NaOH}$$ reacts with $$\pu{0.020 mol}$$ of $$\ce{\ce{CH2ClCOOH}}$$ to give $$\pu{0.020 mol}$$ of $$\ce{\ce{CH2ClCOONa}}$$ or $$\ce{\ce{CH2ClCOO- + Na+}}$$, if you look at the net reaction.

The unreacted amount of $$\ce{\ce{CH2ClCOOH}}$$ in the mixture ($$\pu{0.300 dm3}$$ in total volume) is: $$\pu{0.050 mol} - \pu{0.020 mol} = \pu{0.030 mol}$$.

Therefore, the concentration of acid, $$\ce{\ce{CH2ClCOOH}}$$: $$\frac{\pu{0.030 mol}}{\pu{0.300 dm3}} = \pu{0.10 mol dm-3}$$

Simillarly, the concentration of salt, $$\ce{\ce{CH2ClCOONa}}$$ or $$\ce{\ce{CH2ClCOO-}}$$: $$\frac{\pu{0.020 mol}}{\pu{0.300 dm3}} = \pu{0.067 mol dm-3}$$

Since, $$\ce{\ce{CH2ClCOOH}}$$ is a weak acid, it would ionize in aqueous solution as follows:

$$\ce{CH2ClCOOH (aq) + H2O (aq) -> CH2ClCOO- (aq) + H3O+ (aq)} \tag3$$

Suppose, $$\alpha$$ amount (in $$\pu{mol dm-3}$$) of $$\ce{\ce{CH2ClCOOH}}$$ ionizes at $$\ce{\ce{CH2ClCOOH}}$$, the equilibrium concentrations of $$\ce{\ce{CH2ClCOOH}}$$, $$\ce{\ce{CH2ClCOO-}}$$, and $$\ce{\ce{H3O+}}$$ are $$0.10 - \alpha$$, $$0.067 + \alpha$$, and $$\alpha$$, respectively. $$K_\mathrm{a} = \frac{\ce{[H3O+]}\ce{[A-]}}{\ce{[HA]}}$$ $$\therefore \ \ce{[H+]} = \alpha = \frac{K_\mathrm{a} \times \ce{[HA]}}{\ce{[A-]}} = \frac{1.3 \times 10^{-3} \times (0.10 - \alpha)}{(0.067 + \alpha)} \approx \pu{1.9 \times 10^{-3} mol dm-3}$$

$$\mathrm{pH} = -\log\alpha = -\log (1.9 \times 10^{-3}) = 2.72$$

On the other hand this can be easily solved using the Henderson-Hasselbalch equation as directed in the other answer elsewhere: $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac {\ce{[A-]}}{\ce{[HA]}}\right) = 2.89 + \log \left(\frac {0.067}{0,10}\right) = 2.72$$

$$\ce{CH2ClCOOH}$$ is a weak acid; $$\ce{NaOH}$$ is a strong base. Now, tell me, what do you think happens when you mix a strong base with a weak acid?

## They react.

$$\ce{CH2ClCOOH + NaOH -> CH2ClCOONa + H2O}$$

A neutralisation reaction in aqueous medium depends upon the number of equivalents of acid and base. $$N_1V_1=N_2V_2$$ At this stage, an equivalence point is reached. However, the limiting reagent in the given question is $$\ce{NaOH}$$. At the equivalence point, the solution will contain $$\pu{0.03 mol}$$ of $$\ce{CH2ClCOONa}$$ and $$\pu{0.02 mol}$$ of excess reagent $$\ce{CH2ClCOOH}$$.

Now, $$\ce{CH2ClCOONa}$$ is a salt of a WEAK acid/STRONG Base while $$\ce{CH2ClCOOH}$$ is the WEAK acid. Now, I believe you can refer to the solution given in your book.

Calculate the concentration of the respective constituents in the equilibrium, and then proceed with plugging those values into the Henderson-Hasselbalch equation: $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac {\ce{[A-]}}{\ce{[HA]}}\right)$$ where $$\mathrm{p}K_\mathrm{a}$$ is the acid dissociation constant of $$\ce{CH2ClCOOH}$$.