# Amount of benzoic acid required to alter pH of a NaOH-solution

I am trying to do a practice problem but I can't figure it out:

How many grams of benzoic acid needs to be added to 0.10 l of a NaOH-solution with a concentration of 1.0 mM to reduce the pH of said solution to 4.00? Assume that the added mass of acid does not alter the total volume of solution. $$K_a = 6.5 \cdot 10^{-5}$$

My attempt:

Benzoic acid is a monoprotic acid with a molecular weight of 122.1 $$\,$$ g mol$$^{-1}$$ and I will assume that NaOH has a 100% dissociation.

I choose to divide the problem into two parts; calculating the substance amout of acid needed to neutralize the OH$$^-$$ ions separate from the substance amount needed to lower the pH to 4.

Substance amount needed to neutralize OH$$^-$$, $$n_n$$:

$$\ce{HA_{eq} + H_2O <=> A^- + H^+ + OH^-}$$

$$K_a = \frac{[H^+][A^-]}{[HA]_{eq}}$$

$$[HA]_{eq} = [HA] - [H^+]$$

$$[HA] = \frac{n_n}{V}$$

Deriving an expression for $$n_n$$:

$$n_n = \biggl(\frac{[H^+][A^-]}{K_a} + [H^+]\biggr) \cdot V$$

$$K_a$$ is listed as $$6.5 \cdot 10^{-5}$$, the volume, $$V$$, is $$0.10$$ l and $$[H^+] = [A^-] = [OH^-] = 1.0$$ mM as listed.

Result: Plugging in these values yields the following substance amount: $$n_n =1.6 \cdot 10^{-3}$$ mol.

Substance amount needed to lower the pH to 4, $$n_s$$:

Deriving expression for $$n_s$$ using same method as above:

$$n_s = \biggl(\frac{[H^+][A^-]}{K_a} + [H^+]\biggr) \cdot V$$

In this case $$[H^+] = [A^-] = 1.00 \cdot 10^{-4}$$ based on pH = 4.00. ($$V$$ and $$K_a$$ same as above)

Result: Plugging in these values yields: $$n_s = 2.5 \cdot 10^{-5}$$ mol.

The mass of benzoic acid needed can now be calculated using the substance amount $$n = n_n + n_s$$:

$$m = nM$$

Final result: Plugging in the values for $$n$$ and $$M = 122.1$$ g mol$$^{-1}$$ yields $$m = 0.20$$ g.

This answer is not right and I know I am doing something wrong. Teach me the ways.

• If you keep adding benzoic acid until all of NaOH present is neutralized, your solution will consist of sodium benzoate ions, which will undergo hydrolysis to form a solution with pH > 7. So you have to continue adding benzoic acid. This means that your beaker will now consist of a buffer solution. So, you can use the Henderson-Hasselbalch equation to get the answer directly. – himanshu May 15 '19 at 11:49
• Have you tried pH=pKa + log ( [A-]/[HA]) ? – Poutnik May 16 '19 at 11:27