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:


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.

  • 3
    $\begingroup$ 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. $\endgroup$ – himanshu May 15 at 11:49
  • $\begingroup$ Have you tried pH=pKa + log ( [A-]/[HA]) ? $\endgroup$ – Poutnik May 16 at 11:27

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