# Buffer Solution - Sodium Benzoate

A buffer with a pH of $\mathrm{4.79}$ contains $\mathrm{ 0.39\ M}$ of sodium benzoate and $\mathrm{0.10\ M}$ of benzoic acid. What is the concentration of $\mathrm{[\ce{H} ]}$ in the solution after the addition of $\mathrm{0.058\ mol}$ of $\ce{HCl}$ to a final volume of $\mathrm{1.6\ L}$?

My thought is that $$\mathrm{pH=pK_a+Log_{10}(\frac{[base]}{[acid]})}$$

When we added $\mathrm{0.058\ mol}$ of $\ce{HCl}$, the concentration of both acid and base decreased by $\frac{.058}{1.6}$, so I take that ratio and apply it. I found $\mathrm{pK_a}$ by just applying what's given. Not sure where I am going wrong.

In that equation, the acid and base are a conjugate acid/base pair. We assume HCl quantitatively dissociates and protonates benzoates. The concentration of the acid increases, and the base decreases.

Empirically, the pH should go down in this case. Without seeing your specific calculation, we can't comment where you are going wrong, but the proper flow should look something like:

$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]-[Added H]}{[acid]+[AddedH]})}$$

$$\mathrm{pH=pK_a+Log_{10}(\frac{0.39M - 0.036M}{0.10M + 0.036M})}$$

$$\mathrm{pH=pK_a+Log_{10}(\frac{0.354M}{0.136M})}$$ From your calculated $\mathrm{pKa = 4.20}$, you should get there pretty quickly.

Just applying the difference in the log values of the changed ratios, I get $\mathrm{pH = 4.61}$

The concentration of [H] ion will be $2.45x10^{-5}$ if the pH was 4.61.

• Generally it is favored to show how you arrive to the specific answer, so that we can assist in the understanding of how to solve the problem. – John Snow Dec 16 '14 at 17:39
• pH=-log[H+], then [H+]=antilog(-pH) [H+]=antilog(-4.61)=2.45x10^-5 – Elif Bayındır Özlü Dec 17 '14 at 7:20