# Finding is the concentration of H⁺ in a benzoate buffer solution after adding HCl

Q: A buffer with a pH of 4.44 contains 0.21 M of sodium benzoate and 0.12 M of benzoic acid. What is the concentration of $\ce{[H^+]}$ in the solution after the addition of 0.050 mol of $\ce{HCl}$ to a final volume of 1.3 L?

A: So these are the steps that I have taken for this:

The reaction is: $$\ce{C6H5COOH +H2O -> C6H5COO- + H^+}$$

The $K_ a$ of Benzoic acid is $6.3 × 10^{-5}$

Thus: $K_a=\ce{\frac{[C6H5COO- ][H^+]}{[C6H5COOH]}}$

We dont know the concetrations of $\ce{[H^+]}$ so we solve for that. We then assume at equilibrium $\ce{[C6H5COO- ]}$ is equal.

Then: $6.3×10^{-5}=\ce{\frac{[C6H5COO- ][H+]}{[C6H5COOH]}}$ and $\ce{[H+]}=0.00272$

But this is wrong. I think Im doing something wrong in my problem, but really can't figure out what.

• Welcome to chemistry.SE! Please have a look at here and fix your question's formatting to make it more readable. And well done for asking a homework Q appropriately! – M.A.R. Apr 5 '15 at 0:01
• Much appreciated @MARamezani – jonsca Apr 5 '15 at 0:11
• You missed the information you wrote above. The pH is 4.44, which means the $[H^+]$ is $3.63\times 10^{-5} M$. Also, it is rare that the 2 concentration are equal at equilibrium. – LDC3 Apr 5 '15 at 1:53

Initially, it's true that you have the following equilibrium between sodium benzoate and benzoic acid: $$\ce{C6H5COOH +H2O <=> C6H5COO- + H^+_(\mathrm{aq})}$$ When you add $\ce{HCl}$ to your buffer solution, sodium benzoate reacts with the acid, and the aforementioned equilibrium is displaced to the left: $$\ce{C6H5COO- + H^+_(\mathrm{aq}) <=> C6H5COOH +H2O }$$ Now, the equilibrium constant of this reaction is: $$K=\frac{1}{K_ a}=0.16 \times 10^5>10^3$$ The reaction is quantitative (total). So, we can rewrite it: $$\ce{C6H5COO- + H^+_(\mathrm{aq}) -> C6H5COOH +H2O }$$ Now, we have to construct an ICE table: Assuming that we didn't dilute the solution when we added $\ce{HCl}$ (this point is not clear in your question): $$\ce{[C6H5COOH]_0= }\mathrm {0.12 M}$$ $$\ce{[C6H5COO^-]_0= }\mathrm {0.21 M}$$ $$\ce{[H^+]_0= }(\frac{0.050}{1.3}+10^{-4.4}) \mathrm{M}\approx 0.0385 \mathrm{M}$$ As you can clearly seen$\ce{[H^+]_0}$ is the lowest, and so ion hydronium is the limiting reactant. At the end of the reaction, we have: $$\ce{[C6H5COO^-]_\mathrm{eq}= }\mathrm {(0.21-0.0385) = 0.1715M}$$ $$\ce{[C6H5COOH]_\mathrm{eq}= }\mathrm {(0.12+0.0385) = 0.1585M}$$ $$\ce{[H^+]_\mathrm{eq}=}\epsilon$$ Substituting all these numbers in the expression of $K$ $$K=\frac{ 0.1585}{0.1715\epsilon}=0.16\times 10^5$$ By solving the above equation, we find:

$$\ce{[H^+]}=\epsilon=5.82 \times 10^{-5}\mathrm { M}$$ $$\mathrm{p}\ce{H}=4.23$$

Here is a different approach:

Since we know the initial pH, as well as, the concentration of the benzoic acid "a" and its corresponding base "b", we can calculate the pka for benzoic acid:

pH = pka + log Cb/Ca => 4.44 = pka + log (0.21/0.12) => pka = 4.20.

Adding the strong acid HCl to the buffer, will convert the corresponding base "b" into benzoic acid "a". Suppose we start with 1 liter of the buffer. This solution will contain 0.12 mole of "a" and 0.21 mole of "b". At equilibrium we get: Ca = (0.12 + 0.050)/1.3 = 0.1307 (M) and

Cb = (0.21 - 0.050)/1.3 = 0.1230 (M).

The pH at equilibrium will be: pH = 4.20 + log(0.1230/0.1307). => pH = 4.17