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

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

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

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

The $$K_ \mathrm{a}$$ of benzoic acid is $$6.3 × 10^{-5}$$

But, $$K_\mathrm{a}=\ce{\frac{[C6H5COO-][H+]}{[C6H5COOH]}}$$

We don't know the concentrations of $$\ce{[H+]}$$ so we solve for that. We then assume at equilibrium $$\ce{[C6H5COO-]}$$ is equal to it.

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

But this is wrong. I think I'm 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! Apr 5, 2015 at 0:01
• Much appreciated @MARamezani Apr 5, 2015 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, 2015 at 1:53

Initially, it's true that you have the following equilibrium between sodium benzoate and benzoic acid:

$$\ce{C6H5COOH + H2O <=> C6H5COO- + H3O+_\mathrm{(aq)}}$$

When you add $$\ce{HCl}$$ to your buffer solution, sodium benzoate reacts with the acid, and the aforementioned equilibrium is displaced to the left: $$\ce{C6H5COO- + H3O+_\mathrm{(aq)} <=> C6H5COOH + H2O }$$ 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: $$\ce{C6H5COO- + H3O+_\mathrm{(aq)} -> C6H5COOH + H2O }$$

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): $$\ce{[C6H5COOH]_0} = \pu{0.12 M}$$ $$\ce{[C6H5COO^-]_0} = \pu{0.21 M}$$ $$\ce{[H^+]_0} = (\frac{0.050}{1.3} + 10^{-4.4}) \mathrm{M}\approx 0.0385 \mathrm{M}$$ 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: $$\ce{[C6H5COO^-]_\mathrm{(eq)}} = (0.21-0.0385) = \pu{0.1715 M}$$ $$\ce{[C6H5COOH]_\mathrm{(eq)}} = (0.12+0.0385) = \pu{0.1585 M}$$ $$\ce{[H^+]_\mathrm{(eq)}} = \epsilon$$ 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:

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

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

The given solution is a buffer of the equilibrium:

$$\ce{C6H5COOH + H2O <=> C6H5COO- + H3O+} \tag1$$

The buffer question can be easily solved using Henderson-Hasselbalch equation. In case you don't know how to derive it, we'll do it from the scratch. Following expression can be written from the equation $$(1)$$:

$$K_\mathrm{a}=\ce{\frac{[C6H5COO-][H+]}{[C6H5COOH]}}$$

Take $$\log$$ on both side:

$$\log K_\mathrm{a}=\log\left(\ce{\frac{[C6H5COO-][H+]}{[C6H5COOH]}}\right) = \log [\ce{H+}] +\log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right)$$

Multiply both side by negtive sign:

$$-\log K_\mathrm{a}= -\log [\ce{H+}] -\log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right) \; \Rightarrow \; \mathrm{p} K_\mathrm{a}= \mathrm{pH} -\log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right)$$

After rearranging the last equation, you got:

$$\mathrm{pH} = \mathrm{p} K_\mathrm{a} + \log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right) \tag2$$

The equation $$(2)$$ represent the Henderson-Hasselbalch equation of the equilibrium reaction $$(1)$$. Accordingly, any changes to $$\ce{[C6H5COO-]}$$ and $$\ce{[C6H5COOH]}$$ changes the $$\mathrm{pH}$$ of the solution. For example, let's check the $$\mathrm{pH}$$ of original conditions ($$\mathrm{p} K_\mathrm{a} = -\log (6.3 \times 10^{-5}) = 4.201$$):

$$\mathrm{pH} = \mathrm{p} K_\mathrm{a} + \log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right) = 4.201 + \log\left(\ce{\frac{0.21}{0.12}}\right) = 4.444$$

Let's look at what happens when a strong acid such as $$\ce{HCl}$$ is added to the system. The following reaction would happen (See the $$ICE$$ chart):

$$\begin{array}{lccc} \ce{&C6H5COO- &+ & HCl &-> & C6H5COOH & + & Cl-} \\ \text{I} & 0.21 \times 1.3 && 0.050 && 0.21 \times 1.3 \\ \text{C} & - 0.050 && - 0.050 && + 0.050 \\ \text{E} & 0.21 \times 1.3 - 0.050 && 0.0 && 0.21 \times 1.3 + 0.050 \\ \end{array}$$

After addition of $$\ce{HCl}$$ assuming volume didn't change, $$\ce{[C6H5COO-]} = \frac{0.21 \times 1.3 - 0.050}{1.3} = \pu{0.1715 M}$$, and $$\ce{[C6H5COO-]} = \frac{0.21 \times 1.3 + 0.050}{1.3} = \pu{0.1585 M}$$. Now apply these values to the equation $$(2)$$:

$$\mathrm{pH} = \mathrm{p} K_\mathrm{a} + \log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right) = 4.201 + \log\left(\ce{\frac{0.1715}{0.1585}}\right) = 4.235$$

Hence $$[\ce{H+}] = 10^{-4.235} = 5.82 \times 10^{-5}$$.