Question

What is the [H3O+] and the pH of a benzoic acid-benzoate buffer that consists of 0.17 M C6H5COOH and 0.42 M C6H5COONa? (Ka of benzoic acid = 6.3 × 10^−5). Be sure to report your answer to the correct number of significant figures.

Answer 1

Answer:

$[H_{3}O^{+}]=x M = 2.5\times 10^{-5}M$ and pH = 4.6

Explanation:

Construct an ICE table to calculate changes in concentration at equilibrium.

$C_{6}H_{5}COOH+H_{2}O\rightleftharpoons C_{6}H_{5}COO^{-}+H_{3}O^{+}$

I(M): 0.17                                    0.42                0

C(M): -x                                        +x                  +x

E(M): 0.17-x                                0.42+x             x

So, $\frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}=K_{a}(C_{6}H_{5}COOH)$

or, $\frac{(0.42+x)x}{(0.17-x)}=6.3\times 10^{-5}$

or, $x^{2}+0.4201x-(1.071\times 10^{-5})=0$

So, $x=\frac{-0.4201+\sqrt{(0.4201)^{2}+(4\times 1\times 1.071\times 10^{-5})}}{(2\times 1)}M$

($ax^{2}+bx+c=0\Rightarrow x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a};x< 0.17M$)

So, $x=2.5\times 10^{-5}$M

Hence $[H_{3}O^{+}]=x M = 2.5\times 10^{-5}M$

$pH=-log[H_{3}O^{+}]=-logx=-log(2.5\times 10^{-5})=4.6$