Question

Sodium benzoate (C6H5COONa), the sodium salt of the weak acid benzoic acid, is used as a food preservative. A solution is prepared by dissolving 0.100 mol of sodium benzoate in enough pure water to produce 1.00 L of solution. If the pKa for benzoic acid is 4.20, calculate the pH of the sodium benzoate solution.

Answer 1

Answer:

8.6

Explanation:

Using the expression :

$K_a\times K_b=K_w$

Where, $K_w$ is the dissociation constant of water.

At $25\ ^0C$, $K_w=10^{-14}$

Thus, for benzoic acid , pKa = 4.20

Thus, $K_a=10^{-4.20}=6.31\times 10^{-5}$

$K_b$ for Sodium benzoate can be calculated as:

$K_a\times K_b=K_w$

$6.31\times 10^{-5}\times K_b=10^{-14}$

$K_b=1.58\times 10^{-10}$

The benzoate ion will dissociate as:-

$C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}$

$K_b$ expression is:-

$K_{b}=\frac {\left [ C_6H_5COOH^{+} \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}$

Given that:-

Moles = 0.100 moles

Volume = 1.00 L

Thus, Concentration = 0.100/ 1.00 M = 0.1 M

Considering the ICE table in the image below.

So,

$1.58\times 10^{-10}=\frac{x^2}{0.1-x}$

$1.58\left(0.1-x\right)=10^{10}x^2$

Solving for x, we get that, $x=3.97\times 10^{-6}$

Thus, $[OH^-]=3.97\times 10^{-6}$

Also,

$pOH=-log[OH^-]=-log(3.97\times 10^{-6})=5.4$

Also, pH + pOH = 14  

So, pH = 14 - 5.4 = 8.6