Consider the titration of 0.100 L of 0.200 M oxalic acid with 0.500 M NaOH. What is the pH at the...

Answer: At the endpoint of the titration, the pH of the solution will be 7.2

Explanation:

Oxalic acid is a weak acid having the acid ionization constant value {eq}K_a=5.37 \times 10^{-2} {/eq}

The reaction between oxalic acid and NaOH is as follows:

{eq}H_2C_2O_4\ (aq) + 2 NaOH\ (aq) \rightarrow Na_2C_2O_4\ (aq) + 2 H_2O\ (l) {/eq}

Now for getting an endpoint, a volume (V) of NaOH solution is added:

V = {eq}\frac{0.1\ L \times 0.2\ M}{0.5\ M}=0.04\ L {/eq}

Hence, the total volume {eq}V_t {/eq} of the solution will be:

{eq}V_t=(0.1+0.04)~ L = 0.14~ L {/eq}

Now, at the endpoint, there will be no acid or base left in the mixture. Only {eq}Na_2C_2O_4 {/eq} salt will be left with a concentration (C):

{eq}C=\frac{0.1\ L \times 0.2\ M}{0.14\ L} \simeq 0.143\ M {/eq}

{eq}Na_2C_2O_4 {/eq} is a salt of a weak acid and a strong base. Thus, after salt hydrolysis, the resulting solution will be basic in nature.

The pH of the solution of a salt of a weak acid and a strong base can be calculated with the help of the formula:

{eq}pH=7+\frac{1}{2}pK_a+\frac{1}{2}log_{10}[Salt] {/eq}

{eq}\Rightarrow pH=7+\frac{1}{2}\left ( -log_{10}(5.37 \times 10^{-2}) \right )+\frac{1}{2}log_{10}(0.143)=7.2 {/eq}

Thus at the endpoint of the titration, the pH of the solution will be 7.2