Question

A chemist titrates 220.0 mL of a 0.1917M propionic acid (HC2H5CO2) solution with 0.1787 M KOH solution at 25°C. Calculate the pH at equivalence. The pKa of propionic acid is 4.89.
Round your answer to 2 decimal places.

Answer 1

Answer : The pH at equivalence is, 9.08

Explanation : Given,

Concentration of $HC_2H_5CO_2$ = 0.1917 M

Volume of $HC_2H_5CO_2$ = 220.0 mL = 0.220 L (1 L = 1000 mL)

First we have to calculate the moles of $HC_2H_5CO_2$

$\text{Moles of }HC_2H_5CO_2=\text{Concentration of }HC_2H_5CO_2\times \text{Volume of }HC_2H_5CO_2$

$\text{Moles of }HC_2H_5CO_2=0.1917M\times 0.220L=0.0422$

As we known that at equivalent point, the moles of $HC_2H_5CO_2$ and KOH are equal.

So, Moles of KOH = Moles of $HC_2H_5CO_2$ = 0.0422 mol

Now we have to calculate the volume of KOH.

$\text{Volume of }KOH=\frac{\text{Moles of }KOH}{\text{Concentration of }KOH}$

$\text{Volume of }KOH=\frac{0.0422mol}{0.1787M}$

$\text{Volume of }KOH=0.00754$

Total volume of solution = 0.220 L + 0.00754 L = 0.22754 L

Now we have to calculate the concentration of KCN.

The balanced equilibrium reaction will be:

$HC_2H_5CO_2+KOH\rightleftharpoons C_2H_5CO_2K+H_2O$

Moles of $C_2H_5CO_2K$ = 0.0422 mol

$\text{Concentration of }C_2H_5CO_2K=\frac{0.0422mol}{0.22754L}=0.1855M$

At equivalent point,

$pH=\frac{1}{2}[pK_w+pK_a+\log C]$

Given:

$pK_w=14\\\\pK_a=4.89\\\\C=0.1855M$

Now put all the given values in the above expression, we get:

$pH=\frac{1}{2}[14+4.89+\log (0.1855)]$

$pH=9.08$

Therefore, the pH at equivalence is, 9.08