Question

Calculate the pH of a solution that contains 0.250 M formic acid, HCOOH (Ka =1.8 x 10-4 ), and 0.100M sodium formate, HCOONa after the addition of 10.0 mL of 6.00M NaOH to the original buffered solution volume of 500.0 mL.

Answer 1

Answer : The pH of the solution is 3.5

Explanation :

First we have to calculate the moles of $HCOOH$, $HCOONa$ and $NaOH$.

$\text{Moles of }HCOOH=\text{Concentration of }HCOOH\times \text{Volume of solution in L}$

$\text{Moles of }HCOOH=0.250M\times 0.500L=0.125mol$

and,

$\text{Moles of }HCOONa=\text{Concentration of }HCOONa\times \text{Volume of solution in L}$

$\text{Moles of }HCOONa=0.100M\times 0.500L=0.05mol$

and,

$\text{Moles of }NaOH=\text{Concentration of }NaOH\times \text{Volume of solution in L}$

$\text{Moles of }NaOH=6.00M\times 0.010L=0.06mol$

The equilibrium chemical reaction is:

                         $HCOOH+OH^-\rightleftharpoons HCOO^-+H_2O$

Initial moles      0.125          0.06      0.05

At eqm.          (0.125-0.06)     0      (0.05+0.06)

                        = 0.185                        = 0.11

Given:

$K_a=1.8\times 10^{-4}$

Now we have to calculate the value of $pK_a$.

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (1.8\times 10^{-4})$

$pK_a=4-\log (1.8)$

$pK_a=3.7$

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[HCOONa]}{[HCOOH]}$

$pH=pK_a+\log \frac{[\frac{\text{Moles of HCOONa}}{\text{Volume of solution}}]}{[\frac{\text{Moles of HCOOH}}{\text{Volume of solution}}]}$

As the volume of solution are same. So,

$pH=pK_a+\log \frac{\text{Moles of HCOONa}}{\text{Moles of HCOOH}}$

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

$pH=3.7+\log (\frac{0.11}{0.185})$

$pH=3.5$

Therefore, the pH of the solution is 3.5