Question

A volume of 500.0 mL of 0.160 M NaOH is added to 585 mL of 0.200 M weak acid ( K a = 1.28 × 10 − 5 ) . What is the pH of the resulting buffer? HA ( aq ) + OH − ( aq ) ⟶ H 2 O ( l ) + A − ( aq

Answer 1

Answer : The pH of the resulting buffer is, 5.22

Explanation : Given,

$K_a=1.28\times 10^{-5}$

First we have to calculate the moles of $NaOH\text{ and }HA$

$\text{Moles of }NaOH=\text{Concentration of }NaOH\times \text{Volume of solution}}=0.160M\times 0.500L=0.08mol$

and,

$\text{Moles of }HA=\text{Concentration of }HA\times \text{Volume of solution}}=0.200M\times 0.585L=0.117mol$

The balanced chemical reaction is:

$HA+(aq)+OH^-(aq)\rightarrow H_2O(l)+A^-(aq)$

Moles of HA left = 0.117 mol - 0.08 mol = 0.037 mol

Moles of $A^-$ = 0.08 mol

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:

$pK_a=-\log (1.28\times 10^{-5})$

$pK_a=5-\log (1.28)$

$pK_a=4.89$

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

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

$pH=pK_a+\log \frac{[A^-]}{[HA]}$

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

$pH=4.89+\log (\frac{0.08}{0.037})$

$pH=5.22$

Thus, the pH of the resulting buffer is, 5.22