Question

A 100.00 mL buffer solution is prepared by dissolving 1.22 grams benzoic acid (pKa = 4.20, Molar Mass = 122g/mol) and 2.88 grams sodium benzoate (Molar Mass = 144g/mol) in an appropriate amount of water.

a. What is the pH of the buffer solution?

b. What is the pH of the solution if 10.00 mL of 0.25 M HCl is added?

c. What is the pH of the solution if 5.00 mL of 0.25 M NaOH is added instead of HCl?

Just want some proper explanations too to properly understand :>>

Answer 1

a.)

We have given,

Number of moles of $C_6H_5COOH$ , $n_1$ =$\dfrac{1.22}{122} = 0.01$

Number of moles of$C_6H_5COONa,n_2$ =$\dfrac{2.88}{144} = 0.02$

Concentration of $C_6H_5COOH, M_1$ = 0.1M

Concentration of $C_6H_5COONa, M_2$ = 0.2M

We know, pH in a buffer solution can be calculated as

$pH = pK_a+log\dfrac{M_2}{M_1}\\ pH = pK_a+log\dfrac{0.2}{0.1}\\ pH = 4.20+0.30\\ pH = 4.50$

b.)

Now, 10.00 mL of 0.25 M $HCl$ is added then,

$M_1V_1 = M_2V_2$

$0.25\times10 = M2\times100$

$M_2 = 0.025M$

where $M_2$ is the concentration of solution when $HCl$ is added

then,

$pH = -log[H]^+$

$pH = -log[0.025]$

$pH = 1.60$

c.)

When 5.00 mL of 0.25 M $NaOH$ is added then,

$M_1V_1 = M_2V_2$

$0.25\times5 = M2\times100$

$M2 = 0.0125M$

$pH = -log[H]^+$

$pH = 1.90$

Hence, pOH = $14-1.90$

= $12.1$