Question

A buffer with a pH of 4.31 contains 0.31 M of sodium benzoate and 0.24 M of benzoic acid. What is the concentration of [ H 3 O ] in the solution after the addition of 0.060 mol HCl to a final volume of 1.3 L

Answer 1

Answer: The hydronium ion concentration in the solution is $1.29\times 10^{-4}M$

Explanation:

To calculate the molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles hydrochloric acid solution = 0.060 mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of HCl}=\frac{0.060}{1L}\\\\\text{Molarity of HCl}=0.060M$

The chemical reaction for aniline and HCl follows the equation:

                   $C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-$

Initial:           0.24          0.060              0.31

Final:             0.18          -                     0.37

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})$

$pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of benzoic acid = 4.2

$[C_6H_5COO^-]=0.18M$

$[C_6H_5COOH]=0.37M$

pH = ?

Putting values in equation 1, we get:

$pH=4.2+\log(\frac{0.18}{0.37})\\\\pH=3.89$

To calculate the hydronium ion concentration in the solution, we use the equation:

$pH=-\log[H_3O^+]$

pH = 3.89

Putting values in above equation, we get:

$3.89=-\log[H_3O^+]$

$[H_3O^+]=10^{-3.89}=1.29\times 10^{-4}M$

Hence, the hydronium ion concentration in the solution is $1.29\times 10^{-4}M$