Question

Calculate the pH of a 0.078 M solution of NaOCl (Ka of HClO = 3.5 x 10-8). Enter pH to 3 decimal places.

Answer 1

Answer:

The pH of the solution is 4.282

Explanation:

First we need to write the dissociation reaction

$HCLO_{(aq)} + H_2O_{(l)$  ⇄ $CLO^-_{(aq)} +H_3O^+_{(aq)}$

Also, we need to prepare I C E table

           $HCLO_{(aq)} + H_2O_{(l)$  ⇄ $CLO^-_{(aq)} +H_3O^+_{(aq)}$

I          0.078                               0             0

C         - x                                    +x            +x

E        0.078 - x                            x               x

Thus, pH= - Log[H₃O⁺]

The equilibrium concentration of  H₃O⁺ = x

So, we solve for x

$K_a = \frac{[CLO^-].[H_3O^+]}{[HCLO]} \\\\3.5X10^{-8} = \frac{[x].[x]}{[0.078-x]} \\\\3.5X10^{-8} =\frac{x^2}{0.078-x} \\\\x^2 = 3.5X10^{-8} (0.078-x)\\\\x^2 = 2.73X10^{-9} - (3.5X10^{-8})x\\\\x^2 + (3.5X10^{-8})x - 2.73X10^{-9} =0$

Solving this quadratic equation; x = 5.2232 x 10⁻⁵ M

pH = - Log[H₃O⁺]

pH = - Log[5.2232 x 10⁻⁵]

pH = 4.282 (in 3 decimal places)

Therefore, the pH of the solution is 4.282