Question

Calculate the pH of a solution which contains 9.9ml HCl and 100ml of 0.1M NaOH.

Answer 1

NaOH reacts with HCl :

$NaOH+HCl\to NaCl+H_{2}O$

So, 1 mol NaOH reacts with 1 mol HCl

Moles of HCl in 9.9mL of 1.0M solution = $\frac{9.9}{1000\times 1}$ = 0.0099 mol HCl

Moles of NaOH in 100mL of 0.1M solution = $\frac{100}{1000\times 0.1}$ = 0.01 mol NaOH

On mixing the 0.0099 mol HCl will be neutralised and 0.01 - 0.0099 = $1\times {10}^{-4}$mol NaOH dissolved in 109.9mL solution

Molarity of NaOH solution =$\frac{1\times {10}^{-4}}{0.1099}$= $9.099\times {10}^{-4}M$

Hence,

pOH = -log ( $9.099\times {10}^{-4}$)

pOH = 3.04

So,

pH = 14.00 - pOH

pH = 14.00 - 3.04

$⟹pH=10.96$