A solution was made by mixing 75.00 mL of 0.120 mol L⁻¹ hydrochloric acid with 25.00 mL of 0.200 mol L⁻¹ sodium hydroxide - HSC - SSCE Chemistry - Question 28 - 2012 - Paper 1

Using the same formula for NaOH: $n_{NaOH} = 0.200 \text{mol L}^{-1} \times 0.025 \text{L} = 0.005 \, \text{mol}$

Now, we find the excess reactant: Since HCl is in excess: $n_{excess} = n_{HCl} - n_{NaOH} = 0.009 \, \text{mol} - 0.005 \, \text{mol} = 0.004 \, \text{mol}$

The total volume of the solution after mixing is: $V_{total} = 75.00 \, \text{mL} + 25.00 \, \text{mL} = 100.00 \, \text{mL} = 0.100 \, \text{L}$

The concentration of HCl in the mixed solution is: $C_{HCl} = \frac{n_{excess}}{V_{total}} = \frac{0.004 \text{mol}}{0.100 \, \text{L}} = 0.040 \, \text{mol L}^{-1}$

The pH is then calculated using: $pH = -\log_{10}(C_{HCl}) = -\log_{10}(0.040) \approx 1.40$