What is the pH of the resultant solution after 20.0 mL of 0.20 mol L$^{-1}$ HCl(aq) is mixed with 20.0 mL of 0.50 mol L$^{-1}$ NaOH(aq)? - HSC - SSCE Chemistry - Question 15 - 2021 - Paper 1

Next, calculate the moles of NaOH:

[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.50 , \text{mol L}^{-1} \times 0.020 , \text{L} = 0.010 , \text{mol} ]

Since NaOH is present in excess, we subtract the moles of HCl from the moles of NaOH:

[ \text{Moles of excess NaOH} = 0.010 , \text{mol} - 0.004 , \text{mol} = 0.006 , \text{mol} ]

The total volume of the resulting solution is the sum of the two volumes:

[ \text{Total Volume} = 20.0 , \text{mL} + 20.0 , \text{mL} = 40.0 , \text{mL} = 0.040 , \text{L} ]

Now, find the concentration of excess NaOH:

[ \text{Concentration of NaOH} = \frac{\text{Moles}}{\text{Total Volume}} = \frac{0.006 , \text{mol}}{0.040 , \text{L}} = 0.15 , \text{mol L}^{-1} ]

Calculate the pOH of the solution:

[ \text{pOH} = -\log(0.15) \approx 0.823 ]

Finally, calculate the pH from pOH:

[ \text{pH} = 14 - \text{pOH} = 14 - 0.823 \approx 13.177 ]

Thus, the pH is approximately 13.2, which corresponds to option B.