Calculate the pH of a 0.2 mol L$^{-1}$ solution of hydrochloric acid - HSC - SSCE Chemistry - Question 17 - 2006 - Paper 1

First, we need to calculate the moles of HCl and NaOH:

1. Moles of HCl:

   $\text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.2 \text{ mol L}^{-1} \times 0.050 \text{ L} = 0.01 \text{ mol}$

2. Moles of NaOH:

   $\text{Moles of NaOH} = 0.01 \text{ mol L}^{-1} \times 0.020 \text{ L} = 0.0002 \text{ mol}$

Next, we determine the limiting reactant. The balanced equation for the reaction is:

$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

Since 0.01 mol of HCl is much greater than the 0.0002 mol of NaOH, NaOH is the limiting reactant. After the reaction, the moles of HCl remaining will be:

$\text{Remaining HCl} = 0.01 \text{ mol} - 0.0002 \text{ mol} = 0.0098 \text{ mol}$

Now, the new total volume of the solution is:

$\text{Total Volume} = 50 ext{ mL} + 20 ext{ mL} = 70 ext{ mL} = 0.070 ext{ L}$

The concentration of HCl in the mixed solution becomes:

$\text{[HCl]} = \frac{0.0098 \text{ mol}}{0.070 \text{ L}} \approx 0.140 \text{ mol L}^{-1}$

Finally, we can calculate the pH:

$\text{pH} = -\log_{10}(0.140) \approx 0.85$