Question

500 mL of 0.1M HCl is mixed with 500 mL of 0.02M HA (weak acid) solution. To this solution, 2 g of solid NaOH is added. What is the pH of the resultant solution? (Given} \( K_a \) of HA is \( 4 \times 10^{-10} \), molar mass of NaOH is 40 g/mol, log 2 = 0.3, log 4 = 0.6\bf{)

• A. 5.7
• B. 6.7
• C. 6.4
• D. 4.7

Hint:

In titration problems involving a weak acid and NaOH, always check the limiting reagents and the effect of neutralization on the pH.

Answer 1

The Correct Option is A

- First, calculate the moles of NaOH added: \[ \text{Moles of NaOH} = \frac{2 \text{g}}{40 \text{g/mol}} = 0.05 \text{mol} \] - The total volume of the solution after mixing is: \[ \text{Total volume} = 500 \text{mL (HCl)} + 500 \text{mL (HA)} = 1000 \text{mL} = 1 \text{L} \] - The moles of HCl are: \[ \text{Moles of HCl} = 0.1 \text{M} \times 0.5 \text{L} = 0.05 \text{mol} \] - Since NaOH neutralizes HCl, the remaining moles of NaOH are: \[ \text{Remaining moles of NaOH} = 0.05 \text{mol (NaOH)} - 0.05 \text{mol (HCl)} = 0 \text{mol} \] - The weak acid HA dissociates, and the pH is determined by the concentration of \(\text{H}^+\) ions, which are affected by the dissociation constant \( K_a \). The calculation shows that the pH of the resultant solution is approximately 5.7.