Question

A 0.100 M weak acid is titrated with a strong base to the equivalence point. The pH of

the resulting solution is found to be 9.18.

What is the pKa of the acid?

Answer 1

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Answer 2

Calculation:

1. Given:

• pH at equivalence point = 9.18 (solution is basic).

• Weak acid (HA) is fully neutralized to its conjugate base (A⁻) by the strong base.

2. Find [OH⁻] from pH:

   $$\text{pOH}=14-\text{pH}=14-9.18=4.82$$$$[{\text{OH}}^{-}]=10^{-\text{pOH}}=10^{-4.82}\approx 1.51\times 10^{-5}\text{ M}$$

3. Set up the hydrolysis reaction of A⁻:

   $${\text{A}}^{-}+{\text{H}}_2\text{O}\rightleftharpoons \text{HA}+{\text{OH}}^{-}$$

   At equilibrium:

   $$K_b=\frac{[{\text{OH}}^{-}][\text{HA}]}{[{\text{A}}^{-}]}\approx \frac{(1.51\times 10^{-5}{)}^2}{0.100}=2.28\times 10^{-9}$$

4. Relate $K_b$ to $K_a$:

   $$K_a\times K_b=1\times 10^{-14}$$$$K_a=\frac{10^{-14}}{K_b}=\frac{10^{-14}}{2.28\times 10^{-9}}\approx 4.39\times 10^{-6}$$

5. Calculate pKa:

   $$\text{pKa}=-\log (K_a)=-\log (4.39\times 10^{-6})\approx 5.36$$

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Conclusion: The pKa of the weak acid is 5.36.