$$K_a$$ for $$CH_3COOH$$ is $$1.8 \times 10^{-5}$$ and $$K_b$$ for $$NH_4OH$$ is $$1.8 \times 10^{-5}$$. The pH of ammonium acetate solution will be:

The pH of a salt solution of a weak acid and a weak base is independent of its concentration and is calculated using the formula:

pH=7+$$\ \frac{\ 1}{2}$$(pKa​−pKb​)

Given values:

• Ka​=$$1.8×10^{−5}$$⟹pKa​=−log($$1.8×10^{−5}$$)
• Kb​=$$1.8×10^{−5}$$⟹pKb​=−log($$1.8×10^{−5}$$)

Since Ka​=Kb​, it follows that:

pKa​=pKb​

Substitute these into the pH equation:

pH=7+$$\ \frac{\ 1}{2}\times\ $$(0)pH=7