When 10 mL of 0.1 M acetic acid (pKa = 5.0 ) is titrated against 10mL of 0.1 M

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When 10 mL of 0.1 M acetic acid (pKa = 5.0 ) is titrated against 10mL of 0.1 M ammonia solution (pKb = 5.0),the equivalence point occurs at pH

Options

(a) 5
(b) 6
(c) 7
(d) 9

Correct Answer:

7

Explanation:

pKₐ = -logKₐ : pK(b) = -logK(b),
pH = -1/2[logKₐ + log K(w) – logK(b),
-1/2[-5 + log(1*10⁻¹⁴)-(-5)],
-1/2[-5-14+5]=-1/2(-14)=7.

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Topics: Equilibrium (104)
Subject: Chemistry (2512)

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