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Question
\[\ce{K_a}\] for \[\ce{CH3COOH}\] is 1.8 × 10–5 and \[\ce{K_b}\] for \[\ce{NH4OH}\] is 1.8 × 10–5 . The \[\ce{pH}\] of ammonium acetate will be ______.
Options
7.005
4.75
7.0
Between 6 and 7
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Solution
\[\ce{K_a}\] for \[\ce{CH3COOH}\] is 1.8 × 10–5 and \[\ce{K_b}\] for \[\ce{NH4OH}\] is 1.8 × 10–5 . The \[\ce{pH}\] of ammonium acetate will be 7.0.
Explanation:
The equation for weak acid and weak base is given as
\[\ce{pH = 7 + 21 (pK_a − pK_b)}\]
\[\ce{K_a}\] = 1.8 × 10–5
\[\ce{pK_a = - log (1.8 × 10^{-5})}\] = 5 − log1.8 = 4.744
\[\ce{pK_b = −log (1.8 × 10^{-5)}}\] = 4.744
\[\ce{pH = 7 + 21 (pK_a − pK_b)}\] = 7.
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