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Write Brief Answer · Q17

Q.Calculate the extent of hydrolysis and the pH of 0.1M ammonium acetate. Given that Ka=Kb=1.8×10−5K_a = K_b = 1.8 \times 10^{-5}.

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Step 1. Ammonium acetate is the salt of a weak acid AND a weak base, with Ka=Kb=1.8×10−5K_a=K_b=1.8\times10^{-5}.

Step 2. Degree of hydrolysis (concentration-independent for this salt type): h=KwKaKb=10−14(1.8×10−5)2=10−143.24×10−10=3.086×10−5≈5.56×10−3h=\sqrt{\dfrac{K_w}{K_aK_b}}=\sqrt{\dfrac{10^{-14}}{(1.8\times10^{-5})^2}}=\sqrt{\dfrac{10^{-14}}{3.24\times10^{-10}}}=\sqrt{3.086\times10^{-5}}\approx5.56\times10^{-3}, i.e. about 0.556% hydrolysed.

Step 3. pKa=pKb=−log⁡10(1.8×10−5)=4.7447pK_a=pK_b=-\log_{10}(1.8\times10^{-5})=4.7447 (equal, since Ka=KbK_a=K_b). …

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