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The pH of 0.1 M monobasic acid is 4.50. Calculate the concentration of species H+ , A– and HA at equilibrium. Also, determine the value of Ka and pKa of the monobasic acid. |
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Answer» Given C = 0.1 M pH = 4.50 \(\therefore\) pH = – log [H+] or [H+] = 10–pH = 10–4.50 = 3.16 × 10–5 [H+] = [A–] = 3.16 × 10–5 For the reaction, HA ⇌ H+ + A– Ka = \(\frac{[H^+][A^-]}{[HA]}\) = \(\frac{(3.16\times10^{-15})(3.16\times10^{-5})}{0.1}\) [∵ [HA]equi = 0.1 – (3.16 × 10–5 ) ~ 0.1] = 1.0 × 10–8 \(\because\) pKa = – log [Ka] = – log (1 × 10–8 ) = 8 log 10 = 8 |
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