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Calculate the pH of a solution obtained by mixing of 100 ml of 0.1 M HCI and 100 ml of 0.2 M NH3 ? Kb, for NH3 = 1.8 x 10-5 ? |
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Answer» 100 ml of 0∙1 M HCl + 100 ml of 0∙2 M NH3 Kb for NH3 = 1.8 x 10-5 100 × 0∙1 = 10 m mol of HCl 100 × 0∙2 = 20 m mol of NH3 NH3 (1 mol) + HCl (1 mol) → NH4Cl (1 mol) i.e., 10 m mol 10 m mol 10 m mol NH3 left unreacted = 10 m mol Volume of solution = 200 ml Molarity of NH3 = \(\frac{10}{200}\) M = 0 ∙ 05 M [NH4+] = \(\frac{10}{200}\) M = 0 ∙ 05 M pOH = pKb + log\(\frac{[Salt]}{[Acid]}\) = -log (1.8 x 10-5) + log \(\frac{0.05}{0.05}\) = 4∙74 pH = 14 – pOH = 14 − 4∙74 = 9∙26. |
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