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How much volume of `0.1M Hac` should be added to `50mL` of `0.2M NaAc` solution to have a `pH 4.91`? |
Answer» It is an acidic buffer `pH = pK_(a) + "log" (["Salt"])/(["Acid"])` `4.91 = 4.76 + log [("Salt")/("Acid")]` `:. log [("salt")/("Acid")] = (4.91 - 4.76 ) = 0.15` `(["Salt"])/(["Acid"]) = "Antilog" (0.15) = 1.41` mmoles of `NaAc = 50 xx 0.2 = 10` mmoles of HAc required `= 0.1 M xxV` `[("Salt")/("Acid")] = 1.41 = (10)/(0.1xxV)` `rArr V = (10)/(0.1 xx 1.41) = 70.92mL` `:.` Volume of 0.1M HAc required `= 70.92mL` |
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