$$100$$ mL of $$0.02$$ M benzoic acid (pKa$$ $$=$$ $$4.2) is titrated by using $$0.02$$ M NaOH. pH after $$50$$ mL and $$100$$ mL of NaOH have been added are

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Infinity Learn · Accepted Answer

Millimoles of benzoic acid $$=$$ $$2Millimole$$ of NaOH in first addition $$=$$ $$1Millimoles$$ of NaOH in second addition $$=$$ $$2$$                     $$C6H5COOH+NaOH$$⇌$$C6H5COONa+H2O$$ Initial                      $$2$$                   $$0$$                   $$0$$                $$0$$ After first  addition of NaOH  (2-2)=0$$           $$(2-2) $$=$$ $$0$$        $$2$$               After first addition, mixture is a buffer containing $$1$$ millimole each of $$C6H5COONa$$ and $$C6H5COOH$$∴ $$pH=pKa$$∗$$+log$$⁡$$C6H5COONaC6H5COOH$$⇒$$pKa=4.2After$$ second addition, there are only $$2$$ millimoles of, $$C6H5COONa$$ in $$200$$ mL solution and due to hydrolysis, solution is basic.$$C6H5COO$$−$$=2$$×$$10$$−$$302=1$$×$$10$$−$$2MC6H5COO$$−$$+H2O$$⇌$$C6H5COOH+OH$$− ∴ $$pH=7+pKa2+logC2$$ $$=7+2.1-1=8.1$$.

100 mL of 0.02 M benzoic acid (pKa = 4.2) is titrated by using 0.02 M NaOH. pH after 50 mL and 100 mL of NaOH have been added are

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