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InterviewSolution
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A weak acid of dissociation constant `01^(-5)` is being titrated with aqueous NaOH solution. The pH at the point of one-third neutralization of the acid will beA. 5 log 2 - log 3B. 5 - log 2C. 5 - log 3D. 5 - log 6 |
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Answer» Correct Answer - B On partial neutralization of the weak acid, salt is formed. Hence, it becomes a buffer `pH = pK_(a) + log. (["Salt"])/(["Acid"])` 1/3rd neutralisation of the acid means out of 1 mole of the acid , salt formed = 1/3 mole and acid left = 2/3 mole Hence, `pH = - log(10^(-5))+log.(1//3)/(2//3)` `=5+ log .(1)/(2) = 5 - log2`. |
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