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A weak acid of dissociation constant 10–5 is being titrated with aqueous NaOH solution. The pH at the point of one-third neutralisation of the acid will be (A) 5 + log 2–log 3 (B) 5 –log 2 (C) 5 –log 3 (D) 5 –log 6

A weak acid of dissociation constant 10–5 is being titrated with aqueous NaOH solution. The pH at the point of one-third
neutralisation of the acid will be
(A) 5 + log 2–log 3 (B) 5 –log 2 (C) 5 –log 3 (D) 5 –log 6

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2 Answers

Arun
25750 Points
3 years ago
A weak acid of dissociation constant 10-5 is being titrated with aqueous NaOH solution. The pH at the point of one-third neutralisation of the acid will be. 5 + log 2 - log 3.
Vikas TU
14149 Points
3 years ago
On partial neutralization of weak acid, salt is formed. Hence it becomes buffer
∴pH = pKa + log.|Salt| / |Acid|.
1/3 rd neutralization of the acid means out of 1 mole of the acid, salt formed is =1/3 mole of acid left =2/3 mole
∴pH=−log(10^−5)+log.(1/3) / (2/3)
=5+log.12=5−log2

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