IIT-JEE 2009 · Paper 2 · Chemistry Q14

ChemistryInteger+4 / -1Question 14 of 57

The dissociation constant of a substituted benzoic acid at 25°C is $1.0\times10^{-4}$ . The pH of a 0.01 M solution of its sodium salt is

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Answer and worked solution

Official answer: 8

Hydrolysis of the salt. $K_b=\dfrac{K_w}{K_a}=\dfrac{10^{-14}}{10^{-4}}=10^{-10}$ .

Hydroxide. $[\ce{OH-}]=\sqrt{K_bC}=\sqrt{10^{-10}\times10^{-2}}=10^{-6}$ M, so pOH = 6 and pH = 8.

Answer: 8

IIT-JEE 2009 · Paper 2, Chemistry question 14