Question

The dissociation constant of a substituted benzoic acid at 25°c is 1.0 × 10 –4 . The ph of a 0.01 m solution of its sodium salt is

Answer 1

It is given that, Dissociation constant of  (C6H5COOH) = 1 × 10^-4

We have to find pH of 0.01 M C6H5COONa

C6H5COO⁻ + H2O -------> C6H5COOH + OH⁻

 0.01(1-h)                            0.01 h            0.01 h

Kh= Kw/Ka =(0.01 h2)/(1-h),

⇒ kh = 10⁻¹⁴/10⁻⁴ = 10⁻¹⁰

[OH-] = 0.01 h = 0.01 × 10⁻⁴= 10⁻⁶ M

[H⁺ ][ OH⁻] = 10⁻¹⁴

So, [H⁺] = 10⁻⁸ M

From Arrhenius theorem,

pH = -log[H⁺] = -log(10⁻⁸) = 8

pH = 8

Answer 2

Answer:

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