Question

If Ka of acetic acid is 1.75 x 10-5 then hydrolysis constant
of 0.2 M aqueous solution of sodium acetate will be​

Answer 1

we have to find hydrolysis constant of 0.2 M aqueous solution of sodium acetate if Ka of acetic acid is 1.75 × 10¯⁵.

solution : chemical reaction is ...

CH₃COOH + NaOH ⇒CH₃COONa + H₂O

from formula, Kw = Ka × Kh

⇒Kh = Kw/Ka

where Kh is hydrolysis constant

Kw is ionic product of water.

Ka is acid dissociation constant.

here Ka = 1.75 × 10¯⁵ , Kw = 10¯¹⁴

so, Kh = 10¯¹⁴/1.75 × 10¯⁵ = 5.7 × 10¯¹⁰

Therefore the hydrolysis constant of sodium acetate is 5.7 × 10¯¹⁰.

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