Question

Calculate the pH of a 0.0204 M KOH solution at each temperature.

For 15 (4.57 × 10^–15                                                                                                                                                     15 °C pH=                     

For 45 (3.94 × 10^–14) 45 °C pH=                        

Answer 1

Sol :-

(a). At 15 ⁰C :-

Given ,

[KOH] = [OH^-] = 0.0204 M

It is given that , Ionic product of water (K_w) at 15 ⁰C = 4.57 x 10^-15

Also,

K_w = [H⁺].[OH^-]

4.57 x 10^-15 M² = [H⁺] x 0.0204 M

[H⁺] = 4.57 x 10^-15 M² / 0.0204 M

[H⁺] = 2.24 x 10^-13 M

Also,

pH = - log [H⁺]

So,

pH = - log 2.24 x 10^-13 M

pH = 12.65

Hence, pH of 0.0204 M KOH at 150C = 12.65

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(b). At 45 ⁰C :-

Given ,

[KOH] = [OH^-] = 0.0204 M

It is given that , Ionic product of water (K_w) at 45 ⁰C = 3.94 x 10^-14

Also,

K_w = [H⁺].[OH^-]

3.94 x 10^-14​​​​​​​ M² = [H⁺] x 0.0204 M

[H⁺] = 3.94 x 10^-14​​​​​​​  M² / 0.0204 M

[H⁺] = 1.93 x 10^-12 M

Also,

pH = - log [H⁺]

So,

pH = - log 1.93 x 10^-12 M

pH = 11.71

Hence, pH of 0.0204 M KOH at 450C = 11.71

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