Question

In: Chemistry

Calculate the pH of a 0.0204 M KOH solution at each temperature. For 15 (4.57 ×...

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=                        


Solutions

Expert Solution

Sol :-

(a). At 15 0C :-

Given ,

[KOH] = [OH-] = 0.0204 M

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

Also,

Kw = [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

-------------------------

(b). At 45 0C :-

Given ,

[KOH] = [OH-] = 0.0204 M

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

Also,

Kw = [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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