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What is the ph of 0.1 M solution of H2S? Given ka1=10^-7 ka2= 1.3*10^-14

What is the ph of 0.1 M solution of H2S? Given ka1=10^-7 ka2= 1.3*10^-14

Grade:11

2 Answers

Arun
25750 Points
6 years ago
Dear Vaishnavi
 

Here we see that, K2 is much smaller than K1and so, it is expected that [H3O+ ]-hydronium ion is formed in first dissociation as:

H2S + H2O

Arun (askIITians forum expert)

Regards

 

 

Now, pH = - log [H3O+ ] = -log [ 1.0x 10-4 ] = 4

Or, [H3O+ ] = √(K1.[H2S]) = √(1.0 x 10-7 x 0.1) = 1.0x 10-4 

Or, K1 =  [H3O+ ]2 / [H2S]

 So, K1 =  [H3O+ ]  [HS- ]/  [H2S]

 - + HS+O3 H

Devender Chopra
15 Points
2 years ago
Here as j2 is very very less than k1
pH = - log√k1C
= -log√10-⁷ × 0.1 
=-log√10-⁸
=-log10-⁴
pH = 4
Hope this helps....!!!
 
 
 
 
 
 
       
 
 
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