A buffer solution has 0.25 M CH 3 COOH, 0.15 M CH 3 COONa, [Mn 2+ ] = 0.015 M and is saturated with H 2 S (0.1 M). Given : K a (CH 3 COOH) = 1.8 × 10 –5 , K a (H 2 S) = 9 × 10 –21 , K sp (MnS) = 2.4 × 10 –13 . Concentration of a component of the buffer may have to be increased to start the precipitation of MnS. What would be its new concentration (in mole per litre) ? Report your answer after multiplying by 10. (Report your answer as '0', if the concentration of any component need not be increased to start the precipitation).
Text Solution
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Sol. pK a (CH 3 COOH) = 4.74
[CH 3 COOH] = 0.25 M, [CH 3 COONa] = 0.15 M
[H + ] = 
= 
= 3 × 10 –5 M
H 2 S 
2H + + S –2 ⇒ K a = 
[S 2– ] = 
= 10 –12 M
IP (MnS) = [Mn +2 ] [S –2 ] = 1.5 × 10 –2 × 10 –12 = 1.5 × 10 –14
IP < K sp ⇒ No ppt is formed.
For precipitation of MnS, the minimum concentration of [S 2– ] can be obtained as follows :
[Mn +2 ] [S 2– ] = K sp
1.5 × 10 –2 × [S 2– ] = 2.4 × 10 –13 ⇒ [S 2– ] = 1.6 × 10 –11 M
For this [S 2– ],
[H + ] 2 = 
= 
= 7.5 × 10 –6 M
[H + ] = 
⇒ 7.5 × 10 –6 = 
[CH 3 COONa] = 0.6 M
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