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Upon passing 0.01 moles HCl gas through 100mL of 0.05M Formic acid solution (K_(a)=1.8xx10^(-4)) determine change in pH of solution and [HCOO^(-)] in resulting solution.

Answer»

Solution :INITIAL `pH:alpha=sqrt((K_(a))/(C))=0.06(lt0.1)`
So `[H^(+)]=Calpha=0.06xx0.05=0.003M`
`:.pH=2.52`
final `pH=-log(0.01)/(0.1)=1` :.CHANGE =`1-2.52=-1.52`.
Now `K_(a)=([H^(+)][HCOO^(-)])/([HCOOH])rArr 1.8xx10^(-4)=(0.1xx[HCOO^(-)])/(0.05)`
`rArr [HCOOH^(-)]=9xx10^(-5)M`


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