Question

Ammonium acetate which is 0.01 M, is hydrolysed to 0.001 M concentration .Calculate the change in Ph in 0.001 M solution, if initially `Ph =pK_(a)`.
A. 5
B. 10
C. 100
D. 1

Answer 1

Correct Answer - D
`CH_(3)COO^(-)+NH_(4)^(+)+H_(2)O hArr CH_(3)COOH+NH_(4)OH`
`{:("Initial",0.01M,0.01M,(0.01-0.001)M,(0.01-0.001)M),("At eqm.",0.001M,0.001M,=0.009M,=0.009M):}`
`K_(h)=([CH_(3)COOH][NH_(4)OH])/([CH_(3)COO^(-)][NH_(4)^(+)])=((0.009)^(2))/((0.001)^(2))=10^(2)`
`K_(h)=(K_(w))/(K_(a) xx K_(b))`
`:. K_(b)=(K_(w))/(K_(a)xxK_(h))=(10^(-14))/(K_(a)xx 10^(2))=(10^(-16))/(K_(a))`
`[H^(+)]=sqrt((K_(a)xxK_(w))/(K_(b)))=sqrt((K_(a)xx10^(-14))/(10^(-16)//K_(a)))=10K_(a)`
`(pH)_("Initial")=pK_(a)`
`(pH)_("Final")=-log[H^(+)]=-log(10K_(a))=pK_(a)-1`
Change in pH `=(pH)_("Final")-(pH)_("Inital")`
`=pK_(a)-1-pK_(a)= -1`