Question

To prepare a buffer of pH `8.26`, amount of `(NH_(4))_(2)SO_(4)` to be added into 500mL of `0.01M NH_(4)OH` solution `[pK_(a)(NH_(4)^(+))=9.26]` is:
A. 0.05 mole
B. 0.025 mole
C. 0.10 mole
D. 0.0005 mole

Answer 1

Correct Answer - 2
`pH=pK_(a)+log.(["Base"])/(["Salt"])`
`[Base]=(0.01xx500)/(500)=0.01`
`[NH_(4)^(+)]=(axx2)/(500),` Let a millimole of `(NH_(4))_(2)SO_(4)` are added
`:." "[Sal t]=[NH_(4)^(+)]`.
`pH=9.26+log[(0.01)/(2a//500)]`
`8.26=9.26+log.(0.01xx500)/(2a)`
`a=25" ":." "` Mole of `(NH_(4))_(2)SO_(4)` added `=0.025`