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pH of 0.01 M `(NH_(4))_(2) SO_(4)` and 0.02 M `NH_(4)OH` buffer `(pK_(a)` of `NH_(4)^(+)=9.26)` is
A. `4.74+log 2 `
B. `4.74 -log2`
C. `9.26+log 2`
D. `9.26+log 1`

1 Answer

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Correct Answer - D
`pK_(a)(NH_(4)^(+))=9.26`
`pK_(b)(NH_(3))=14-9.26=4.74`
`[NH_(4)]=2xx[(NH_(4))_(2)SO_(4)]`
`=2xx0.01=0.02 M`
`pOH =pK_(b)+log . ([NH_(4)^(+)])/([NH_(4)OH])`
`pOH =4.74 +log . (0.02)/(0.02)=4.74-log 1`
`pH =14-pOH =1404.74+log 1`
`=9.26+log 1`

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