Calculate the weight of (NH(4))(2)SO(4) ...

Calculate the weight of `(NH_(4))_(2)SO_(4)` which must be added to `500mL` of `0.2M NH_(3)` to yield a solution of `pH = 9.35. K_(a)` for `NH_(3) = 1.78 xx 10^(-5)`.

Text Solution

Verified by Experts

`pOH =- log K_(b) + "log" (["Salt"])/(["Base"])`
`pOH =- log K_(b) + "log" ([NH_(4)^(o+)])/([NH_(4)OH])`
`because [NH_(4)^(o+)]` is obtained from salt `(NH_(4))_(2)SO_(4)`.
`because pH = 9.35`, therefore, `pOH = 14 - 9.35 = 4.65`
`:.` Mmol of `NH_(4)OH` in solution `= 0.2 xx 500 = 100`
Let a millimoles of `NH_(4)^(o+)` are added in solution.
`:. [NH_(4)^(o+)] = (a)/(500), [NH_(4)OH] = (100)/(500)`
`4.65 =- log (1.78 xx 10^(-5)) + "log" (a//500)/(100//500)`
`4.65 = 4.496 + "log" (a)/(100) :. a = 79.51`
`:. mmol of (NH_(4))_(2)SO_(4)` added `= (a)/(2) = (79.51)/(2) = 39.755`
`:. (W)/(132) xx 1000 = 39.755 :. W_((NH_(4))_(2)SO_(4)) = 5.248 g`

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Calculate the amount of (NH_(4))SO_(4) in grams which must be added to 500 ml of 0.200 M NH_(3) to yield a solution with pH = 9.35 ( K_(b) for NH_(3) = 1.78 xx 10^(-5) )

10.56 gm

15 gm

12.74 gm

16.25 gm

The amount of (NH_(4))_(2)SO_(4) in grams which must be added to 500 ml of 0.2 M NH_(2) to yield a solution og p^(H) = 9.35 ( K_(b) for NH_(3) = 1.78 xx 10^(-5)) will be

`7.248 g`

`6.248 g`

`5.248 g`

`52.48 g`

Calculate the amount of (NH_(4))_(2)SO_(4) in grams which must be added to 500 ml of 0.2 M NH_(3) to yield a solution of pH=9 , K_(b) for NH_(3)=2xx10^(-5)

`3.248 g`

`4.248 g`

`1.320 g`