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)`

Options

A. `10.56 g`
B. `5.21 g`
C. `12.74 g`
D. `16.25` g

Answer

B

Explanation

`pOH =pK_(a) +log. (["salt"])/(["base"])`
`14-9.35 = - log (1.78 xx 10^(-5))+ log. (["salt"])/([500 xx 0.2])`
`log. (["salt"])/([100]) = - 0.1 `
[salt] = `0.79 xx 100 = 79` millimoles = `[NH_(4)^(+)] `
We get 2 moles of `NH_(4)^(+) ` from a mole of `(NH_(4))_(2)SO_(4)`
Molecular mass of `(NH_(4))_(2)SO_(4)=132`
Thus ,wt. of salt
`=1/2 xx 79 xx 132 xx 10^(-3) g . = (10.42)/2 "g . 5.21 g "`