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. `5.35`
B. `6.6`
C. `1.67`
D. `10.2`

Answer

B

Explanation

`P^(H) = 14 - P^(kh) - "log" (N_(S)V_(S))/(N_(b)V_(b))`
`4.74 = 14 - 9.26 - "log" (N_(S)V_(S))/(0.2 xx 0.5)`
`4.74 + 9.26 = 14 - "log" NsVs - "log"10^(-1)`
`13 = 14 - "log" NSVs + 1`
`13 = 15 - "log" NsVs`
`"log" NsVs = 15 - 13 = 2`
`NsVs = 100`
`(wt)/(66) xx (1000)/(Vs) xx Vs = 100`
`wt = 6.6 g`