How many moles of `NH_(4)C1` should be added to `200mL` solution of `1.18 M NH_(4)OH` to have a `pH` of `9.60. K_(b)` of `NH_(4)OH= 2 xx 10^(-5)`

Answer

A

Explanation

It is basic buffer.
`pH = 9.6, pOH = 14 - 9.6 = 4.4`
`K_(b) = 2 xx 10^(-5), pK_(b) =- log (2xx10^(-5)) =- 0.3 +5 = 4.7`
`pOH = pK_(b) + "log"(["Salt"])/(["Base"])`
`4.4 = 4.7 + "log"(["Salt"])/((0.18))`
`- 0.3 = log["Salt"] - log (0.18) = log ["Salt"] + 0.7447`
`- 0.3 - 0.7447 = log ["Salt"]`
`:. log ["Salt"] =- 1.0447`
`["Salt"] = "Antilog" (-1-0.447+1-1)`
`= "Antilog" (bar(2).9553) = 9.55 xx 10^(-2)M`.
Moles of salt `= (9.55 xx 10^(-2) xx 200)/(1000)`
`= 1.91 xx 10^(-2)= 0.019` mol.