Calculate the `pH` of a buffer by mixing `0.15` mole of `NH_(4)OH` and `0.25` mole of `NH_(4)CI` in a `1000mL` solution `K_(b)` for `NH_(4)OH = 2.0 xx 10^(-5)`

Explanation

`0.15 mol` of `NH_(4)OH` and `0.25mol` of `NH_(4)CI` in `1000mL (1.0L)` solution given a basic buffer.
`pH = 14 - pOH`
Now using Henderson's equation.
`pOH = pK_(b) + "log" (["salt"])/(["base"])`
`rArr pOH =- log (2.0 xx 10^(-5)) = log ((0.25//1)/(0.15//1)) = 4.92`
`rArr pH = 14 - 4.92 = 9.08`