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

Explanation

`0.15` moles of `NH_(4) OH " and " 0.25 " mole of " NH_(4) Cl " in 1000 mL " (1.0 L)`solution give a basic buffer .
`pH = 14 - pOH`
Now using Henderson 's equation :
`pOH = pK_(b) + log_(10).(["salt"])/(["base"])`
`rArr pOH = - log_(10) (2.0 xx10^(-5))+log_(10)((0.25//1)/(0.15//1))`
`= 5 - log_(10) 2 + log_(10)5 - log_(10) 3 = 4.92`
` rArr pH = 14 - 4.92 = 9.08`