Question

To prepare a buffer of pH 8.26 amount of (NH4)2SO4 to be added into 50 mL of 0.01 M NH4OH solution is [pKa(NH+4) = 9.26]

Answer 1

Let (NH4)2SO4 to be added to 500 mL solution = x mol

∴[NH+4]=4xM

pH=8.26

pOH=14−8.26=5.74

pKb=14−9.26=4.74

pOH=pKb+log[NH+4][NH4OH]

⇒5.74=4.74+log4x0.01

log4x×102 = 1

⇒4x×102=1

⇒x=140 mol = 0.025 mol