Question

Calculate the pH when 0.020 mole of HBr is added to 200.0 mL of a solution containing 0.60 M NH3 and 0.60 M NH4Cl, assuming no volume change. (Kb of NH3 =1.8x10-5). Report your answer to three significant figures.

Answer 1

$pOH = pKb + log(\frac{salt}{base}).$

$pH + pOH = 14.$

Kb= 1.8*10^-5

pOH=(1.8*10^-5)+log(0.6/0.1)= 6.00

Therefore; pH=14-6 =8