Question

General Chemistry

Calculate the pH of a 0.50 M NH₄Cl solution (for NH₃, K_b = 1.8 x 10^-5). Write the net ionic equation for the reaction.

Answer 1

Given NH3 Kb = 1.8 x 10^-5 and c = 0.5M

NH4Cl after hydrolysis will give acidic medium solution as it is made of weak base NH4OH and strong acid HCl. The pH relation can be given as:   

pH = 7 - 1/2[pKb + logc] ------------------(1)

Where in this case pKb of ammonia = -logKb = -log1.8 x 10^-5

pKb = - (-4.74)

pKb = 4.74

Concentration of NH4Cl given as "c" = 0.5M

Put all these values in above equation (1):

pH = 7 - 1/2[4.74 + log0.5]

pH = 7 - 1/2[4.74 +(-0.30)]

pH = 7 - 1/2[4.74 – 0.30)]

pH = 7 - 1/2[4.44]

pH = 7 – 2.22

pH = 4.78

Net ionic equation:

NH4+(aq) + H2O(l) <--> NH4OH(l) + H+(aq)