Question

if a solution is made by mixing 100.0 mL of 0.500 M NH3 with 100.0 mL of 0.100 M HCl, calculate its pH value (K b for NH3 = 1.8 x 10 –5 )

Answer 1

NH3 + HCl ---------------> NH4Cl + HOH

the mole ratio between NH3 and HCl is 1 : 1

moles of NH3 =0.100L * 0.500 mol/L = 0.05 mol

moles of HCl = 0.100 L * 0.100mol/L = 0.01 mol

so the reaction proceeds untill 0.01 mol of NH3 has reached then stops , this leaves an excess NH3

0.05 - 0.01 = 0.04 mol

NH3 reacts with water to produce NH+ OH- ions in 1 : 1 ratio

Kb = [ NH+ ] [OH-] / [NH3]

1.8*10^-5 = x^2 / 0.04

x = 8.49*10^-4 = [OH-]

pOH = -log(8.49*10^-4 ) = 3.07

pH = 14 - 3.07 = 10.93

pH = 10.93