Please could anyone help me to give me solution of this problem:
Calculate the pH of a solution obtained by mixing 50 mL of NH3 0.1 M (Kb = 1.8•10–5 mol/L) and 25 mL of HCl 0.2 M
Note the correct spelling of chemistry.
NH3 + HCl ==> NH4Cl
mols NH3 = M x L = 0.1 x 0.05 = 0.005.
mols HCl 0.2 x 0.025 = 0.005.
You see this is an exact neutralizing solution; you have no HCl or NH3 left and have formed 0.005 mols NH3 in 75 mL solution so (NH4Cl) = 0.005/0.075 = approx 0.067 M NH4Cl. The pH of this solution is that due to the hydrolysis of the NH4^+.
....NH4^+ + H2O ==> H3O^+ + NH3
I...0.067............0.......0
C....-x..............x.......x
E...0.067-x..........x.......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.067-x)
Solve for x = (OH^-) and convert to pH.