Question

After diluting 30 ml of 0.100M NH4OH(Kb=1.75x10-5) to 100 ml, the solution is titrated with 0.100 M HCl. Calculate the pH value of the solution: a) at the start of the titration b) two-thirds of the way to the equivalence point c) at the equivalnece point d) when 40.0 ml of the HCl have been added. Make a graph of oH versus Va.

Answer 1

a) Initiallt there are only NH4OH is present, hence -

            [OH^-] = (K_b*C)^0.5 = (1.75*10^-5*0.03)^0.05 = 7.24*10^-4 M

Hence , pOH = -log [OH^-] = -log ( 7.24*10^-4) = 3.1, pH = 10.9

b) two-thirds of the way to the equivalence point:

   After this point, [Salt] = 0.03*2/3 = 0.02

   [Base] = 0.03 - 0.02 = 0.01

Hence - pOH = pKb + log [0.02 /0.01]

                         = 4.75 + 0.30 = 5.05

Hence - pH = 9 (pH + pOH = 14)

c) At equilance point , there are no NH4OH. Only NH4Cl will be left. Hence -

              [H⁺] = (0.03*5.62*10^-10)^0.5 = 4.10*10^-6 M

Hence - pH = 5.40

d) [HCl] at this point = 0.001 mol *1000 / 140 = 0.00714 mol/L

pH = -log [H⁺] = -log (0.00714) = 2.14