Calculate the pH of a solution obtained by diluting 1 mL of 0.10 M weak monoacidic base to 100 mL at constant temperature if Kb of the base is 1 × 10-5.

M1 V1=M2 V21×0.10=M2×100M2=0.001=10-3BOH(g)⇌B++OHC        0 0C(1−α) Cα CαKb=Cα×CαC(1-α)Kb=Cα2  (∵1-α≈1)α=Kb/COH-=Cα=KbC×C=KbC=10-5×10-3=10-4∴pH+pOH=14∴pH=14-4=10