Calculate the pH of a solution obtained by mixing 5 mL of 0.1 M NH_(4) OH with 250 mL of 0.1 M NH_(4) Cl solution . K_(b) for NH_(4)OH=1.8xx10^(-5).

Calculate the pH of a solution obtained by mixing 5 mL of 0.1 M NH_(4)

1.	Calculate the pH of a solution obtained by mixing 5 mL of 0.1 M NH_(4) OH with 250 mL of 0.1 M NH_(4) Cl solution . K_(b) for NH_(4)OH=1.8xx10^(-5).
Answer» Solution :5 mL of 0.1 M `NH_(4)OH = 5 xx 0.1` MILLIMOLE = 0.5 millimole 250 mL of 0.1 M `NH_(4)Cl= 250 xx 0.1` millimole = 25 MILLIMOLES Total VOLUME of solution after mixing = 255 mL `:. `[Salt]=`[NH_(4)Cl] = (25)/(255) M` [Base]`=[NH_(4)OH]=(0.5)/(255) M` `POH = pK_(B) + log. (["Salt"])/(["Base"])=-log. (1.8xx10^(-5))+log.(25//255)/(0.5//255) = (5-0.2553) + 1.6990= 6.4437` `pH = 14-6.4437 = 7.5563`