b) Calculate the pH of a buffer solution consisting of 0.04 M CH_3 COOH and 0.4M CH_3 COONa. What is the change in the pH after adding 0.01 mol of HCI to 500 ml of the above buffer solution. Assume that the addition of HCI causes neglible change in the volume . Given (K_(s)=1.8xx10^(-5)).

b) Calculate the pH of a buffer solution consisting of 0.04 M CH_3 COO

1.	b) Calculate the pH of a buffer solution consisting of 0.04 M CH_3 COOH and 0.4M CH_3 COONa. What is the change in the pH after adding 0.01 mol of HCI to 500 ml of the above buffer solution. Assume that the addition of HCI causes neglible change in the volume . Given (K_(s)=1.8xx10^(-5)).
Answer» Solution :(b)pH of BUFFER `[H_3 O^+]=` ANTILOG of `(-pH)` `underset(0.4-alpha)(CH_3 COOH_((aq))) hArr underset(0.4)( CH_3 COOH_((aq))^(-))+underset(alpha)(H_((aq))^(+)1)` `underset(0.4)(CH_3 COOH_((aq))) to underset(0.4)( CH_3 COOH_((aq))^(-))+underset(0.4)(Na_((aq))^(+))` `[H^+]=(K_(a)[CH_3 COOH])/([CH_3 COO^-])` `[CH_3 COOH]=0.4 - alpha cong 0.4` `[CH_3 COO]=0.4+alpha cong 0.4` `THEREFORE [H^(+)] = (K_(a)(0.4))/((0.4))` `[H^+]=1.8xx10^(-5)` `therefore pH=-log (1.8xx10^(-5)) =4.74` Addition of 0.01 mol HCI to 500 ml of buffer Added `[H^+]=(0.01 mol)/(500 mL )=(0.01 mol)/((1)/(2)L)` `=0.02M` `CH_3 COOH_((aq)) hArr CH_3 COO_((aq))^(-)+H_((aq))^(+)` `underset(0.4-alpha)(CH_3 COOH_((aq))) hArr underset(0.4)( CH_3 COOH_((aq))^(-))+underset(alpha)(H_((aq))^(+)1)` `underset(0.4)(CH_3 COOH_((aq))) to underset(0.4)( CH_3 COOH_((aq))^(-))+underset(0.4)(Na_((aq))^(+))` ` underset((0.02))(CH_3 COO^(-))+underset(0.02)(HCI) to underset(0.02)(CH_3 cOOH)+underset(0.02)(CI^-)` `[CH_3 COOH]=0.4-alpha +0.02` `=0.42 - alpha cong 0.42` `[CH_3 COO^-] =0.4+alpha -0.02` `=0.38+alpha cong 0.38` `[H^+]=((1.8xx10^(-5))(0.48))/((0.38))` `[H^+]=1.9xx10^(-5)` `=5-log1.99` `=5-0.30` `=4.70`