Question

1) Calculate the pH of a solution produced by mixing 100.0 mL of 0.10 M HCl(aq) and 100.0 mL of 0.20 M NaOH(aq).

2) Calculate the pH of an aqueous solution made by mixing 100.0 mL of 0.40 M NH₄Cl and 50.0 mL of 0.40 M NaOH. The pK_a of NH₄⁺ is 9.24.

Answer 1

is a is strong 400 strong a NaOH strong acid acid and Her an base. and a car), will completely neutralize Now, 100 ml 0-10 M Ha (aq), will completely me 0.10 m NaOH (aq) i.e. 50 ml 0.20 m Naolt (ore) 1100 me 0.10 m Naoh = 50 me 0-20 on Naon I found out by applying the rule. Vis, = V2 S2 Then 50 in me 0.20m Naot will remain, which will be (100+50+50) = 200 me aqueous solut present 2xo-20 e same rule then, in that solution, concentration of 50x 0. 20 Nach M 200 applied. I = [OH] = 0.05 it Then, pH = 14-DOH = 14 - [-log [oHe] ] [o pot = -log [on]] = 8/4 + dog (0-05) = (14 - 1.301) = 12.699 ~ 12.70 (2. (base For this problers, calculation of pH is mady using Henderson's equation which is pH= 14 - POH = 14 - [pk 6+ log (Salt) NaOH + NHACH - NHGOH + Nace 50 ml 50ml 100 ml 0-40 0.40m 0.40m (salt) (base) strong (base) weak m M 0.80 N solutim = 50+100 Now, in final solution [salt] = [NHA CA] = 50x0.40 tso 3 = 0.40 [ base] - SNHOH) = 100 x 0.40 150 3 since, NagH bas. total volume of reacted reacted completely? compl = 000 m = 150 ml and pkb = 14 - pka = 14- 9.24 = 4.76 Given Then, pH = 14 - (pkb + log [salt) =14 - 4.76 - log 2 20/3) = 9.54 1 [pka (M)=9-24) I

Therefore, (1) pH = 12.70 (strong base involved; higher pH).

(2) pH = 9.54 (base involved, is weak; lower pH).