Question 7 of 10 Map Learning macmillan Calculate the pH for each of the followi
ID: 505203 • Letter: Q
Question
Question 7 of 10 Map Learning macmillan Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.110 M KOH(aq), with 0.110 M HBr(aq). Number Note: Enter your answers with two decimal places. (a) before addition of any HBr Number (b) after addition of 13.5 mL of HBr Number (c) after addition of 21.5 mL of HBr Number (d) after the addition of 35.0 mL of HBr Number (e) after the addition of 42.5 mL of HBr Number (f) after the addition of 50.0 mL of HBr Previous Give Up & View solution Check Answer Next Exit HintExplanation / Answer
Solution:-
(a) When no HBr is added then the pH will be because of the strong base KOH only.
[OH-] = 0.110 M
pOH = - log(0.110)
pOH = 0.96
pH = 14 - 0.96
pH = 13.04
(b) Initial mmol of KOH = 35.0 x 0.110 = 3.85 mmol
mmol of HBr added = 13.5 x 0.110 = 1.485 mmol
excess mmol of KOH = 3.85 - 1.485 = 2.365 mmol
total volume = 35.0 ml + 13.5 ml = 48.5 ml
concentration of excess KOH or OH- = 2.365mmol/48.5ml = 0.0488 M
pOH = - log 0.0488 = 1.31
pH = 14 - 1.31
pH = 12.69
(c) mmol of HBr added = 21.5 x 0.110 = 2.365 mmol
excess mmol of KOH = 3.85 - 2.365 = 1.485 mmol
Total volume = 35.0 ml + 21.5 ml = 56.5 ml
[OH-] = 1.485mmol/56.5ml = 0.0263 M
pOH = - log 0.0263 = 1.58
pH = 14 - 1.58
pH = 12.42
(d) mmol of HBr added = 35.0 x 0.110 = 3.85
mmol of KOH are also 3.85, Since we have equal mmol of strong acid and strong base so the
reaction will be complete, solution will be neutral and pH would be 7.00
(e) mmol of HBr added = 42.5 x 0.110 = 4.675 mmol
excess mmol of HBr = 4.675 - 3.85 = 0.825 mmol
total volume = 77.5 ml
[H+] = 0.825mmol/77.5ml = 0.0106 M
pH = - log 0.0106
pH = 1.97
(f) mmol of HBr added = 50.0 x 0.110 = 6.05 mmol
excess mmol of HBr = 6.05 - 3.85 = 2.2 mmol
Total volume = 35.0 ml + 50.0 ml = 85.0 ml
[H+] = 2.2 mmol/85.0 ml = 0.0259 M
pH = - log 0.0259
pH = 1.59
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