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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 Hint

Explanation / 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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