Determine the pH during the titration of 10.3 mL of 0.138 M hydrochloric acid by

Question

Determine the pH during the titration of 10.3 mL of 0.138 M hydrochloric acid by 5.95×10-2 M potassium hydroxide at the following points:

(1) Before the addition of any potassium hydroxide

(2) After the addition of 12.0 mL of potassium hydroxide

(3) At the equivalence point

(4) After adding 31.1 mL of potassium hydroxide

2. Determine the pH during the titration of 21.2 mL of 0.369 M perchloric acid by 0.551 M potassium hydroxide at the following points:

(1) Before the addition of any potassium hydroxide

(2) After the addition of 7.10 mL of potassium hydroxide

(3) At the equivalence point

(4) After adding 17.0 mL of potassium hydroxide

Explanation / Answer

A 1)Concentration of HCl = 0.138 M

pH of HCl before addition of KOH = -log [H+] = -log (0.138) = 0.86

2) pH after addition of 0.012 lts (12ml) of KOH =

Moles of KOH = conc x vol = 5.95 x 10-2 x 0.012 = 0.0714 x 10-2 = 0.000714

Initial Moles of HCl = 0.0103 x 0.138 = 0.0014214

moles of HCl after addition of KOH = 0.0014214 - 0.000714 = 0.0007074

Volume of titrant solution after addition = 10.3+12 = 22.3 ml= 0.0223 lts

Molarity of HCl after addition of KOH = 0.0007074/0.0223 = 0.0317

pH after addition of 12 ml of KOH = -log [H+] = -log(0.0317) = 1.499

3) pH at equivalence point is 7 (titration between strong acid and strong base the moles of acid = moles of base)

4) pH after addition of 0.0311 lts (31.1 ml) of KOH =

Moles of KOH = conc x vol = 5.95 x 10-2 x 0.0311 = 0.1850 x 10-2 = 0.001850

Initial Moles of HCl = 0.0103 x 0.138 = 0.0014214

moles of OH- remaining = 0.001850-0.0014214= 0.0004286

Total volume of titrant solution = 0.0311 + 0.0103 = 0.0434

Molarity of KOH in titrant solution = 0.0004286/0.0434 = 0.00988

pOH = -log [OH-] = -log 0.00988 = 2.00

So pH = 14-2.00 = 12.00

B) 1) pH of HClO4 initially = -log [H+] = -log (0.369) = 0.433

2) Ater addition of 7.10 ml of KOH =

pH after addition of 0.0710 lts (7.10 ml) of KOH =

Initial moles of HClO4 = 0.369 x 0.0212 = 0.007786

Moles of KOH = conc x vol = 0.0551 x 0.0710 = 0.0039121

moles of HClO4 left after titration = 0.007786-0.0039121 = 0.0038739

Volume of titrant solution after addition = 21.2+7.10 = 28.3 ml= 0.0283 lts

Molarity of HClO4 after addition = 0.0038739/0.0283 = 0.13689

pH = -log [H+] = -log (0.13689) = 0.864

3) pH at equivalelnce point is 7 (strong acid vs strong base the moles of acid = moles of base )

4)

pH after addition of 0.0170 lts (17.0 ml) of KOH =

Moles of KOH = conc x vol = 0.551  x 0.017 = 0.009367

Initial moles of HClO4 = 0.369 x 0.0212 = 0.007786

moles of OH- remaining = 0.009367-0.007786= 0.001581

Total volume of titrant solution = 0.017 + 0.0212 = 0.0382

Molarity of KOH in titrant solution = 0.001581/0.0382= 0.04139

pOH = -log [OH-] = -log 0.04139= 1.38

So pH = 14-1.38 = 12.62

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