Consider the titration of a 36.0 mL sample of 0.170 M HBr with 0.200 M KOH. Dete

Question

Consider the titration of a 36.0 mL sample of 0.170 M HBr with 0.200 M KOH. Determine each of the following:

Part A

the initial pH

Express your answer using three decimal places.

Part B

the volume of added base required to reach the equivalence point

Express your answer in milliliters.

Part C

the pH at 12.0 mL of added base

Express your answer using three decimal places.

Part D

the pH at the equivalence point

Express your answer as a whole number.

Part E

the pH after adding 5.0 mL of base beyond the equivalence point

Express your answer using two decimal places

Explanation / Answer

Part A : intial pH

HBr is a strong acid, disscoiates completely

[H+] = 0.17 M

pH = -log[H+] = 0.769

Part B. Volume of base required to reach equivalence point

equivalence point

moles of acid = 0.17 M x 0.036 L = 6.12 x 10^-3 mols

moles of base = 6.12 x 10^-3 mols

volume of base added = 6.12 x 10^-3 mols/0.2 M = 0.0306 L = 30.6 ml

Part C. after 12 ml of 0.2 M KOH is added

moles of acid = 0.17 M x 0.036 L = 6.12 x 10^-3 mols

moles of base added = 0.2 M x 0.012 L = 2.4 x 10^-3 mols

excess acid in solution = 3.72 x 10^-3 mols

molar concentration of acid = 3.72 x 10^-3/(0.036 + 0.012) = 0.0775 M

pH = -log(0.0775) = 1.111

Part D. Equivalence point

moles of acid present = moles of base added

moles of acid = 0.17 M x 0.036 L = 6.12 x 10^-3 mols

moles of base = 6.12 x 10^-3 mols

volume of base added = 6.12 x 10^-3 mols/0.2 M = 0.0306 L = 30.6 ml

A solution of strong acid and base forms salt.

pH = 7

Part E. after 5 ml excess base is added

Total volume of solution = 0.036 + 0.0356 = 0.0716 L

moles of base = 0.2 x 0.005 = 1 x 10^-3 mols

molar concentration of base in solution = 1 x 10^-3/0.0716 = 0.014 M

KOH is a strong base, dissociates completely,

[OH-] = 0.014 M

pOH = -log[OH-] = 1.854

pH = 14 - pOH = 12.15

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