In: Chemistry
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
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