9. a) (5 points) Determine the pH at the equivalence point when 0.05 M KOH is ti
ID: 861868 • Letter: 9
Question
9. a) (5 points) Determine the pH at the equivalence point when 0.05 M KOH is titrated with 50.0 ml of 0.01 M benzoic acid HC7H5O2. The Ka for benzoic acid is 6.3 x10-5 .
FINAL ANSWER:
pH = 8.06
b) (5 points) What volume of KOH would you need to add to 25.0 ml of 0.01 M benzoic acid so that the resulting solution was pH 5.50?
FINAL ANSWER
4.8 ml
c) (2 points ) Someone spilled concentrated sulphuric acid on the floor in the chemistry laboratory. You want to neutralize the acid, would it be better to pour concentrated NaOH solution or to sprinkle solid sodium bicarbonate over the acid? Explain your choice.
Explanation / Answer
A- + H2O <--> HA + OH-
Kb = Kw/Ka
Kb = 1x10^-14 / 6.3x10^-5 = 1.59x10^-10
milli moles of acid = molarity X volume = 0.01 X 50 = 0.5
0.5 millimol = mL x 0.05 M NaOH
mL NaOH = 10mL
Total volume = 50 + 10 = 60 mL (0.06 L)
C6H5COO- at equivalence point = 0.5 X 10^-3 / 0.06 L = 0.0083 M
Let X = [OH-]
1.59x10^-10 = X^2 / (0.0083 - X)
assume X << 0.0083
X = 1.31 10^-6
pOH = 5.88
pH = 8.11
2.
pH= pKa + log [salt / acid]
5.5 = 4.2 + log [salt / 0.01]
1.3 = log[salt / acid]
19.95 = salt / acid
Salt = 19.95 X 0.01 = 0.1995
Volume = moles / molarity = 0.1995 / 0.05 = 3.99mL
3. Add solid sodium bicarbonate
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