A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH.
1 answer:
Answer:
1) pH = 4.51
2) pH = 4.87.
3) pH = 12.32
Explanation:
1) the Ka of propanoic acid is 1.34 X 10⁻⁵
Therefore pKa = 4.87
When we add 5 mL of 0.300 M NaOH the moles of base added is
moles = molarity X volume
moles = 0.300 X 5mL = 1.5 mmoles
moles of acid present = molarity X volume = 0.165 X 30.0 = 4.95 mmoles
on addition of 1.5 mmoles of base the moles of acid neutralized = 1.5mmole
This will result in formation of salt of the acid
the moles of salt formed = 1.5 mmoles
the moles of acid left = 4.95 - 1.5 = 3.45 mmol
this acid and its salt mixture results in formation of a buffer
the pH of buffer is calculated as:
pH = pKa + log [salt] / [acid]
pH = 4.87 + log [1.5/3.45] = 4.51
2) at half equivalence point the moles of acid becomes equal to moles of salt formed thus the pH of solution will become equal to the pKa of acid
pH = 4.87.
3) the moles of based added due to addition of 20.0 mL = molarity X volume
moles = 0.300 X 20 = 6mmol
This will completely neutralize the acid (4.95 mmol)
after neutralization the moles of base left = 6-4.95 = 1.05 mmol
Total volume of solution = volume of acid + volume of base =30+20=50
concentration of hydroxide ion (due to excess base) =
[OH⁻]=0.021
pOH = -log[OH⁻]=1.68
pH = 14-pOH = 12.32
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