A30.0-ml sample of 0.165 m propanoic acid is titrated with 0.300 m koh.

1. calculate the ph at 5 ml of added base.

2. calculate the ph at one-half of the equivalence point. (equivalence point is 8.96)

3. calculate the ph at 20 ml of added base.

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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