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

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.