Question

Suppose 10.0 mL of 2.00 MNaOH is added to (a) 0.780 L of pure water and (b) 0.780 L of a buffer solution that is 0.682 Min butanoic acid (HC4H7O2) and 0.674 Min butanoate ion (C4H7O2–). Calculate the pH of (a) and (b) before and after the addition of the NaOH. Assume volumes are additive. (Ka, HC4H7O2= 1.5 × 10-5)

Answer 1

For the mixture:

• (a) Pure water: pH 7.00 (initial) to 12.41 (final)
• (b) Buffer solution: pH 4.81 (initial) to 4.82 (final)

How to solve for the mixture?

Calculating pH before and after NaOH addition:

Initial Solutions:

(a) Pure water:

pH = 7.00 (neutral)

[H⁺]= [OH⁻]

= 10⁻⁷ M

(b) Buffer solution:

[HC₄H₇O₂] = 0.682 M

[C₄H₇O₂⁻] = 0.674 M

Use Henderson-Hasselbalch equation:

pH = pKa + log([A⁻]/[HA])

pH = 4.81 (slightly acidic)

NaOH Addition:

10.0 mL of 2.00 M NaOH adds 0.02 mol OH⁻

In both cases, OH- will react with H+ to form water:

H⁺ + OH⁻ → H₂O

Calculating pH after NaOH addition:

(a) Pure water:

Excess OH⁻ will increase [OH⁻]

New [OH⁻] = 0.02 mol / 0.790 L

= 0.0253 M

New pOH = -log(0.0253)

≈ 1.59

pH = 14 - pOH

= 12.41 (strongly basic)

(b) Buffer solution:

Excess OH- will react with some HC₄H₇O₂ to form C₄H₇O₂⁻

Use ICE table to track changes: (attached)

Substitute into Henderson-Hasselbalch equation:

4.81 = 4.81 + log((0.674+x)/(0.682-x))

Solve for x:

x ≈ 0.0077 M

New [HC₄H₇O₂] = 0.682-x

≈ 0.6743 M

New [C₄H₇O₂⁻] = 0.674+x

≈ 0.6817 M

New pH = 4.81 + log(0.6817/0.6743)

≈ 4.82 (slightly acidic, minimal change)

Answer 2

a) pH will be 12.398

b) pH will be 4.82.

Step-by-step explanation:

a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles

The total volume after addition of pure water = 0.780+0.01 = 0.79 L

The new concentration of /NaOH will be:

`molarity=(molesofsolute)/(volumeofsolution)=(0.02)/(0.79)=0.025M`

the [OH⁻] = 0.025

pOH = -log [OH⁻] = 1.602

pH = 14 -pOH = 12.398

b) The buffer has butanoic acid and butanoate ion.

i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:

`pH=pKa+log([salt])/([acid])`

pKa=
`-logKa=-log(1.5X10^(-5))=4.82`

ii) on addition of base the pH will increase.