Question

What is the pH of a weak acid strong base titration after each of the following additions? Ka for HF is 3.5 x 10-4.

a. 10.0 mL of 0.250 M NaOH added to 15.0 mL of 0.400 M HF

b. 25.0 mL of 0.250 M NaOH added to 35.0 mL of 0.1785 M HF

c. 20.0 mL of 0.250 M NaOH added to 20.0 mL of 0.350 M HF

Please show your work so I can better understand this. thanks! :)

Answer 1

a)

we have:

Molarity of HF = 0.4 M

Volume of HF = 15 mL

Molarity of NaOH = 0.25 M

Volume of NaOH = 10 mL

mol of HF = Molarity of HF * Volume of HF

mol of HF = 0.4 M * 15 mL = 6 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.25 M * 10 mL = 2.5 mmol

We have:

mol of HF = 6 mmol

mol of NaOH = 2.5 mmol

2.5 mmol of both will react

excess HF remaining = 3.5 mmol

Volume of Solution = 15 + 10 = 25 mL

[HF] = 3.5 mmol/25 mL = 0.14M

[F-] = 2.5/25 = 0.1M

They form acidic buffer

acid is HF

conjugate base is F-

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.4559

we have below equation to be used:

This is Henderson–Hasselbalch equation

pH = pKa + log {[conjugate base]/[acid]}

= 3.4559+ log {0.1/0.14}

= 3.3098

Answer: 3.31

b)

we have:

Molarity of HF = 0.1785 M

Volume of HF = 35 mL

Molarity of NaOH = 0.25 M

Volume of NaOH = 25 mL

mol of HF = Molarity of HF * Volume of HF

mol of HF = 0.1785 M * 35 mL = 6.2475 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.25 M * 25 mL = 6.25 mmol

We have:

mol of HF = 6.2475 mmol

mol of NaOH = 6.25 mmol

6.2475 mmol of both will react

excess NaOH remaining = 0.0025 mmol

Volume of Solution = 35 + 25 = 60 mL

[OH-] = 0.0025 mmol/60 mL = 0 M

we have below equation to be used:

pOH = -log [OH-]

= -log (4.167*10^-5)

= 4.3802

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.3802

= 9.62

Answer: 9.62

c)

we have:

Molarity of HF = 0.35 M

Volume of HF = 20 mL

Molarity of NaOH = 0.25 M

Volume of NaOH = 20 mL

mol of HF = Molarity of HF * Volume of HF

mol of HF = 0.35 M * 20 mL = 7 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.25 M * 20 mL = 5 mmol

We have:

mol of HF = 7 mmol

mol of NaOH = 5 mmol

5 mmol of both will react

excess HF remaining = 2 mmol

Volume of Solution = 20 + 20 = 40 mL

[HF] = 2 mmol/40 mL = 0.05M

[F-] = 5/40 = 0.125M

They form acidic buffer

acid is HF

conjugate base is F-

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.4559

we have below equation to be used:

This is Henderson–Hasselbalch equation

pH = pKa + log {[conjugate base]/[acid]}

= 3.4559+ log {0.125/0.05}

= 3.85

Answer: 3.85