Question

A) Calculate the pH of a buffer that is 0.225M HC₂H₃O₂ and 0.162M KC₂H₃O. The Ka for HC₂H₃O₂ is 1.8x10^-5

B) A 1.00L buffer solution is 0.250M in HF and 0.250M in NaF. Calculate the pH of the solution after the addition of 0.100 moles of solid NaOh. Assume no volume chane uon the addition of base. Ka for HF= 3.5x10^-4

^C) A 100.0ml sample of 0.20M HF is titrated with 0.10M KOH. Determine the pH of the solution after the addition of 100.0mL of KOH. The Ka of HF is 3.5x10^-4

Answer 1

A)

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {0.162/0.225}

= 4.602

Answer: 4.60

B)

mol of NaOH added = 0.1 mol

HF will react with OH- to form F-

Before Reaction:

mol of F- = 0.25 M *1.0 L

mol of F- = 0.25 mol

mol of HF = 0.25 M *1.0 L

mol of HF = 0.25 mol

after reaction,

mol of F- = mol present initially + mol added

mol of F- = (0.25 + 0.1) mol

mol of F- = 0.35 mol

mol of HF = mol present initially - mol added

mol of HF = (0.25 - 0.1) mol

mol of HF = 0.15 mol

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.456

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

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

= 3.456+ log {0.35/0.15}

= 3.824

Answer: 3.82

C)

Given:

M(HF) = 0.2 M

V(HF) = 100 mL

M(KOH) = 0.1 M

V(KOH) = 100 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.2 M * 100 mL = 20 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 100 mL = 10 mmol

We have:

mol(HF) = 20 mmol

mol(KOH) = 10 mmol

10 mmol of both will react

excess HF remaining = 10 mmol

Volume of Solution = 100 + 100 = 200 mL

[HF] = 10 mmol/200 mL = 0.05M

[F-] = 10/200 = 0.05M

They form acidic buffer

acid is HF

conjugate base is F-

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.456

use:

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

= 3.456+ log {5*10^-2/5*10^-2}

= 3.456

Answer: 3.46