Question

Given a solution of 0.125 M CH₃COOH (aq) and a solution of 0.150 M NaCH₃COO (aq), how would you prepare 100.0 mL of a buffer solution with a pH of 4.400? pK_a = 4.740 for CH₃COOH (aq).

Select one:

a. Mix 68.6 mL of the CH₃COOH (aq) with 31.4 mL of the NaCH₃COO (aq)

b. Mix 27.6 mL of the CH₃COOH (aq) with 72.4 mL of the NaCH₃COO (aq)

c. Mix 72.4 mL of the CH₃COOH (aq) with 27.6 mL of the NaCH₃COO (aq)

d. Mix 50.0 mL of the CH₃COOH (aq) with 50.0 mL of the NaCH₃COO (aq)

QUESTION 2

If 15.9 grams of NH₄Cl (s) are added to 250.0 mL of 0.350 M NH₃ (aq) solution, what is the resulting pH? Assume no change in volume upon adding the NH₄Cl (s). pK_a = 9.25 for NH₃ (aq).

Select one:

a. 8.55

b. 8.38

c. 9.06

d. 8.72

Answer 1

Given a solution of 0.125 M CH₃COOH (aq) and a solution of 0.150 M NaCH₃COO (aq), how would you prepare 100.0 mL of a buffer solution with a pH of 4.400? pK_a = 4.740 for CH₃COOH (aq).

Solution :-

pH= pka + log ([base]/[acid])

4.40 = 4.74 + log ([base]/[acid])

4.40 – 4.74 = log ([base]/[acid])

-0.34 = log ([base]/[acid])

Antilog [-0.34] = ([base]/[acid])

0.457 = ([base]/[acid])

So we need to check the ratio of the concentration of the base to acid which gives 0.457

Lets find the new molarities of the each solution :-

a. Mix 68.6 mL of the CH₃COOH (aq) with 31.4 mL of the NaCH₃COO (aq)

solution :- New concentrations at total volume

[CH3COOH] = 0.125 M * 68.6 ml / 100 ml = 0.08575 M

[CH3COO-] = 0.150 M * 31.4 ml / 100 ml = 0.0471 M

Ratio = 0.0471 / 0.08575 = 0.459

b. Mix 27.6 mL of the CH₃COOH (aq) with 72.4 mL of the NaCH₃COO (aq)

[CH3COOH] = 0.125 M * 27.6 ml / 100 ml = 0.0345 M

[CH3COO-] = 0.150 M * 72.4 ml / 100 ml = 0.1086 M

Ratio = 0.1086 / 0.0345 = 3.14

c. Mix 72.4 mL of the CH₃COOH (aq) with 27.6 mL of the NaCH₃COO (aq)

Solution

[CH3COOH] = 0.125 M * 72.4 ml / 100 ml = 0.0905 M

[CH3COO-] = 0.150 M * 27.6 ml / 100 ml = 0.0414 M

Ratio = 0.0414 / 0.0905 = 0.457

This solution gives the needed ratio so the answer is option C

QUESTION 2

If 15.9 grams of NH₄Cl (s) are added to 250.0 mL of 0.350 M NH₃ (aq) solution, what is the resulting pH? Assume no change in volume upon adding the NH₄Cl (s). pK_a = 9.25 for NH₃ (aq).

Solution :-

Lets first calculate the moles of the NH4Cl

15.9 g NH4Cl * 1 mol / 53.491 g = 0.29725 mol NH4Cl

Molarity of the NH4Cl = 0.29725 mol /0.250 L = 1.189 M

Now lets calculate the pH

pH= pka + log ([base]/[acid])

    = 9.25 + log [0.350 /1.189]

    = 8.72

So the answer is option ‘d’