Buffer
Solutions
Titration -
Conceptual
Titration -
Computational
Solubility
Equilibria
Entropy
100

Which of the following solutions would make the best buffer system?

a. 0.10 M NaCl and 0.10 M HCl

b. 1.0 M CH3CH2COOH and 1.0 M NaCH3CH2COO

c. 0.10 M NaOH and 0.10 HCl

d. 1.0 M HNO3 and 1.0 NaNO3

e. 1.0 M HCN and 1.0 M NaCl

b. 1.0 M CH3CH2COOH and 1.0 M NaCH3CH2COO

100

What determines pH after the equivalence point of a weak base/strong acid titration?

The concentration of excess strong acid determines pH after the equivalence point.

100

What is the pH of a 2.0 L solution of 0.30 M HCN (Ka = 6.2 x 10-10) after addition of 722.0 mL 0.90 M NaOH?

pH = 12.26

100

The solubility of barium carbonate is 0.0024 g/100 mL at 20 ᵒC. What is the solubility product, Ksp, for barium carbonate at this temperature? MM of BaCO3 is 197.34 g/mol.

Ksp = 1.48 x 10-8

100

The second law of thermodynamics tells us…

a. the entropy of the universe is constant

b. entropy is neither created nor destroyed

c. the universe proceeds toward a state of lower entropy

d. the universe proceeds toward a state of higher entropy

e. the entropy of the universe may increase or decrease

d. the universe proceeds toward a state of higher entropy

200

What is the pKa for the acid HA if a solution of 0.65 M HA and 0.85 M NaA has a pH of 4.75?

pKa = 4.63

200

What is the equivalence point of a titration, and how does the pH at the equivalence point differ between types of titrations?

Equivalence point: when the amount of titrant added is exactly enough to fully neutralize the analyte. pH at equivalence is neutral for strong acid/strong base or strong base/strong acid titration, >7 for a weak acid/strong base titration, and <7 for a weak base/strong acid titration.

200

What is the pH when 20.0 mL of 0.20 M HCl is added to 60.0 mL of 0.133 M hydrazine (N2H4, Kb = 1.3 x 10-6)?

pH = 8.11

200

Calculate the solubility of strontium fluoride, SrF2, in pure water. Ksp = 2.6 x 10-9

s = 8.7 x 10-4 M

200

Which of the following is true?

a. the reverse of a spontaneous reaction is also spontaneous

b. a non-spontaneous process can be caused to occur

c. a highly spontaneous process always occurs rapidly

d. a spontaneous process moves a system out of equilibrium

e. a spontaneous process always occurs without any external influence

b. a non-spontaneous process can be caused to occur

300

Which of the following solutions has the greatest buffering capacity?

a. 0.528 M CH3COOH and 0.197 M NaCH3COO

b. 0.365 M CH3COOH and 0.527 M NaCH3COO

c. 0.817 M CH3COOH and 0.862 M NaCH3COO

d. 0.124 M CH3COOH and 0.131 M NaCH3COO

c. 0.817 M CH3COOH and 0.862 M NaCH3COO

300

You are given the volume and concentration of a monoprotic weak acid, and the volume and concentration of the strong base added to it. How can you select the best strategy to calculate pH?

The most straightforward way is to determine how many moles of weak acid are present, how many moles of base are required to fully neutralize the acid, and compare this to the moles of base added.

300

400.0 mL of 0.640 M formic acid (CH2O2, Ka = 1.8 x 10-4) is titrated with 0.10 M NaOH. What is the pH of the solution after 310.0 mL of base has been added?

pH = 2.88

300

Calculate the solubility of magnesium sulfate, MgSO4, when placed into a 0.10 M MgCl2 solution. Ksp = 5.9 x 10-3

s = 4.2 x 10-2 M

300

Under which of the following conditions would one mole of argon (Ar) have the lowest entropy, S?

a. 14 ᵒC and 5 L

b. 87 ᵒC and 15 L

c. 87 ᵒC and 5 L

d. 14 ᵒC and 15 L

a. 14 ᵒC and 5 L

400

The pH of blood is 7.35. It is maintained in part by the buffer system composed of carbonic acid (H2CO3) and the bicarbonate (HCO3-) ion. What is the ratio of [bicarbonate]/[carbonic acid] at this pH? The Ka of carbonic acid is 4.2 x 10-7.

[A-]/[HA] = 9.397

400

What are the four main strategies to solve for pH over the course of a titration?

Starting point: ICE table for dissociation of weak acid or base

Buffer region: calculate change in conjugate pair concentrations --> Henderson-Hasselbalch

Equivalence point: ICE table for dissociation of conjugate acid or base

Overshoot region: calculate concentration of excess titrant --> -log for pH

400

A 50.0 mL sample of an aqueous H2CO3 solution is titrated with 0.345 M NaOH. The second equivalence point is reached with 83.5 mL of base. What was the initial concentration of the H2CO3?

[H2CO3]initial = 0.288 M

400

Which of the following salts would be most soluble in a solution with a pH of 4?

a. NH4F

b. NaCl

c. LiNO3

d. CaCO3

e. KBr

d. CaCO3

400

In which of the following pairs will the entropy of the first substance be greater than that of the second?

a. 1 mole of F2(g); 1 mole of Cl2(g)

b. 1 mole of I2(s); 1 mole of I2(g)

c. 1 mole of CaCO3(s); 1 mole of CaO(s) plus 1 mole of CO2(g)

d. 1 mole of H2(g) at 25ᵒ C; 1 mole of H2(g) at 50ᵒ C

e. 1 mole of O3(g); 1 mole of O2(g)

e. 1 mole of O3(g); 1 mole of O2(g)

500

Buffer solutions with the component concentrations shown below were prepared. Which of them should have the lowest pH?

a. [CH3COOH] = 0.25 M, [CH3COO-] = 0.25 M            

b. [CH3COOH] = 0.75 M, [CH3COO-] = 0.75 M            

c. [CH3COOH] = 0.75 M, [CH3COO-] = 0.25 M            

d. [CH3COOH] = 0.25 M, [CH3COO-] = 0.75 M

e. [CH3COOH] = 1.00 M, [CH3COO-] = 1.00 M

c. [CH3COOH] = 0.75 M, [CH3COO-] = 0.25 M

500

At what point in a titration does pH = pKa, and why is this true?

pH = pKa at the half equivalence point, because half of the analyte has been converted to its conjugate form and the ratio inside the log of the Henderson-Hasselbalch equation reduces to 1, making the log equal to 0 (pH = pKa + log(1) --> pH = pKa).

500

Calculate the pH of the solution resulting from the addition of 381.0 mL 0.90 M Ca(OH)2 to 600.0 mL of a 0.340 M citric acid (C6H8O7, pKa1 = 3.13, pKa2 = 4.76, pKa3 = 6.4) solution.

pH = 12.88

500

The Ksp for magnesium palmitate (Mg(C16H31O2)2, MM = 535.1 g/mol) is 4.8 × 10-12 at 50C and 3.3 × 10-12 at 25C. A 20 mL solution saturated with magnesium palmitate at 50C is cooled to 25C. How much of the magnesium palmitate, in mg, precipitates out of solution?

0.131 mg

500

Calculate ΔSᵒ for the reaction
SiCl4(g) + 2Mg(s) <---> 2MgCl2(s) + Si(s)

Substance     ||     Sᵒ (J/K*mol)

SiCl4(g)     ||     330.73

Mg(s)     ||     32.68

MgCl2(s)     ||     89.62

Si(s)     ||     18.83

ΔSᵒ = -198.02 J/K