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Chapter 16 Test Bank Chang 12e

1. Which of the following compounds is most likely only slightly soluble for the reason given?

a. NaF because fluoride is the anion of a weak acid b. CdS because of the relatively high ionic charges c. CsBr because of the relatively large atomic masses d. BaCl2 because of the stoichiometric relationship between the cation and

anions

Answer: (b) The position of the elements on the periodic table indicates a +2 and -2 charge on the cation and anion respectively. Ions with relatively high ionic charges tend to be only slightly soluble and this compound has ions with the largest charge. LO 1.9; SP 4.4, 6.4; EK 1.C.1; Level Moderate

2. A solution contains Mg(OH)2, Mg2+, OH-, and H3O+ ions. What is the best description of this solution?

a. a solution containing a weak acid b. a solution containing a weak base c. a solution containing a buffer d. a solution containing a common ion

Answer: (b) The presence of the undissociated magnesium hydroxide along with hydroxide ions indicates a solution equilibrium involving a weak base. LO 2.2; SP 7.2; EK 2.A.3; Level: Moderate

3. Calcium phosphate is a sparingly soluble salt in water. Which of the following gives

a plausible explanation of this observation? a. Calcium salts are generally insoluble due to their full s sublevel. b. Calcium minerals are ionic thereby forming substances such as bones and

teeth which do not dissolve. c. The hydration of both calcium and phosphate ions leads to a decrease in the

entropy of the solution. d. Phosphate ions are the anion of the weak triprotic phosphoric acid indicating

an equilibrium state that highly favors reactants.

Answer (c): The relatively high charge on both the cation (2+) and anion (3-) would lead to highly organized hydrated structures. This greatly decreases the entropy of the water and gives an overall decrease in the entropy (-ΔS) of the system. LO 2.15; SP 1.5, 6.4; EK 2.B.3, 5.B.1; Level: Difficult

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4. Magnesium carbonate is dissolved in various solutions. The solubility of the magnesium carbonate (in g/ 100 mL of solution) is given in the table at the right. What is the correct interpretation of these data?

a. Salts do not dissolve as readily in solutions containing alkaline earth cations (Group 2).

b. Salts do not dissolve as readily in solutions that contain solutions with common ions.

c. Salts dissolve more readily dissolve in solutions containing alkali metal cations (Group I).

d. Salts dissolve more readily in solutions containing sodium ions.

Answer: (b) The solubility of slightly soluble salts is reduced by the presence of a common ion. LO 6.23; SP 5.1, 6.1; EK 6.C.3; Level: Moderate

5. The Ksp values for four silver salts are given in the table

at the right. Which aqueous solution would contain the greatest concentration of silver ions?

a. silver chloride b. silver carbonate c. silver bromide d. silver iodide

Answer: (b) AgCl has the largest Ksp value, but only contains one mole of silver ions per mole of salt. However, there are two moles of silver ions per mole of Ag2CO3. Estimating the square-root and cube-root of AgCl and Ag2CO3 respectively, gives a higher concentration of silver ions (approximately 20 times higher) in the silver carbonate. LO 6.21; SP 2.3, 4.4; EK 6.C.3; Level: Difficult

6. Which of the following would constitute a buffer solution?

a. an aqueous solution containing HC2H3O2 and NaC2H3O2 b. an aqueous solution containing HC2H3O2and H2CO3 c. an aqueous solution containing HC2H3O2 and NH3 d. a HC2H3O2 solution and water

Solution Solubility (g/100 mL) Mg(NO3)2 0.041 Na2SO4 0.052 KNO3 0.048 NaCl 0.051

Compound Ksp AgCl 1.6 X 10-10

Ag2CO3 8.1 X 10-12 AgBr 7.7 X 10-13 AgI 8.3 X 10-17

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Answer: (a) A buffer can be made by mixing a weak acid and a salt containing the anion of the weak acid. LO 6.20; SP 6.4; EK 6.C.2;Level: Basic

7. Which of the following reactions would occur when 0.55 M hydrochloric acid is

added to a solution containing carbonic acid (H2CO3) and sodium bicarbonate (NaHCO3)?

a. HCl + H2O H+(aq) + Cl-(aq) b. NaHCO3 + H2O Na+(aq) + HCO3

-(aq) c. HCO3

- H+(aq) + CO32-(aq)

d. HCO3- + H3O+ H2CO3(aq)

Answer: (d) The solution containing carbonic acid and sodium bicarbonate is a buffer. The addition of the hydrochloric acid will cause a reaction between the hydronium ion and the common acid anion (HCO3

-) shifting the equilibrium toward production of H2CO3, thus using some of the added hydronium ions and buffering the pH change. LO 6.20; SP 6.2; EK 6.C.2; Level: Difficult

8. A student prepares two solutions. Solution A is 0.20 M aqueous ammonia and

solution B is 0.20 M aqueous ammonia with 5.0 g of ammonium nitrate added. Which of the following observations would she make when using these solutions during her experiment?

a. The addition of 10.0 mL of hydrochloric acid turns both solutions pink. b. The addition of 10.0 mL of hydrochloric acid turns both solutions clear. c. Adding 10.0 mL of hydrochloric acid lowers the pH more in solution A than

adding 10.0 mL of hydrochloric acid to solution B. d. It requires 10.0 mL of hydrochloric acid to completely neutralize solution A

but it only requires 8.0 mL of hydrochloric acid to completely neutralize solution B.

Answer: (c) Solution B is a buffer which would resist changes in pH upon the addition of an acid or a base. Thus the pH would change more drastically in solution A (unbuffered) if the same amount of acid is added to each of the two solutions. (LO 6.20; SP 6.2, 6.4; EK 6.C.2; Level: Moderate

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9. Copper(I) bromide has a Ksp value of 4.2 X 10-8. Which of the following mixtures would form a precipitate?

a. 2.0 X 10-8 M CuNO3 mixed with 2.0 X 10-8 M NaBr b. 4.0 X 10-8 M CuNO3 mixed with 2.0 X 10-8 M NaBr c. 4.0 X 10-6 M CuNO3 mixed with 2.0 X 10-6 M NaBr d. 4.0 X 10-4 M CuNO3 mixed with 2.0 X 10-4 M NaBr

Answer: (d) The reaction quotient, Q = [Products] / [Reactants]. In the case of slightly soluble salts, the [Reactants] term is omitted since the reactant is a solid and does not change in concentration. Thus Q = [Reactants]. When Q > Ksp, the reaction shifts to produce a solid, forming a precipitate and lowering the concentration of products. Q is calculated as [Cu+][NO3

-]. The Q values for a-d are: 4 X 10-16, 8 X 10-16, 8 X 10-12, and 8 X 10-8. Only the last one (d) has a value greater than Ksp. LO 6.4; SP 2.3, 6.1; EK 6.A.3; Level: Moderate

10. Which of the following salts will have the greatest solubility?

a. Copper(I) bromide (CuBr) Ksp = 4.2 X 10-8 b. Calcium carbonate (CaCO3) Ksp = 8.7 X 10-9 c. Zinc sulfide (ZnS) Ksp = 3.0 X 10-23 d. Cadmium sulfide (CdS) Ksp = 8.0 X 10-28

Answer: (b) The solubility will be greatest when the Ksp value is the highest. 4.2 X 10-8 is the highest Ksp value. LO 6.21; SP 2.3; EK 6.C.3; Level: Basic

11. A 0.40 M solution of a weak monobasic base, X–, is found to have an equilibrium

concentration of HX of 2.0 × 10–3 M. What are the Kb value and the range of the pH for the solution?

Kb value pH

range a. 4.0 × 10–

6 23

b. 1.6 × 10–

6 34

c. 1.0 × 10–

5 1112

d. 5.0 × 10–

3 1213

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Answer: (c) Learning Objective: 6.16 Science Practice: 2.2, 6.4 Essential Knowledge: 6.C.1 Difficulty Level: medium

Name Formula pKa value Methanoic acid CH2O2 3.75 Ethanoic acid C2H4O2 4.76 Propanoic acid C3H6O2 4.86 Butanoic acid C4H8O2 4.83

12. Which one of the acids listed in the table will have the lowest concentration of un-ionized acid in solution? Assume that you have equal volumes of equal molar solutions.

a. Methanoic acid b. Ethanoic acid c. Propanoic acid d. Butanoic acid

Answer: (c) Learning Objective: 6.16 Science Practice: 2.2, 6.4 Essential Knowledge: 6.C.1 Difficulty Level: medium

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Name Formula of acid Name of conjugate base

Ka value

Formic acid HCOOH Formate 1.8 × 10–4 Acetic acid CH3COOH Acetate 1.8 × 10–5 Propionic acid

CH3CH2COOH Propionate 1.4 × 10–5

Butyric acid CH3CH2CH2COOH Butyrate 1.5 × 10–5 13. Given in the table above is information on the first four acids of a related family of

organic acids. In a 45.0 mL sample of a 0.05 M solution of each acid, which conjugate base would have the highest concentration?

a. Formate ion b. Acetate ion c. Propionate ion d. Butyrate ion

Answer: (a) Learning Objective: 6.16 Science Practice: 2.2, 6.4 Essential Knowledge: 6.C.1 Difficulty Level: medium

Name Formula of acid Name of conjugate base

Ka value

Formic acid HCOOH Formate 1.8 × 10–4 Acetic acid CH3COOH Acetate 1.8 × 10–5 Propionic acid

CH3CH2COOH Propionate 1.4 × 10–5

Butyric acid CH3CH2CH2COOH Butyrate 1.5 × 10–5 14. A student is conducting an investigation into the kinetics of the reaction between

lead metal and each of the acids above. If the student uses four identical pieces of lead metal and places each one into an equal molar, equal volume sample of each acid, in what order will the lead metal be consumed. Assume that the lead metal in each case is the limiting reagent.

a. formic acid, acetic acid, propionic acid, butyric acid b. formic acid, acetic acid, butyric acid, propionic acid c. butyric acid, propionic acid, acetic acid, formic acid d. propionic acid, butyric acid, acetic acid, formic acid

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Answer: (b) Learning Objective: 4.1, 6.12 Science Practice: 1.4, 4.2, 5.1, 6.4 Essential Knowledge: 4.A.1, 6.C.1 Difficulty Level: medium

15. The diagram above shows the ionization of a weak acid, HY. Calculate the percent

ionization of HY. Oxygen is represented by the red spheres, hydrogen by the small gray spheres. For simplicity, water molecules are omitted from the diagram.

a. 12.5% b. 20.0% c. 33.3% d. 50.0%

Answer: (b) Learning Objective: 6.11 Science Practice: 1.1, 1.4, 2.3 Essential Knowledge: 6.C.1 Difficulty Level: easy

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16. Each of the diagrams above shows the ionization of an acid in water. Which one of

the diagrams correctly shows the ionization of sulfuric acid? Oxygen is represented by the red spheres, hydrogen by the small gray spheres. For simplicity, water molecules are omitted from the diagram.

a. Diagram (a) b. Diagram (b) c. Diagram (c) d. Diagram (d)

Answer: (a) Learning Objective: 6.11 Science Practice: 1.1, 1.4, 2.3 Essential Knowledge: 6.C.1 Difficulty Level: easy

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17. The diagrams above represent solutions of three salts, NaA, NaB, and NaC. Arrange

the three anions (A–, B–, and C–) in order of increasing Kb values.

a. A– > B– > C– b. C– > B– > A– c. B– > A– > C– d. C– > A– > B–

Answer: (d) Learning Objective: 6.11 Science Practice: 1.1, 1.4, 2.3 Essential Knowledge: 6.C.1 Difficulty Level: easy

18. H2X is a weak, diprotic acid. Which of the diagrams above shows the correct

diagram for the ionization of H2X? Water molecules have been omitted for clarity.


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Answer: (b) Learning Objective: 6.11, 6.12 Science Practice: 1.1, 1.4, 2.3, 6.4 Essential Knowledge: 6.C.1 Difficulty Level: easy 19. HX is a weak monoprotic acid with a Ka value of 4.5 × 10–5. If you start with a 0.20

M solution, what is the percent ionization?

a. 1.5% b. 3.0% c. 4.5% d. 6.0%

Answer: (a) Learning Objective: 6.12, 6.16, 6.19 Science Practice: 1.4, 2.2, 2.3, 5.1, 6.4 Essential Knowledge: 6.C.1, 6.C.2 Difficulty Level: hard 20. HX is a weak monoprotic acid. If you start with a 0.250 M solution and HX ionizes

20.0%, what is the Ka value?

a. 1.25 × 10–2 b. 2.50 × 10–2 c. 3.75 × 10–2 d. 5.00 × 10–2

Answer: (a) Learning Objective: 6.12, 6.16, 6.19 Science Practice: 1.4, 2.2, 2.3, 5.1, 6.4 Essential Knowledge: 6.C.1, 6.C.2 Difficulty Level: hard

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Use the figures and information below to answer questions 21 through 24.

A 30.0 mL sample of a weak base, B, is titrated with HCl and the pH of the resulting solution is measured with a pH meter and graphed as a function of the volume of 0.100 M HCl added, as shown above on the left. The diagrams on the right represent solutions at various stages in the titration of the weak base, B, with HCl. Answer choices A, B, C, and D correspond to these four diagrams. For simplicity, water molecules and Cl– ions have been omitted. 21. Which diagram corresponds to point 1 on the titration curve?


Answer: (c) Learning Objective: 6.13 Science Practice: 5.1, 6.4 Essential Knowledge: 1.E.2 Difficulty Level: medium

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22. Which diagram corresponds to point 2 on the titration curve?


Answer: (a) Learning Objective: 6.13 Science Practice: 5.1, 6.4 Essential Knowledge: 1.E.2 Difficulty Level: medium 23. Which diagram corresponds to point 3 on the titration curve?


Answer: (d) Learning Objective: 6.13 Science Practice: 5.1, 6.4 Essential Knowledge: 1.E.2 Difficulty Level: medium 24. Which diagram corresponds to point 4 on the titration curve?


Answer: (b) Learning Objective: 6.13 Science Practice: 5.1, 6.4 Essential Knowledge: 1.E.2 Difficulty Level: medium

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Use the figures and information below to answer questions 25 through 27.

A 30.0 mL sample of a weak acid, HA, is titrated with NaOH and the pH of the resulting solution is measured with a pH meter and graphed as a function of the volume of 0.100 M NaOH added, as shown above on the left. The diagrams on the right represent solutions at various stages in the titration of the weak acid, HA, with NaOH. Answer choices A, B, C, and D correspond to these four diagrams. For simplicity, water molecules and Cl– ions have been omitted. 25. Which diagram corresponds to point 2 on the titration curve?


Answer: (d) Learning Objective: 6.13 Science Practice: 5.1, 6.4 Essential Knowledge: 1.E.2 Difficulty Level: medium

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26. Which diagram corresponds to point 3 on the titration curve?


Answer: (b) Learning Objective: 6.13 Science Practice: 5.1, 6.4 Essential Knowledge: 1.E.2 Difficulty Level: medium 27. What is the approximate Ka value for the weak acid, HA?

a. 8 × 10–2 b. 8 × 10–3 c. 8 × 10–4 d. 8 × 10–5


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28. Which of the diagrams above represents the most effective buffer system? The

hydrated proton is shown as a hydronium ion and water molecules are omitted for simplicity.



Acid Ka value Acetic acid (HC2H3O2)

1.8×10–5

Ammonium (NH4+) 5.6×10–10

29. A student is asked to prepare a buffer solution with a pH of 4.75. The student is

give equal molar solutions and plans on mixing equal volumes of two solutions to make the buffer. Given the information in the box above, of which set of solutions is the correct set to make the desired buffer solution?

a. HC2H3O2 and NaC2H3O2 b. NH4Cl and HCl c. NH4Cl and NH3 d. NH3 and NaOH

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Answer: (a) Learning Objective: 6.18 Science Practice: 2.3, 4.2, 6.4 Essential Knowledge: 6.C.2 Difficulty Level: medium 30. An insoluble compound, M2X is tested to determine its Ksp value. It is found that

the molar solubility of M2X is 2.0 × 10–4 mol L–1. What is the Ksp value for M2X?

a. 8.0 × 10–8 b. 3.2 × 10–11 c. 4.0 × 10–12 d. 8.0 × 10–12

Answer: (b) Learning Objective: 6.23 Science Practice: 5.1, 6.4 Essential Knowledge: 6.C.3 Difficulty Level: hard 31. An unknown compound is determined to have a ΔHsoln of -74.84 kJ/mol in water at 25 oC.

Which of the following correctly describes the likely solubility of this compound in terms of both enthalpic and entropic changes?

a. Nearly insoluble, since ΔHsoln and ΔS are negative. A decrease in enthalpy and entropy both favor formation of a solution.

b. Very soluble, since ΔHsoln and ΔS are negative. A decrease in enthalpy and entropy both favor formation of a solution.

c. Nearly insoluble, since ΔHsoln is negative and ΔS is positive. A decrease in enthalpy and an increase in entropy both favor formation of a solution.

d. Very soluble, since ΔHsoln is negative and ΔS is positive. A decrease in enthalpy and an increase in entropy both favor formation of a solution.

Answer: (d) Learning Objective: 6.24 Science Practice: 1.4, 7.1 Essential Knowledge: 6.C.3 Difficulty Level: hard

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Free Response 1:

A student titrates 25.0 mL of an unknown base with 0.10 M HCl. During the titration the pH is monitored and the collected data is recorded. These data are shown in the table at the right.

a. Use the information provided to draw a titration curve showing the pH as a function of the volume of added HCl. Be certain to label your axes.

b. Identify the equivalence point on your graph and justify your selection of this particular point.

c. Use the data to determine the Kb value for the weak base. Be certain to show the mathematical steps you take to arrive at the answer. Report your final answer to the correct number of significant digits.

d. The student has three indicators that she could use for this experiment. The indicators (with their endpoints) are: Bromophenol Blue (3.0 – 4.6), Methyl Red (4.2 – 6.3), and phenolphthalein (8.3 – 10.0). Which indicator would be appropriate for this titration? Justify your selection.

e. Determine the (i) molarity and the (ii) % ionization of the original weak base solution (before titrating). Report your answers to the correct number of significant digits.

Volume HCl added (mL) pH

0.0 11.13 5.0 9.86 10.0 9.44 12.5 9.26 15.0 9.08 20.0 8.66 22.0 8.39 24.0 7.88 25.0 5.28 26.0 2.70 28.0 2.22 30.0 2.00 35.0 1.70 37.5 1.61 40.0 1.52 45.0 1.40 50.0 1.30

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Scoring Rubric:

a. Student graphs should resemble the example at right. One point is awarded for correctly graphing the data and one point for correctly and completely labelling both x- and y- axes with units. Subtract one point for graphing volume as a function of pH. Two points total.

b. The equivalence point is shown on the graph. This point is chosen because it is the midpoint of where the pH changes rapidly. This rapid change in pH shows that the weak base has been completely neutralized with neither the hydronium ions (H3O+) nor base molecules (NH3 in this case) being in excess. Therefore, the addition of any more HCl rapidly lowers the pH of the solution. One point is awarded for correctly identifying the equivalence point and one point for providing a correct justification. To gain the justification point the student must include the reasoning behind the rapid change in pH. It is NOT sufficient to simply state that the equivalence point is the mid-point of the rapidly changing pH. Two points total.

c. Since the initial concentration of the unknown base is not given, we will have to use the point which is half-way to equivalence (half-equivalence point). A generic equation for the neutralization reaction could be represented as:

Base + H+ Base-H+

The conjugate acid (Base+) can react with water. We can represent this reaction as:

Base-H+(aq) + H2O ↔ Base(aq) + H3O+(aq)

We can write the equilibrium expression for this reaction as:

Ka = [Base] [H3O+] .

[Base-H+]

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At the half-equivalence point (12.5 mL), half of the base has been neutralized meaning that [Base] = [Base-H+], simplifying this expression to:

Ka = [H3O+]

The [H3O+] can be calculated from the pH when 12.5 mL of acid have been added. And [H3O+] = 10-9.26 = 5.50 X 10-10. The question asks for the Kb value, so we need to calculate that from the equation where Ka X Kb = Kw. (5.50 X 10-10) (Kb) = 1 X 10-14. Kb = 1.8 X 10-5. We could simplify the calculations by taking a few short cuts if we know that pH X pOH = 14. Then at half-equivalence point pOH = 14 – 9.26 = 4.74. Kb at this point = [OH-] so Kb = 10-4.74 = 1.8 X 10-5. While this simplifies the calculations, I don’t feel it lends itself as much to understanding the underlying equilibrium. Additionally, the student can simply state that one-half equivalence point pH = pKa and solve it that way. Again the chemical principles are not illustrated by this solution, but it is certainly valid and with time being a factor, a normal way for the student to proceed. Two points are awarded for showing the correct mathematical steps to determine the Kb value. While student solutions may vary significantly, generally there will be one point given for using the pH to determine concentrations and one point for using the concentrations to determine the Kb value. NOTE: it is NOT necessary to show the equilibrium reaction equations to obtain the point. One point is awarded for the correct answer. Subtract a total of one point for any math errors. Also subtract one point for reporting the final answer using significant digits other than 2 ± 1 (i.e. other than 1-3). Three points total.

d. Methyl red would be the appropriate indicator. A useful indicator changes color at the equivalence point. That is you want to match the end point (color change) to the equivalence point (neutralization). The equivalence point occurs in the 4-7 pH range. Methyl red is the only indicator given that changes color in this range. One point for identifying methyl red as the appropriate indicator along with the correct justification. One point total.

e. (i) The molarity can be calculated from the Kb expression. In the reaction

Base + H2O(l) ↔ Base-H+(aq) + OH-(aq)

The Kb = [Base-H+] [OH-] / [Base]. We also know that [Base-H+] = [OH-]. We can calculate the [OH-] from the pH where pOH = 14 – 11.13 = 2.87. [OH-] = 10-2.87 = 1.35 X 10-3. The expression becomes:

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1.8 X 10-5 = [1.35 X 10-3][1.35 X 10-3] .

[Base]

Solving this for [Base] we calculate Molarity = 0.10 M (ii) The % ionization is calculated as [OH-] / [Base] X 100%

% Ionization = 1.35 X 10-3 X 100% = 1.4%

(0.100) One point is awarded for the correct calculation of the molarity and one point is awarded for the correct calculation of percent ionization. Subtract one point for any one math error UNLESS a point for a math error was subtracted previously. Subtract one point for the reporting the final answers using significant digits other than 2 ± 1 UNLESS a point was subtracted in part ‘c’ for an incorrect number of significant digits. Two points total.

LOs: 1.4, 1.20, 6.5, 6.6, 6.12, 6.13, 6.16, 6.19; SP: 1.1, 1.4, 2.1, 2.2, 4.1, 4.4, 5.3, 6.1, 6.2; EK: 1.A.3, 1.E.2, 6.A.3, 6.C.1, 6.C.2 Levels: Basic, Moderate, Difficult

Free Response 2:

A student is asked to determine the identity of the base in the bottle shown. The bottle must contain one of the following bases. Formula Ksp value Formula Ksp value Be(OH)2 6.92 × 10–22 Cu(OH)2 4.8 × 10–20 Cd(OH)2 7.2 × 10–15 Fe(OH)2 4.9 × 10–17 Co(OH)2 5.9 × 10–15 Ni(OH)2 5.5 × 10–16

The student makes a saturated solution of the unknown base and uses a pH meter to determine the pH of the resulting solution, which was 7.66.

(a) Determine the [OH–] in the solution.

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(b) What is the identity of the unknown base? Answers:

(a) pOH + pH = 14 pOH = 14 – 7.66 = 6.34 6.34 710 [ ] [ ] 10 [ ] 4.6 10pOH OH OH OH− − − − − −= ⇒ = ⇒ = ×

(b) M(OH)2 Δ M2+ + 2OH– Ksp = [M2+][OH–]2 Ksp = [2.3 × 10–7][4.6 × 10–7]2 = 4.8 × 10–20 M(OH)2 is Cu(OH)2

Learning Objective: 6.23 Science Practice: 5.1, 6.4 Essential Knowledge: 6.C.3 Difficulty Level: hard Free Response 3: An analytical chemist is asked to determine whether or not a sample of a weak organic acid, represented by B–, contains any inert impurities. She decides the first test will involve a 0.124 M solution. Two tables are shown below, one listing properties of imidazole and the second listing the results of the investigation.

Table One: Properties of Imidazole Molar mass 68.077 g mol–1 Appearance white or pale yellow solid pKb value 8.75 Solubility in water Soluble Density 1.23 g/cm3, solid

Table Two: Properties of B–

Mass of base used 2.754 g Volume of flask used to make solution 250.00 mL Amount of solution tested 50.00 mL pH of solution tested 9.137

(a) Calculate the molarity of the solution that was made. (b) If the sample contained inert impurities, how would that affect the pH of the

solution? Explain your reasoning. (c) Does the sample contain an inert impurity? Justify your answer with

calculations.

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Answer:

(a)

1 mol2.754 g × 68.077 g = 0.1618

250.00 mLM

(b) If the sample contained an inert impurity, then the mass that was measured would be too high for the amount of B– in it (since it would also have the mass of the impurity in it). This would make the molarity appear too high resulting in too high of a pH.

(c) Yes it does. R B– + HOH HB + OH– I 0.1618 0 0 C –X

+X +X

E ≈ 0.1618

X

X

9

29 5

5

10 1.78 10

1.78 10 1.696 100.1618

log1.696 10 4.771 9.229

bpKbK

X X OH

pOH pH

− −

− − −

−

= = ×

= × ⇒ = × =

= − × = ⇒ =

Since the measure pH does not equal the calculated pH from the apparent molarity made up, the solution must contain an inert impurity.

Learning Objective: 6.12, 6.16, 6.19 Science Practice: 1.4, 2.2, 2.3, 5.1, 6.4 Essential Knowledge: 6.C.1, 6.C.2 Difficulty Level: hard

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