BCIT Winter 2013 Chem 0012 Exam #2 Name: ___________________ Attempt all questions in this exam. Read each question carefully and give a complete answer in the space provided. Part marks given for wrong answers with partially correct reasoning/calculations. Constants and equations are attached at the back. Data sheets and tables are provided. Total points = 30 Page 1 of 7
Section I: Multiple choice (15 points total, 1 point each) Choose the BEST answer to the following questions. 1. Which of the following is a suitable term for representing solubility? a. grams b. moles c. molarity d. millilitres per second 2. What is observed when H2SO4 is added to a saturated solution of CaSO4? a. The pH increases b. The [Ca2+] increases c. Bubbles of H2 gas are formed d. Additional CaSO4 precipitates 3. There are 10.0 mL of four 0.10 M solutions. One solution each of Clˉ, Brˉ, IO3ˉ, and BrO3ˉ. Equal moles of AgNO3 are added to each of the four solutions. A precipitate forms in all but one solution. Which solution does not form a precipitate? a. 0.10 M Clˉ b. 0.10 M Brˉ c. 0.10 M IO3ˉ d. 0.10 M BrO3ˉ 4. Which of the following expressions represents [Fe3+] in a saturated Fe(OH)3 solution? a.
b.
c.
d.
5. Which of the following substances has the lowest solubility? a. BaS b. CuS c. FeS d. ZnS Page 2 of 7
6. Solid NaCl is added to a saturated AgCl solution. How have [Ag+] and [Clˉ] changed when equilibrium has been reestablished? [Ag+] [Clˉ] a. increased increased b. increased decreased c. decreased increased d. decreased decreased 7. A compound has a solubility of 7.1x10‐5 M at 25°C. the compound is a. CuS b. AgBr c. CaCO3 d. CaSO4 8. What is the concentration of OHˉ ions in 250 mL of 0.20 M Sr(OH)2? a. 0.050 M b. 0.10 M c. 0.20 M d. 0.40 M 9. Which of the following represents the dissociation equation of a salt in water? a. KCl(s)  K+(aq) + Clˉ(aq) b. Ca2+(aq) + SO42ˉ(aq)  CaSO4(s) c. HCl(aq) + KOH(aq)  KCl(aq) + H2O() d. 2Na(s) + 2H2O()  2NaOH(aq) + H2(g) 10. A 25.0 mL sample of H2SO4(aq) is titrated with 15.5 mL of 0.50 M NaOH. What is the concentration of H2SO4(aq)? a. 0.0078 M b. 0.16 M c. 0.31 M d. 0.62 M Page 3 of 7
11. Which of the following salt solutions will be acidic? a. KClO4 b. NH4Br c. NaHCO3 d. Na2C2O4 12. Which of following 1.0 M solutions will have the greatest electrical conductivity? a. HI b. H2S c. HCN d. H3PO4 13. The relationship is the a. Ka for H3P2O7ˉ b. Kb for H3P2O7ˉ c. Ka for H2P2O7²ˉ d. Kb for H2P2O7²ˉ 14. What is the pH at the transition point for an indicator with a Ka of 2.5x10‐4? a. 2.5x10‐4 b. 3.60 c. 7.00 d. 10.40 15. Consider two buffer solutions; compare the two buffers in regards to their pH and their ability to resist a pH change as acid is added. Buffer A: 0.50M NH4Cl and 0.25M NH3 Buffer B: 0.0050M NH4Cl and 0.0100M NH3 Higher pH Can absorb more acid a. Buffer A Buffer A b. Buffer A Buffer B c. Buffer B Buffer A d. Buffer B Buffer B Page 4 of 7
Section II: Written problems (15 points total). 16. The solubility of Zn(OH)2 is 4.2x10‐6 M. What is the value of Ksp for Zn(OH)2? (3 points) Zn(OH)2(s) Zn2+(aq) + 2 OH‐(aq) [Zn2+] = 4.2x10‐6 M [OH‐] = 2(4.2x10‐6 M) = 8.4x10‐6 M Ksp = [Zn2+][OH‐]2 = (4.2x10‐6)(8.4x10‐6)2 = 3.0x10‐16 17. a. Write the net ionic equation for the reaction between aqueous Pb(NO3)2 and aqueous NaCl. (1 point) Pb2+(aq) + 2Cl‐(aq)  PbCl2(s) b. Show with calculations whether a precipitate will form when 15.0 mL of 0.050 M Pb(NO3)2 is added to 35.0 mL of 0.085 M NaCl. (4 points) Ksp = [Pb2+][Cl‐]2 = 1.2x10‐5 15.0
15.0
35.0
15.0
0.050
35.0
0.0150
0.085
35.0
0.0595
Q = [Pb2+][Cl‐]2 = (0.015)(0.0595)2 = 5.3x10‐5 > Ksp Therefore a precipitate will form Page 5 of 7
18. What is the pH of a 0.50 M H2S solution? (4 points) H2S(aq)
H+(aq)
I
0.50
0
0
C
-x
x
x
E
0.50 - x
x
x
9.1 10
9.1 10
9.1 10
+ HS-(aq)
0.50
0.50
0.50
2.13 10
,
| |
0.5
0.50
0.50
,
0
3.67
2.13 10
19. The following two titrations were done: Titration A: a strong acid was titrated with a strong base Titration B: a weak acid was titrated with a strong base a. How does the pH at the equivalence point of the two titrations compare? Circle the correct answer (1 point) i.
pH titration A < pH titration B ii.
pH titration A = pH titration B iii.
pH titration A > pH titration B b. Explain your answer to part a. (2 points) The titration of a strong acid by a strong base will result in a salt that does not hydrolysis water. So at the equivalence point the pH = 7. The titration of a weak acid by a strong base will result is a salt that is a weak base. So at the equivalence point the pH > 7. Therefore the pH at equivalence point of a weak acid with a strong base titration is greater the pH at equivalence point of a strong acid with a strong base titration Page 6 of 7
Equations and Constants pH = ‐log[H⁺] pOH = ‐log[OHˉ] pH + pOH = 14 at 25°C Kw = 1.0x10‐14 at 25°C Ka Kb = Kw pX = ‐log(X) [H⁺] = 10‐pH [OHˉ] = 10‐pOH X = 10‐pX  
 A 

pH  pK a  log
 HA 
The solution to the quadratic equation ax² + bx +c =0 is √
4
2
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