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This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 6.0 points What is the concentration of SO^24 − in 2.0 M H 2 SO 4? Ka1 is strong and Ka2 = 1. 2 × 10 −^2.
- 2 × 10 −^2 M correct
- 0 × 10 −^1 M
- 0 × 10 −^1 M
- 0 × 10 −^2 M
- 0 × 10 −^7 M
Explanation:
002 6.0 points If 5.00 mL of 0.500 M NaOH is added to 100 mL of 0.0200 M HCl, what is the pH of the final solution?
- 11.7 correct
Explanation: VNaOH = 5.00 mL MNaOH = 0.500 M VHCl = 100 mL MHCl = 0.0200 M Find number of moles present:
mol of NaOH = (0.005)(0.5 mol/L)
= 0.0025 mol
mol of HCl = (0.1 L)(0.02 mol/L)
= 0.002 mol
Write the equation for neutralization with- out spectator ions: H+^ + OH−^ −→ H 2 O
Initial 0. 002 0. 0025 Change − 0. 002 − 0. 002 Final 0 0. 0005
[OH−] = 0.0005 moles but REMEMBER to account for the new volume!!! 105 mL contains 0.0005 moles of OH−, so 1 L will contain( 1 L 0 .105 L
(0.0005 mol/L)
= 0.00476 M OH−
pOH = −log(0.00476) = 2. 32222 pH = 14 − pOH = 14 − 2 .32222 = 11. 68
003 6.0 points Which equation does not require any assump- tions in order to provide an accurate solution?
- [H+] = Ca
- [OH−] = Kb
Cb Ca
- All require assumptions. correct
- [OH−] = (Kb Cb)^1 /^2
- [H+] = (Ka Ca)^1 /^2 Explanation: All of these equation assume, at the very least, that autoprotolysis is negligible.
004 6.0 points How many equations must be solved in order to accurately calculate all of the unknown concentrations formed at equilibrium when solid NaHSO 3 and solid CaSO 3 are dissolved in water?
- 7 correct
- 6
- 4
- 8
Explanation: There are 7 unknowns: H 2 SO 3 , HSO− 3 ,
SO^23 − , Na+, Ca2+, H+, OH−.
005 6.0 points What is the [OH−] of 8 × 10 −^9 M NaOH?
- 608 × 10 −^8 M
- 8 × 10 −^9 M
- 15096 × 10 −^7 M
- 0408 × 10 −^7 M correct
- 0016 × 10 −^7 M
Explanation: NaOH is a strong base, so it completely dissociates. This adds OH−^ ions to solution, which initially will violate the equilibrium ex- pression for the dissociation of water. Then some of the OH−^ will be neutralized with the H+^ until the expression is obeyed again.
Kw = [H+][OH−] 10 −^14 = (10−^7 M − x) · (10−^7 M + 8 × 10 −^9 M − x) = ([OH−] − 8 × 10 −^9 M)[OH−] [OH−] = 1. 0408 × 10 −^7 M
006 6.0 points What will be the pH at the equiva- lence( point of a titration of 0.4 M HIO Ka = 2. 0 × 10 −^11
with 0.4 M NaOH?
- 2
- 12 correct
Explanation: Equal volumes of the titrant and analyte will be used to reach the equivalence point. Regardless of the starting volume, at the equivalence point the volume will be twice the starting value and all of the hypoiodous acid (HA) will have been converted to hy- poiodite (A−). The solution will be 0.2 M hypoiodite.
Kb =
Kw Ka
10 −^14
2. 0 × 10 −^11
= 5. 0 × 10 −^4
[OH−] = (Kb Cb)^1 /^2
=
5. 0 × 10 −^4 · 0. 2
= 10−^2
pOH = 2 pH = 12
007 6.0 points Carbonic acid (H 2 CO 3 ) is a diprotic acid with Ka1 = 4. 2 × 10 −^7 and Ka2 = 4. 8 × 10 −^11. The ion product for water is Kw = 1. 0 × 10 −^14. What is the [H 3 O+] concentration in a saturated carbonic acid solution that is 0. molar?
- 5 × 10 −^4 M
- 2 × 10 −^7 M
- 2 × 10 −^4 M correct
- 7 × 10 −^20 M
- 4 × 10 −^2 M Explanation:
008 6.0 points Consider the polyprotic acid H 3 AsO 4 (arsenic acid), which has Ka values, in no particu- lar order, of 10−^7 , 10−^2 , and 10−^12 for the equilibria shown below:
α(H 2 SO 3 ) and α(SO^23 − ) at this pH? For H 2 SO 3 , pKa1 and pKa2 are 1.81 and 6.91, respectively.
- α(H 2 SO 3 ) = 0.085 and α(SO^23 − ) = 0.
- α(H 2 SO 3 ) = 0.415 and α(SO^23 − ) ∼ 0
- α(H 2 SO 3 ) = 0.17 and α(SO^23 − ) ∼ 0 cor- rect
- α(H 2 SO 3 ) = 0.17 and α(SO^23 − ) ∼ 1
- α(H 2 SO 3 ) ∼ 0 and α(SO^23 − ) = 0.
Explanation:
013 6.0 points 100 mL of 0.50 M BaCl 2 is added to 100 mL of 0.50 M HCl. Which of the following equations is the correct charge balance for this system?
- [Ba2+] + [H+] = 3[Cl−]
- [Ba2+] = 2[OH−]
- 2[Ba2+] + [H+] = [OH−] + [Cl−] correct
- 2[Ba2+] + [H+] = 2[OH−] + 2[Cl−]
- 2[Ba2+] + [H+] = 2[OH−] + [Cl−]
- [Ba2+] + [H+] = [OH−] + [Cl−]
- 2[Ba2+] = [Cl−]
Explanation:
014 6.0 points Which of the following pairs of solutions would result in a buffer upon mixing?
- 146 mL of 0.4 M perchloric acid; 23 mL of 0.4 M CsCl
- 3 .2 L of 0.6 M NaCl; 4 .3 L of 0.3 M KClO 4
- 1 .1 L of 0.9 M chlorous acid; 2.2 L of 0 .3 M RbOH correct
- 1 L of 0.3 M HCl; 2 L of 0.2 M CsOH
- 5 L of 0.1 M pyridine; 1 L of 0.5 M HI Explanation: A buffer prepared by a neutralization re- action requires a weak acid mixed with less strong base or a weak base mixed with less strong acid. The only pair of solutions which satisfies this constraint is 1.1 L of 0.9 M chlor- ous acid; 2.2 L of 0.3 M RbOH.
015 6.0 points Which of the following compounds is the MOST soluble (in moles/liter)? Compound CaC 2 O 4 CuCO 3 Ag 2 CrO 4 MnS AgCl
Ksp
1 × 10 −^9
5 × 10 −^10
0 × 10 −^12
3 × 10 −^13
6 × 10 −^10
CuCO 3
Ag 2 CrO 4 correct
MnS
AgCl
CaC 2 O 4 Explanation:
016 6.0 points What is the final pH of a solution containing 100 mL of 0.2 M HX and 300 mL of 0.1 M NaX after 0.01 mol of NaOH is added? The pKa is 3.00.
- 3.60 correct
Explanation: Initially (100 mL) (0.2 M) = 20 mmol HA (300 mL) (0.1 M) = 30 mmol A− Now add the impurity: 0.01 mol of NaOH = 10 mmol OH−:
HA ⇀↽ H+^ + A−
ini 20 ≈ 0 30 ∆ − 10 + fin 10 40
Thus
A−
HA
= 4 and
pH = 3.0 + log(4) = 3. 60206.
017 6.0 points A solution initially contains 1 M Ag+^ and 0.1 M Pb2+. If NaCl is added to the solu- tion, which cation will precipitate first, and how many orders of magnitude separate the concentrations of Cl−^ ions at which the pre- cipitations start? AgCl: Ksp ≈ 10 −^10 PbCl 2 : Ksp ≈ 10 −^5
- Both will precipitate at the same time.
- Pb2+; 6
- Pb2+; 8
- Ag+; 8 correct
- Ag+; 6
Explanation: For AgCl,
Ksp = [Ag+] [Cl−] 10 −^10 = (1 M) [Cl−] [Cl−] = 10−^10 For PbCl 2 ,
Ksp = [Pb2+] [Cl−]^2 10 −^5 = (0.1 M) [Cl−]^2 [Cl−] = 10−^2 So Ag+^ precipitates first, and 8 orders of magnitude separate the concentrations which cause precipitation.
018 6.0 points For a weak diprotic acid system (H 2 A) which of the following statements are true? I) H 2 A and A^2 −^ form a buffer. II) Ka1 > Ka III) [A^2 −] > [HA−] in acidic solution IV) HA−^ is amphiprotic.
- II only
- I and IV only
- II and IV only correct
- I and III only Explanation: H 2 A and HA−^ or HA−^ and A^2 −^ will form buffers, but H 2 A and A^2 −^ will not form a buffer. For polyprotic acids, protons become harder to remove (dissociate) as the ion be- comes more charged. Therefore Ka1 > Ka for this system. In acidic solution (access pro- tons) all ions will be protonated. HA−^ is amphiprotic because it can accept a proton or it can donate a proton.
019 6.0 points Calculate the pH of 0.586 M NaH 2 AsO 4 (aq). pKa1 = 2.25, pKa2 = 6.77, and pKa3 = 11.6.
- 51 correct
- None of these
Explanation:
