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Version 171 – Quiz 3 – laude – (51155) 1
This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 5.0 points What is the pH of an aqueous solution that is 0.12 M C 6 H 5 NH 2 (Kb = 4. 3 × 10 −^10 ) and 0.018 M C 6 H 5 NH 3 Cl?
- 5.46 correct
Explanation:
002 5.0 points A buffer solution contains 0.25 M NaNO 2 (aq) and 0.80 M HNO 2 (aq) (pKa = 3.37). What is the pH after 0.10 mol HBr is added to 1.00 L of this buffer?
- 2.59 correct
Explanation:
003 5.0 points Which solution has the highest pH?
- 0.1 M C 6 H 5 NH 3 Cl, Kb C 6 H 5 NH 2 = 4. 2 × 10 −^10
- 0.1 M NH 4 Cl, Kb NH 3 = 1. 8 × 10 −^5
- 0.1 M NH 2 (OH) 2 Cl, Kb NH 2 OH = 6. 6 × 10 −^9 4. 0.1 M CH 3 NH 3 Cl, Kb CH 3 NH 2 = 5. 0 × 10 −^4 5. 0.1 M NaCl, Kb ≈ zero correct Explanation:
004 5.0 points 600 mL of a 0.6 M solution of HClO is titrated with a solution of 0.9 M NaOH. What is the pH after adding 400 mL of the NaOH solu- tion? The ionization constant for HClO is
6 × 10 −^8.
10.5 correct
Explanation:
005 5.0 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
Version 171 – Quiz 3 – laude – (51155) 2
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
006 5.0 points What is the pH at the half-stoichiometric point for the titration of 0.88 M HNO 2 (aq) with 0.10 M KOH(aq)? For HNO 2 , Ka =
3 × 10 −^4.
3.37 correct
Explanation:
007 5.0 points Which of the following solutions will produce a buffer? I) 20 mL of 0.5 M (CH 3 ) 3 NHCl + 50 mL of 0.1 M (CH 3 ) 3 N II) 20 mL of 0.5 M HNO 2 + 50 mL of 0.1 M NaOH III) 20 mL of 0.5 M HCl + 50 mL of 0.1 M NH 3 IV) 20 mL of 0.5 M HClO 2 + 50 mL of 0. M CH 3 COOH
V) 20 mL of 0.5 M NH 4 Cl + 50 mL of 0. M NaOH
- I, II, and V only correct
- I, II, IV, and V only
- II and IV only
- II only
- I, II, III, and V only Explanation: A buffer contains a weak acid or weak base, plus the salt of that weak acid or base; or, a mixture which will have this composition after any acid-base reactions occur. You may have to calculate the number of moles of each species to determine the composition after any acid-base reaction.
008 5.0 points The value of Ksp for CuI is 1 × 10 −^12. What is the solubility for CuI?
- 9 × 10 −^4 g/L correct
- 0 × 10 −^12 g/L
- 4 × 10 −^4 g/L
- 0 × 10 −^6 g/L
- 0 × 10 −^6 g/L Explanation: Ksp = 1 × 10 −^12
CuI → Cu+^ + I−
Let x = molar solubility, so
Ksp = [Cu+] [I−] = 1 × 10 −^12 = x^2
x =
(1 × 10 −^12 ) = 1. 0 × 10 −^6 mol/L
molecular weight of CuI = 63 + 127 = 190 g/mol.
solubility = (1. 0 × 10 −^6 mol/L) (190 g/mol) = 1. 9 × 10 −^4 g/L
