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001 6.0 points For a given triprotic acid H 3 A, which of its pKa values usually will be the largest?
- Not enough information is given.
- pKa
- pKa
- pKa3 correct
- pKa
- pKa
Explanation: Typically, for polyprotic acids each succes- sive deprotonation event is less spontaneous than the preceding event, and so has a smaller Ka and larger pKa associated with it. Conse- quently, the smallest Ka and largest pKa will correspond to the final deprotonation, in this case the third.
002 6.0 points How many equations are needed to fully determine an aqueous system initially con- taining the strong electrolyte NH 4 NO 2?
- 6 correct
- 4
- 7
- 2
Explanation: The system would have an unknown value for [H+], [OH−], [NH+ 4 ], [NH 3 ], [NO− 2 ], and [HNO 2 ].
003 6.0 points At slightly below room temperature, the
solubility product constant for Zn(OH) 2 is
2 × 10 −^17. What is the molar solubility of Zn(OH) 2 in water at this temperature?
- 2 × 10 −^6 M
- 9 × 10 −^7 M
- 0 × 10 −^6 M correct
- 0 × 10 −^3 M
- 8 × 10 −^9 M
Explanation:
004 6.0 points Which of the following pairs of solutions would result in a buffer upon mixing?
- 71 mL of 0.6 M nitric acid; 112 mL of 0 .3 M KNO 3
- 2 L of 0.4 M methylamine; 1 L of 0.9 M HCl
- 1 .1 L of 0.9 M chlorous acid; 2.2 L of 0 .3 M RbOH correct
- 3 .2 L of 0.6 M NaCl; 4 .3 L of 0.3 M KClO 4
- 1 L of 0.3 M HCl; 2 L of 0.2 M CsOH
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.
005 6.0 points For a solution labeled “0.10 M H 3 PO 4 (aq),”
- [PO^34 − ] = 0.10 M.
- [H 2 PO− 4 ] is greater than 0.10 M.
- [H+] = 0.10 M.
- [H+] is less than 0.10 M. correct
- [H+] = 0.30 M.
Explanation:
006 6.0 points Consider five generic acids (HA, HB, HC, HD, and HE) that have the following ionization constants.
Ionization Acid Constant Ka value HA 3. 6 × 10 −^3 HB 8. 3 × 10 −^4 HC 2. 6 × 10 −^6 HD 9. 3 × 10 −^6 HE 7. 3 × 10 −^7
Which of the following anions will be the WEAKEST base?
- D−
- B−
- A−^ correct
- C−
- E−
Explanation: The larger the Ka, the greater the dissoci- ation and the stronger the acid. In general, a conjugate base is opposite in strength from its parent acid strength. HA is the strongest acid listed (largest Ka), therefore A−^ is the weakest base.
007 6.0 points What is the pH of 1 M Na 3 A if pKa1 = 2, pKa2 = 6, and pKa3 = 10 for the triprotic acid H 3 A?
- 12 correct
Explanation:
008 6.0 points The solubility product constant of Ag 2 CrO 4 is 9. 0 × 10 −^12. What is the molar solubility of Ag 2 CrO 4 in a solution in which the silver ion concentration is maintained at 2. 0 × 10 −^3 M by addition of AgNO 3?
- 3 × 10 −^6 correct
- 0 × 10 −^3
- 6 × 10 −^7
- 5 × 10 −^9
- 3 × 10 −^4 Explanation:
009 6.0 points Consider the aqueous solution initially with H 3 PO 4 , NH 3 , and NH 4 H 2 PO 4. A system of equations is needed to determine the equilib- rium concentrations. Which of the following equations correctly balances the charge for the system?
- [H+] + [NH 4 +] = [OH−] + [H 2 PO 4 −]
- [NH 4 +] = [H 2 PO 4 −] + 2 [HPO 42 −]
- [H+] + [NH 4 +] = [OH−] + [H 2 PO 4 −]
- 2 [HPO 42 −] + 3 [PO 43 −] correct
- [H+] = [OH−]
- [H+] + [NH 4 +] = [OH−] + [H 2 PO 4 −]
- [HPO 42 −] + [PO 43 −] Explanation:
the two compounds, you can expect that they will react in a one-to-one fashion. So our first order of business will be to determine how many moles of each compound we have.
For HClO 4 , we have
30 .0 mL
( 1 L
1000 mL
)( (^0) .200 mol
1 L
= 0.006 mol HClO 4
Likewise, for NaOH, we have
60 .0 mL
( 1 L
1000 mL
)( (^0) .150 mol 1 L
= 0.009 mol NaOH So when HClO 4 and NaOH react, all of the HClO 4 will be consumed (it’s the limiting reagent) and
0 .009 mol − 0 .006 mol = 0.003 mol
will remain. This 0.00300 mol excess of NaOH will determine the pH of this solution. The solution now is
30.0 mL + 60.0 mL = 90 mL
and, since NaOH is a strong base (i.e., it’s completely dissociated), it contains 0.003 mol OH−. [OH−] is then
[OH−] =
0 .003 mol 0 .09 L
= 0.0333333 M
which means that the pOH of this solution is
pOH = −log[OH−] = −log(0.0333333) = 1. 47712
However, we wanted pH. We can use the equa- tion that relates pH to pOH to get pH
pH + pOH = 14 pH + 1.47712 = 14 pH = 12. 5229
014 6.0 points The figure below shows the titration of a monoprotic weak acid with a strong base.
b
b
I
II
Volume of base added
pH
The pH at point I is equal to the and the pH at point II is 7.
- pH of the acid; greater than
- pH of the acid; less than
- pKa of the acid; greater than correct
- pKa of the acid; less than
- pKa of the acid; equal to
Explanation:
015 6.0 points A rather cavalier general chemistry student uses the equation H+^ = (Ka · Ca)^0.^5 whenever he needs to calculate a pH. Which of the solutions described below is the only one for which this equation will yield a pH close to the true value
- A 10−^1 M solution of carbonic acid with pKa 1 = 4 and pKa 2 = 10. correct
- A 10−^1 M solution of nitric acid.
- A 10−^7 M solution of hydrochloric acid.
- A 10−^4 M solution of acetic acid with a pKa = 4.
Explanation: The equation H+^ = (Ka · Ca)^0.^5 is best ap- plied to fairly concentrated solutions of weak acids.
016 6.0 points A buffer solution is made from 10 mL of 0. 1 M acetic acid and 20 mL of 0.2 M sodium acetate. To the solution 5 mL of 0.8 M HNO 3 is added. What is the pH of the resulting solution? Ka for acetic acid is 1. 8 × 10 −^5.
- 51
- 52
- 34
- 096
- 79 correct
Explanation: HNO 3 : (0.8 M)(5 mL) = 4 mmol will react with acetate Neutralization:
HAc ⇀↽ H+^ + Ac− 10 mL 20 mL 0 .1 M 0 .2 M 1 mmol 4 mmol 4 mmol −4 mmol 5 mmol 0 mmol
Equilibrium re-established: 5 mmol 35 mL
= 0.143 mol/L
Ka HAc ⇀↽ H+^ + Ac− 0 .143 M − −
- 143 − x x x
Ka =
x^2 CHA − x Assumption:
x = [H+] =
Ka CHA
=
(1. 8 × 10 −^5 )(0.143)
= 0.00160437 M
pH = −log[H+] = −log(0.00160437) = 2. 7947
017 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.
- 0 × 10 −^1 M
- 0 × 10 −^7 M
- 2 × 10 −^2 M correct
- 0 × 10 −^1 M
- 0 × 10 −^2 M
Explanation:
018 6.0 points Calculate the pH of 0.1 M Na 2 HAsO 4 (aq). pKa1 = 2.25, pKa2 = 6.77, and pKa3 = 11.6.
- 9.18 correct
- None of these
Explanation: Initially the salt dissociates into two Na+ and HAsO^24 − ions. Na+^ is an extremely weak acid and does not affect the equilib- rium. There are three equilibria to consider for the anion but as we start with HAsO^24 − , the second and third dissociations are most important; we use these to calculate the pH:
pH =
(pKa2 + pKa3)
=
019 6.0 points Consider the following Ksp values for lead salts:
