Download Solubility Equilibria-Labrake - Principles of Chemistry I - Homework | CH 302 and more Quizzes Chemistry in PDF only on Docsity!
This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points In an ideal solution of two compounds with indentical intermolecular forces, the solution will have a lower free energy than the unmixed components because
- ∆Hsolution < 0
- ∆Ssolution > 0 correct
- ∆Hsolution > 0
- ∆Gsolution > 0
Explanation: If the intermolecular forces are identi- cal then ∆Hsol = 0. The mixture will have a higher entropy, so ∆S > 0. Since ∆G = ∆H − T ∆S, and ∆H = 0, then ∆G = −T ∆S < 0.
002 10.0 points Which of the following statements about col- ligative properties of solutions is FALSE?
- Osmosis is a colligative property.
- Colligative properties are identical for all solvents correct
- Colligative properties arise from the con- centration of the solute but not the inter- molecular forces of the solute
- The higher the concentration of solute in the solution, the lower the vapor pressure of the solvent.
- Colligative properties assume ideal solu- tions
Explanation: Colligative properties, which include osmo- sis, vapor pressure lowering, melting and boil- ing point elevations, depend only on the num-
ber of solute particles present in solution, not on their properties. However, different solvents have very dif- ferent colligative properties
003 10.0 points Red blood cells contain both Na+^ ions and K+^ ions as well as some water. If we place some red blood cells into beaker full of pure water, what will happen to them?
- nothing
- they will shrivel and collapse
- they will wiggle around rapidly
- they will swell and burst correct
Explanation: This illustrates the process of osmosis. Wa- ter can move through the membrane of the blood cells in either direction, but because there is less water inside the cells than outside in the pure water, there will be a net flow of water inside the cells, so they swell and burst.
004 10.0 points Two aqueous solutions are separated by a semipermeable membrane. Solution A = 0.34 M KCl Solution B = 0.64 M KCl Which of the following statements is TRUE?
- For a reverse osmosis process, pressure must be applied to solution A.
- K+^ ions flow from solution B to solution A.
- Pressure must be applied to solution A to create an equilibrium flow of water molecules.
- There is a net flow of H 2 O molecules from solution B to solution A.
- There is a net flow of H 2 O molecules from solution A to solution B. correct
- Cl−^ ions flow from solution B to solution A.
Explanation: The semipermeable membrane is only per- meable to water (the solvent), so only H 2 O can pass from solution A to solution B (or vice versa). As solution A is less concentrated than solution B, H 2 O molecules will sponta- neously travel from solution A to solution B in order to try and make the concentrations of the two solutions equal.
005 10.0 points In the following reaction, 123 g of CO 2 is collected.
H 2 SO 4 + Na 2 CO 3 → Na 2 SO 4 + CO 2 + H 2 O What is the molarity of the sulfuric acid solution if 10.3 L were used in the reaction? The molar mass of H 2 SO 4 is 98.08 g/mol, the molar mass of Na 2 CO 3 is 106.01 g/mol, the molar mass of Na 2 SO 4 is 142.07 g/mol, the molar mass of CO 2 is 44.01 g/mol and the molar mass of H 2 O is 18/01 g/mol.
- 1.78 M
- 0.271 M correct
- 2.65 M
- 0.0145 M
- 0.542 M
- 0.453 M
- 9.52 M
Explanation: Find the number of moles of CO 2 :
(123 g CO 2 ) ×
1 mol CO 2 44 .01 g CO 2 = 2.79482 mol CO 2 Find the number of moles of H 2 SO 4 used:
(2.79482 mol CO 2 ) ×
1 mol H 2 SO 4 1 mol CO 2 = 2.79482 mol H 2 SO 4
Then the molarity of the H 2 SO 4 is
2 .79482 mol H 2 SO 4 10 .3 L
= 0.271342 M H 2 SO 4
006 10.0 points An equilibrium in which processes occur con- tinuously, with NO NET change, is called
- static equilibrium.
- dynamic equilibrium. correct
- hetergeneous equilibrium.
- homogeneous equilibrium.
Explanation: In dynamic equilibrium the forward and re- verse processes of the system continue but be- cause their rates are equal there is no change in the composition of the system.
007 10.0 points What is the net ionic equation for the reaction between aqueous solutions of Na 3 PO 4 and CuSO 4?
- 2 Na+^ + SO^24 − → Na 2 SO 4
- Cu2+^ + PO^34 − → CuPO 4
- No reaction occurs since no precipitate is formed.
- 3 Cu2+^ + 2 PO^34 − → Cu 3 (PO 4 ) 2 correct Explanation: Both copper(II) sulfate and sodium phos- phate are soluble. This is a metathesis reac- tion with the products copper(II) phosphate (Cu 3 (PO 4 ) 2 ) and sodium sulfate (Na 2 SO 4 ). Na 2 SO 4 is soluble but Cu 3 (PO 4 ) 2 is insoluble and will form a precipitate. The formula unit equation is
3 CuSO 4 (aq) + 2 Na 3 PO 4 (aq) → Cu 3 (PO 4 ) 2 (s) + 3 Na 2 SO 4 (aq).
In the total ionic equation soluble com- pounds are written as their ions:
Assume the molar solubility of aluminum carbonate (Al 2 (CO 3 ) 3 ) is represented as x. Which of the following expressions correctly expresses the relationship between the molar solubility of aluminum carbonate and the sol- ubility product constant (Ksp) for this com- pound?
- Ksp = 6 x^5
- Ksp = 108 x^5 correct
- Ksp = 36 x^5
- Ksp = 12 x^3
- Ksp = 54 x^5
- Ksp = 108 x^6
- Ksp = 27 x^4
- Ksp = 5 x^2
Explanation:
013 10.0 points Determine the concentration of strontium ions in saturated solution of strontium sul- fate (SrSO 4 ) if the Ksp is 2. 35 × 10 −^7.
Correct answer: 0.000484768 mol/L.
Explanation: Ksp of SrSO 4 = 2. 35 × 10 −^7 [Sr2+] =?
SrSO 4 (s) ⇀↽ Sr2+(aq) + SO^24 − (aq)
[Sr2+] = [SO^24 − ] = x
Ksp = [Sr2+] [SO^24 − ]
- 35 × 10 −^7 = x^2 x =
2. 35 × 10 −^7
= 0.000484768 mol/L = [Sr2+]
014 10.0 points
The Ksp equation for sodium bicarbonate (NaHCO 3 ) should be written as
- Ksp = [Na+] [H+] [CO− 3 2 ]^2.
- Ksp = [Na+] [H+] [C4+] [O^2 −]^3.
- Ksp = [NaH+2] [CO^23 − ].
- Ksp = [Na+] [H+] [CO^23 − ]. correct Explanation:
015 10.0 points The value of Ksp for SrSO 4 is 2. 8 × 10 −^7. What is the solubility of SrSO 4 in moles per liter?
- 3 × 10 −^4 correct
- 6 × 10 −^7
- 4 × 10 −^7 Explanation: Ksp = 2.8 × 10 −^7
SrSO 4 ⇀↽ Sr2+^ + SO^24 −
Let [Sr2+] = [SO^24 − ] = x.
Ksp = [Sr2+] [SO^24 − ]
- 8 × 10 −^7 = x^2 x =
2. 8 × 10 −^7
= 5. 3 × 10 −^4
So one liter will contain 5. 3 × 10 −^4 moles of each ion; that means it will contain 5. 3 × 10 −^4 moles of the salt.
016 10.0 points Determine the molar solubility of some salt (AB 2 ) if Ksp = 2. 56 × 102.
- 10 M
- 0.005 M
- 4 M correct
- 1 M
5. 0.1 M
Explanation: The equation for Ksp = x(2x)^2 = 4x^3
x =
Ksp
. Plugging in numbers gives
you 3.97 M, or about 4 M.
017 10.0 points BaF 2 (s) is slightly soluble in water at 25◦C. If I put excess BaF 2 (s) into water, and measure the equilibrium [Ba2+] concentration to be 1.06 × 10 −^2 M, what is Ksp?
- between 1 × 10 −^7 and 1 × 10 −^6
- between 1 × 10 −^3 and 1 × 10 −^2
- between 1 × 10 −^4 and 1 × 10 −^3
- between 1 × 10 −^6 and 1 × 10 −^5 correct
- between 1 × 10 −^5 and 1 × 10 −^4
Explanation:
018 10.0 points Rank following salts from least to most solu- ble: BiI Ksp = 7. 7 × 10 −^19 Cd 3 (AsO 4 ) 2 Ksp = 2. 2 × 10 −^33 AlPO 4 Ksp = 9. 8 × 10 −^21 CaSO 4 Ksp = 4. 9 × 10 −^5
- CaSO 4 < Cd 3 (AsO 4 ) 2 < BiI < AlPO 4
- Cd 3 (AsO 4 ) 2 < BiI < AlPO 4 < CaSO 4
- CaSO 4 < BiI < AlPO 4 < Cd 3 (AsO 4 ) 2
- AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 < CaSO 4 correct
- Cd 3 (AsO 4 ) 2 < AlPO 4 < BiI < CaSO 4
Explanation: Molar solubility can be approximated by taking the nth^ root of the Ksp where n is the number of ions in the salt. Doing so results in
approximate molar solubilities of 10−^10 , 10−^7 , 10 −^11 and 10−^3 for bismuth iodide, cadmium arsenate, aluminum phosphate and calcium sulfate, respectively. Arranging these from least to greatest produces: AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 < CaSO 4.
019 10.0 points The compound MX 3 has a molar solubility of 0.00211 M. What is the value of Ksp for MX 3?
- 98 × 10 −^11
- 54 × 10 −^7
- 78 × 10 −^10
- 76 × 10 −^8
- 35 × 10 −^10 correct Explanation: [MX 3 ] = 0.00237 M
MX 3 → M3+^ + 3 X^1 −
Ksp = (x) (3x)^3 = 27x^4 = 3 (0.00211)^4 = 5. 35 × 10 −^10.
020 10.0 points Calculate the solubility product constant for calcium carbonate, given that it has a solubil- ity of 5. 3 × 10 −^5 g/L of water.
- 89 × 10 −^26
- 11 × 10 −^14
- 06 × 10 −^13
- 81 × 10 −^13 correct
- 49 × 10 −^19 Explanation:
021 10.0 points Consider the Ksp values of the following salts
Correct answer: 2. 96429 × 10 −^17 mol/L.
Explanation: [Ag+] = 2. 8 [I−] =?
Ksp = 8. 3 × 10 −^17
AgI(s) ⇀↽ Ag+(aq) + I−(aq)
Ksp = [Ag+] [I−]
[I−] =
Ksp [Ag+]
=
8. 3 × 10 −^17
= 2. 96429 × 10 −^17 mol/L
026 10.0 points What would be the molar solubility of Li 3 PO 4 (Ksp = 2. 37 × 10 −^4 ) in a 1 M LiCl solution?
- 24 × 10 −^1
- 37 × 10 −^4 correct
- 44 × 10 −^2
- 54 × 10 −^2
Explanation:
(^ molar^ solubility^ = Ksp [Li+]^3
2. 37 × 10 −^4
(1)^3
= 2. 37 × 10 −^4
