Unit 3 Flashcards
When 250 mL of 0.36 M Sr(OH)2 are added to 750 mL of water, the resulting ion concentrations are
Select one:
a. [Sr 2+] = 0.12 M and [OH−] =0.12 M
b. [Sr 2+] = 0.12 M and [OH−] =0.24 M
c. [Sr 2+] = 0.090 M and [OH−] =0.090 M
d. [Sr 2+] = 0.090 M and [OH−] =0.180 M
D.
[Sr+2] = 0.090M
[OH-] = 0.18M
The correct answer is: [Sr 2+] = 0.090 M and [OH−] =0.180 M
Which of the following occurs when equal volumes of 0.20 M MgS and 0.20 M ZnSO4
are mixed?
Select one:
a. A precipitate does not form.
b. A precipitate of ZnS forms.
c. A precipitate of MgSO4 forms.
d. Precipitates of MgSO4 and ZnS form.
b
Consider the following equilibrium:
When Br-1(aq) is added to a saturated solution of AgCl:
Select one:
a. more AgCl dissolves and its solubility product constant increases.
b. more AgCl precipitates and its solubility product decreases.
c. more AgCl dissolves and its solubility product remains constant.
d. more AgCl precipitates and its solubility product remains constant.
C.
ksp for AgCl = 1.8 x 10-10
ksp for AgBr = 5.4 x 10-13
As the ksp for AgBr is smaller, it will precipitate AgBr, decreasing the concentration of Ag+. This stresses the system, causes the reaction to shift to the right and increases the solubility of AgCl. The ksp remains constant (only temperature changes ksp)
Which of the following could be used to precipitate both Mg 2+ and Ca 2+ from hard water?
Select one:
a. lithium sulphate
b. sodium phosphate
c. potassium sulphide
d. ammonium chloride
B.
Both ions will precipitate with PO4-3
What is the maximum [Ag+] that can exist in 0.20 M NaBrO3 ?
Select one:
a. 1.1 × 10 −5 M
b. 5. 3 × 10 −5 M
c. 2.7 × 10 −4 M
d. 7.3 × 10 −3 M
C.
Look at what forms when ions are in solution: BrO3- breaks off and pairs up with the Ag+.
ksp of AgBrO3 is 5.3 x 10-5
When equal volumes of 2.0 M Pb(NO3)2 and 2.0 M KCl are mixed,
Select one:
a. a precipitate forms because trial ion product <ksp></ksp>
<p>b. a precipitate forms because trial ion product >ksp</p>
<p>c. a precipitate does not form because trial ion product <ksp>
<p>d. a precipitate does not form because trial ion product >ksp</p></ksp></p>
</ksp>
B.
Pb(NO3)2(aq) + 2KCl(aq) ↔ PbCl2(s) + 2KNO3(aq)
Lead (II) Chloride will precipitate
Consider the following equilibrium:
Which of the following reagents, when added to the equilibrium system, would cause more
CaCO3 to dissolve?
Select one:
a. KNO3 (s)
b. CaCO3 (s)
c. H2C2O4 (s)
d. Na2CO3 (s)
C.
Not B, because adding more solid, will just be adding to the solid pile.
To increase the amount of CaCO3 that dissolved, precipitate either the Ca+2 or the CO3-2 which will stress the system, causing the reaction to shift right and allowing more to dissolve to reduce the stress.
At 25° C, the maximum [Zn2+] that can exist in 0.250 M Na2S is
Select one:
a. 5. 0 × 10 −26 M
b. 2.0 × 10 −25 M
c. 8. 0 × 10 −25 M
d. 4.5 × 10 −13 M
Given molar concentration of Na2S , so look at the NEW compound’s Ksp (ZnS):
For ZnS ksp = 2.0 x 1025
The ion concentrations in 2.00 L of 0.32 M K3PO4 are …
Select one:
a. D
b. E
c. C
d. B
e. A
A.
K3PO4 –> 3K+ + PO43-
[PO43-] = 0.32M
[K+] = 0.32(3) = 0.96M
Which of the following saturated solutions has the greatest [CO32-] ?
Select one:
a. CaCO3
b. MgCO3
c. SrCO3
d. BaCO3
B,
Ksp Values
SrCO3: 5.6 x 10-10
MgCO3: 6.8 x 10-6
CaCO3: 5.0 x 10-9
BaCO3: 2.6 x 10-9
MgCO3 has the highest Ksp value.
In a saturated solution of zinc hydroxide, at 40°C, the [Zn2+] = 1.8 x 10-5 M. The ksp of Zn(OH)2 is
Select one:
a. 6.5 x 10-10
b. 5.8 x 10-15
c. 1.8 x 10-14
d. 2.3 x 10-14
D.
Zn(OH)2 –> Zn2+ + 2OH-
ksp = (1.8 x 10-5)(3.6 x 10-5)2 = 2.3 x 10-14
In an experiment, a student mixes equal volumes of 0.0020 M Pb2+ions with 0.0040 M I-ions. The trial ion product (TIP) is……..
Select one:
a. 3.2 x 10-8
b. 1.3 x 10-7
c. 3.0 x 10–6
d. 8.0 x 10-6
e. 4.0 x 10-9
E.
When a student mixes equal volumes, the solution is diluted by half.
0.001(0.0020)2 = 4 x 10-9
Consider the following solubility equilibrium:
PbCl 2(s) ⇌ Pb(aq)2++ 2Cl(aq)–
A student adds NaCl(s) to a saturated solution of PbCl2. When equilibrium is reestablished, how have the concentrations changed from the original equilibrium?
Select one:
a. [Pb2+] increased and [Cl–] decreased.
b. Pb2+] and [Cl–] both remain unchanged.
c. [Pb2+] and [Cl–] both decreased.
d. [Pb2+] decreased and [Cl–] increased. Correct!
e. [Pb2+] and [Cl–] both increased
D.
Adding Chloride ion will increase the concentration of Chloride ion, stressing the equilibrium and causing a shift of the equation to the left. This will cause a decrease in the concentration of Lead ion.
The solubility of SnS is 3.2 x 10-3M. The value of Kspis …
Select one:
a. 1.0 x 10-5
b. 6.4 x 10-3
c. 3.2 x 10-3
d. 5.7 x 10-2
e. 7.2 x 10-2
A.
SnS –> Sn2+ + S2-
ksp = (3.2 x 10-3)2 = 1.0 x 10-5
A 3.0 L solution of NiCl2 is found to have a chloride ion concentration of 0.60 M. The concentration of nickel(II) ions in this solution is ….
Select one:
a. 0.60 M
b. 0.30 M
c. 1.8 M
d. 0.90 M
e. 1.2 M
B.
NiCl2 –> Ni2+ + 2Cl-
[Cl-] = 0.6M
When equal volumes of 0.20 M ZnSO4and 0.20 M Sr(OH)2 are combined, …
Select one:
a. a precipitate of only Zn(OH)2 forms.
b. a precipitate of only SrSO4 forms.
c. precipitates of both SrSO4 and Zn(OH)2 form.
d. no precipitate forms.
C.
Both compounds have low solubility.
The solubility of PbI2 will increase with the addition of
Select one:
a. Pb(NO3)2
b. water.
c. PbI2
d. heat.
e. NaNO3
D.
NOT A, because This will create an common ion effect, stressing the system, causing the reaction to shift left, More PbI2 will precipitate
A 200.0 mL solution contains 0.050 mol of Ba(NO3)2. The [NO3–] is….
Select one:
a. 0.12 M
b. 0.050 M
c. 0.50 M
d. 0.25 M
e. 0.10 M
C.
Ba(NO3)2 –> Ba2+ + 2NO3-
At 25°C, the solubility of an unknown compound is 7.1 x 10–5M. The compound is ….
Select one:
a. CuI
b. MgCO3
c. CaSO4
d. CaCO3
e. AgI
D.
Ksp = (7.1 x 10-5)2 = 5.0 x 10-9
The Ksp of CaCO3 is 5.0 x 10-9
Consider the following equilibrium:
**needs pictur3e
Which of the following reagents, when added to the equilibrium system, would cause more
CaCO3 to dissolve?
Select one:
a. KNO3 (s)
b. CaCO3 (s)
c. H2C2O4 (s)
d. Na2CO3 (s)
C.
To increase the amount of CaCO3 that dissolved, precipitate either the Ca+2 or the CO3-2 which will stress the system, causing the reaction to shift right and allowing more to dissolve to reduce the stress.
A solution of AgNO3 is slowly added to a mixture containing 0.10 M Br−, Cl −, I− , and IO3− . The precipitate which forms first is
Select one:
a. AgI
b. AgCl
c. AgBr AgBr
d. AgIO3
A.
AgI has the smallest value for ksp so it will precipitate first.
AgI ksp = 8.5 x 10-17
AgBr ksp = 5.4 x 10-13
AgCl ksp = 1.8 x 10-10
AgIO3 ksp = 3.2 x 10-8
The molar solubility of iron (II) sulphide is
Select one:
a. 3. 6 × 10 −37 M
b. 3. 0 × 10 −19 M
c. 6.0 ×10 −19 M
d. 7. 7 ×10 −10 M
D.
FeS(s) ↔ Fe+2(aq) + S-2(aq)
ksp = [Fe+2][S-2]
The solubility of manganese(II) sulphide is 1.7 × 10 −7 M at 25° C. The solubility product constant is
Select one:
a. 2.9 × 10 −14
b. 1. 7 × 10 −7
c. 3. 4 × 10 −7
d. 4.1 × 10 −4
A.
MnS(s) ↔ Mn+2(aq) + S-2(aq)
At 25° C, the maximum [Zn2+] that can exist in 0.250 M Na2S is
Select one:
a. 5. 0 × 10 −26 M
b. 2.0 × 10 −25 M
c. 8. 0 × 10 −25 M
d. 4.5 × 10 −13 M
C.
For ZnS ksp = 2.0 x 1025
What is the maximum [Ag+] that can exist in 0.20 M NaBrO3 ?
Select one:
a. 1.1 × 10 −5 M
b. 5. 3 × 10 −5 M
c. 2.7 × 10 −4 M
d. 7.3 × 10 −3 M
C.
ksp of AgBrO3 is 5.3 x 10-5
When equal volumes of 2.0 M Pb(NO3)2 and 2.0 M KCl are mixed,
Select one:
a. a precipitate forms because trial ion product <ksp></ksp>
<p>b. a precipitate forms because trial ion product >ksp </p>
<p>c. a precipitate does not form because trial ion product <ksp>
<p>d. a precipitate does not form because trial ion product >ksp</p></ksp></p>
</ksp>
B.
Pb(NO3)2(aq) + 2KCl(aq) ↔ PbCl2(s) + 2KNO3(aq)
Lead (II) Chloride will precipitate
Which of the following could be used to precipitate both Mg 2+ and Ca 2+ from hard water?
Select one:
a. lithium sulphate
b. sodium phosphate
c. potassium sulphide
d. ammonium chloride
B.
Both ions will precipitate with PO4-3
The solubility of magnesium carbonate is
Select one:
a. 4.6 × 10 −11 M
b. 3.4 × 10 −6 M
c. 6.8 × 10 −6 M
d. 2.6 × 10 −3 M
D.
ksp = 6.8 x 10-6
MgCO3 ↔ Mg+2 + CO3-2
ksp = [Mg+2][CO3-2]
6.8 x 10-6 = x2
x = 2.6 x 10-3M
When 250 mL of 0.36 M Sr(OH)2 are added to 750 mL of water, the resulting ion concentrations are
Select one:
a. [Sr 2+] = 0.12 M and [OH−] =0.12 M
b. [Sr 2+] = 0.12 M and [OH−] =0.24 M
c. [Sr 2+] = 0.090 M and [OH−] =0.090 M
d. [Sr 2+] = 0.090 M and [OH−] =0.180 M
D.
[Sr+2] = 0.090M
[OH-] = 0.18M
The correct answer is: [Sr 2+] = 0.090 M and [OH−] =0.180 M
Which of the following occurs when equal volumes of 0.20 M MgS and 0.20 M ZnSO4
are mixed?
Select one:
a. A precipitate does not form.
b. A precipitate of ZnS forms.
c. A precipitate of MgSO4 forms.
d. Precipitates of MgSO4 and ZnS form.
B
Trial ksp = [Zn+2][S-2]
Trial ksp = [0.20][0.20]
Trial ksp = 0.04
ksp ZnS = 2.0 x 10-25
Trial ksp>> ksp therefore a precipitate forms
Consider the following equilibrium:
**needs picture**
Sodium chloride is added to a saturated solution of AgCl. The amount of solid AgCl will
Select one:
a. increase as the equilibrium shifts to the left.
b. decrease as the equilibrium shifts to the left.
c. increase as the equilibrium shifts to the right.
d. decrease as the equilibrium shifts to the right.
A.
Adding sodium chloride adds chloride ions…this creates a common ion effect…this stresses the reaction and causes the reaction to shift to the left and increase the amount of AgCl solid.
In a saturated solution, the rate of dissolving is
Select one:
a. equal to zero.
b. equal to the rate of crystallization.
c. less than the rate of crystallization.
d. greater than the rate of crystallization.
B.
As with all equilibrium equations, when at equilibrium, the forward rate equals the reverse rate. In the case of solubility equilibrium, the rate of dissolving (forward rate) equals the rate of crystallization (the reverse rate)
What concentration of Calcium Sulphate can dissociate in a 2.00 M solution of Sodium Sulphate?
NaSO4^2 is soluble and must be the supplier of SO4^2- ions (so have to use it’s 2.00 M value)
CaSO4<−−>Ca2++SO42−CaSO4 <–> Ca^2+ + SO4^2-CaSO4<−−>Ca2++SO42−
Ksp = [Ca^2+][SO4^2-]
7.1 x 10^-5 = [Ca2+] [2.00]
[Ca2+] = 3.6 x 10^-5 M
Using the Ksp and the concentration of one ion, the minimum concentration of the other ion just able to start precipitation can be calculated.
How much KIO3 mass is needed to just start precipitation, with 0.010 M of Cu(NO3)2?
1.1 g KIO3. ( I still think it should be double, due to the 6.9 x 10^-8 = [Cu^2+][IO3-]^2 step.
Cu(IO3)2 <—> Cu^2+ + 2IO2^1-
2.58 x 10^-3 x 2.0 L x 1 mol IO3^1- / 1 mol KIO3 x 214 g/1 mol = 1.1 grams KIO3
What molarity of SrCO3 can dissolve in a 0.500 M solution of Na2CO3.
Ksp = 5.6 x 10^-10 (SrCO3)
Solubility table states that Na2CO3 is soluble while SrCO3 isn’t, meaning that the CO3^2- ions come from the Na2CO3 molecules.
5.6 X 10^-10 = [Sr^2+][0.500 M]
= 1.0 x 10^-9
The concentration of SrCO3 capable of dissolving is this amount.
**remember, the molarity of the two molecules can’t go above 5.6 x 10^-9
Choose a cation (Ag+, Pb2+, Cu+, Ca2+, Sr2+, Ba2+) for a 0.20 M solution that contains Br- and a second solution that contains SO4^2- ions, that would precipitate them one by one.
- Use Sr2+ or Ba2+ to precipitate SO4^2-
- Ag+ or Pb2+ or Ca+ to precipitate the SO4^2-
A longer question: But what is the order in which Ba^2+, Fe^2+ and Ag^2+ can be precipitated so that only one at a time is filtered out (via precipitation)?
Can use Na2SO4, NaOH, or NaCl.
**hint: break into ions 1st
- Cl- first. Precipitates Ag^2+
- SO4^2-. Precipitates Ba2+
- OH- last. Precipitates Fe^2+.
Solubility equilibrium can only exist in a :
a) dilute solution
b) saturated solution
c) concentrated solution
d) ionic solution
B
Which of the following saturated solutions will have the lowest [Pb2+]?
a) PbI2
b) PbCl2
c) Pb(IO3)2
d) Pb(NO3)2
C) it has the lowest Ksp at 3.7 x 10^-13
meaning its the salt with the lowest solubility
Explain what a saturated solution is?
These solutions have the maximum concentration of ions and the concentrations of the ions will satisfy the Ksp expression
BaSO4<−−>Ba2++SO42−
Will adding BaSO4 dec/inc the solubility?
No change at all. Adding the more of the same solid will not change anything.
BaSO4<−−>Ba2++SO42−
Will Ba(NO3)2 inc/dec the solubility?
Yes, Ba^2+ will decrease the solubility
BaSO4<−−>Ba2++SO42−
Will adding (NH4)2SO4 inc/dec solubility?
SO4^2- ion will decrease solubility
BaSO4<−−>Ba2++SO42−
Will KNO3 dec/inc the solubility?
Will more precipitate form (on the left side of reaction)
No, will not decrease solubility
Explain the Common Ion Effect?
If there is a common ion and I add more of it, it will decrease the solubility. The soluble ion will decrease the solubility of the insoluble ion (low-soluble ion).
PbCl2<−−>Pb2++2Cl−
(solid <–> ions)
What happens to the solubility when NaCl is added to the already saturated solution:
What happens to the Cl- and the Pb2+?
Does solubility of PbCl2 inc/dec?
Does more precipitate form in the beaker?
Decreases, shifts to the left.
Solubility decreases. Too much Cl-.
Precipitate will increase in the beaker.
A+heat<−−>B+C
Exothermic or endothermic?
When heat is increased, does the solubility increase or decrease?
endothermic
increases, because more of the solid dissolves
A<−−>D+C+heat
Is it exothermic or endothermic?
and
Solubility increases or decreases when temperature increased?
exothermic
decreases
BaSO4<−−>Ba2+ + SO42−
Solubility:
If more reactant is added, solubility ____ and if reactant is negated, then solubility ______
increases, as there are more ions on the right side
decreases, as the concentrations of the ions decrease to replace the solid BaSO4 removed
Will a precipitate form?
- 0 mL of 0.018 M Na2C2O4
- 0 of 0.030 M Mg(NO3)2
and Ksp = 8.6 x 10^-5 of MgC2O4
Yes, a precipitate will form.
Use Dilution Formula C1V1 = C2V2 2.2 x 10^-2==C
Ktrial = [Mg^2+][C2O4^2-]
Ktrial > Ksp
Will a precipitate form?
500 mL of 2.0 x 10^-3 of Pb(NO3)2
300 mL of 0.11 M KBr
*hint: look up solubilities on table
Need to use Dilution formula:
C1V1 = C2V2 for each ion.
Then do this:
Ktrial = [Pb2+][Br-]^2
[1.25 x 10^-3][0.04125]^2
=2.1269 x 10^-6
Ktrial < Ksp
Will a precipitate form if there is 3.0 x 10^10- M of Cu(NO3)^2 and 2.0 x 10^11- M of Na2S mixed?
*hint = look up solubilities of ions 1st
Ktrial > Ksp
Ktrial = 1.5 x 10^-21 > Ksp = 6.0 x 10^-37
Precipitate will form
Which of the following compounds could be used to precipitate the ion Ca^2+
- H2S
- AgNO3
- Na2CO3
- CaCl2
the CO3^2- ion will precipitate the calcium ion in the water
List four common insoluble compounds…
sulphites SO3^2-
phosphates PO4^3-
carbonates CO3^2-
almost all hydroxides OH-
Compounds are soluble if the negative ion is soluble, such as ….(name 2)
Nitrates NO3^-1
Acetates CH3COO-
Compounds are soluble if there is a positive ion such as….(list 8 ions)
Li+, Na+, K+,Rb+, Cs+, Fr+, NH4+, H+
If the Ksp is 8.7 x 10^-9 at 25 C for CaCO3, what is the dissociation expression ? molar solubility?
Ksp = [Ca^2+][CO3^2-]
9.3 x 10^-5 M
What is the Ksp expression for the following:
Fe(OH)3<−−>Fe3++2OH−
Ksp = [Fe^3+][OH-]^3
What is the Ksp expression for
Ag2CO3<−−>2Ag++CO32−
(solid <–> aq + gas)
Ksp = [Ag+]^2[CO3^2-]
What’s the Ksp expression for the following:
CaCO3 <−−>Ca2+ + CO32−
(solid <–> aq + aq)
Ksp = [Ca^2+][CO3^2-]
Find the Ksp from the molar solubility, if there was 0.031 grams of solid BaF2 left over from evaporation, WITH 50 mL only.
hint: grams > moles > M (Ksp after)
1. Find molar solubility: 3.54 x 10^-3 M.
2. Ksp = [Ba^2+][F-]^2 = 1.77 x 10^-7
A 200.0 mL sample of saturated copper(II)iodate Cu(IO3)2 is evaporated to dryness.
What is the mass of the residue?
*hint: look up Ksp on the Solubility Table 1st. and it’s M–>moles–>mass
It’s a 1:2 ratio
Ksp= [Cu^2+][IO3^-]^2
2.58 x 10^-3 M. 0.21 g Cu(IO3)2.
Calculate the molar solubility if Ksp = 8.9 x 10 ^-17
Ag3PO4<−−>3Ag+ + PO43−
Let s represent the solubility and it’s a 1:3:1 ratio
Ksp = [Ag+]^3 [PO4^3-]
4.3 x 10^-5 M (do this Q and see if it’s correct..)
Calculate the molar solubility if the Ksp = 2.1 x 10^-13.
Mn(OH)2<−−>Mn2+ + 2OH−
A 1:1:2 ratio, 2.1 x 10^-13 = (s)(2s)^2
3.74 x 10^-5 M
Calculate the molar solubility of:
PbSO4<−−>Pb2++SO42−
if Ksp = 1.8 x 10^-8.
Since it’s a 1:1:1 ratio of the moles,
1.8 x 10^-8 = [s][s],
Let s represent the ions solubility and solve for s.
1.34 x 10^-4 M
What is molar solubility?
the number of moles that dissolves in 1.0 L to create a saturated solution
Saturation only occurs at equilibrium.
Ksp represents the solubility of the whole molecule and the concentrations on the right-hand side represent the molar solubility.
TOGETHER they represent the Ksp.
**it helps to type out the whole answer**
Ksp = [part of the molecule][other part of molecule]
The equation for the equilibrium constant for DISSOCIATION is…
Ksp = products / reactants
IF Keq is the constant for equilibrium, the constant for dissociation or solubility is ..
Ksp
A precipitate is…
the insoluble product that forms in a chemical reaction.
DEFINE what solubility is a measure of?
measure of the solute dissolved in a solvent at a temperature (solute is the lesser-quantity substance)
Consider the following:
2 NO2 (g) N2O4 (g) keq = 1. 20
A 1.0 L flask is filled with 1.4 mol NO2 and 2.0 mol N2O4 . To reach equilibrium, the reaction proceeds to the
Select one:
a. left as Trial K eq > K eq Trial k eq < k eq and this results in a shift to the right
b. left as Trial K eq < K eq
c. right as Trial k eq < k eq
d. right as Trial k eq > k eq
Answer: C. Trial Keq < actual Keq
2C6H14+1902<−−>12CO2+14H2O+7713kJ2 C6H14 + 19 02 <–> 12 CO2 + 14H2O + 7713kJ2C6H14+1902<−−>12CO2+14H2O+7713kJ
Min. enthalpy favours the _____
Max entropy favours the _____
Equilibrium, completion, or no reaction at all?
products (exo)
products (21 vs. 26 moles)
completion
Either entropy or enthalpy must dominate a reaction (min/max), but:
if min. enthalphy favours the products and max entropy favours the products,
what happens?
Goes to completion, with no equilibrium
