Download Review Unit 8 Test (Chp 15,17)

Name: [KEY] __________________________
[KEY]
1.
Date:______________ Period:______
Review Unit 8 Test (Chp 15,17): Chemical and Solubility Equilibrium
[KEY]
Describe dynamic chemical equilibrium in terms of concentrations and rates.
The forward rate of reaction equals the reverse rate of reaction while concentrations of reactants and products are constant
2.
What does the value of K, as K >1 or K < 1 , describe about a reaction at equilibrium?
________________________________________________________________________________
K = [products] K>1 has more products
________________________________________________________________________________
[reactants] K<1 has more reactants
3.
The equilibrium-constant expression for the reaction
Ti(s) + 2 Cl2(g)
TiCl4(l)
is given by
(A)
(B)
(C)
(D)
(E)
4.
[Cl2 (g)]–2
No pure (l) or (s) in K expression b/c they are
not concentrations. Only (g) and (aq) allowed.
Products over reactants so in this case, it would
be 1/[Cl2]2 or [Cl2]–2
Which of the following changes to a reaction system in equilibrium would affect the value of the
equilibrium constant, Keq, for the reaction? (Assume that all other conditions are held constant.)
(A)
Adding more of the reactants to the system
(B)
Adding a catalyst for the reaction to the system
(C)
Changing the temperature of the system
(D)
Changing the pressure on the system
(E)
Removing some of the products from the system
1
(∆H < 0 , exothermic , produces heat forward)
HgO(s) + 4 I– + H2O
5.
6.
HgI42– + 2 OH– + heat
Which of the following changes will increase the concentration of HgI42– in the equilibrium reaction
above?
(A)
Increasing the concentration of OH¯
(B)
Adding 6 M HNO3
(C)
Increasing the mass of HgO present
(D)
Increasing the temperature
(E)
Adding a catalyst
Adding H+ will react with OH– thereby removing
a product causing a shift to the right to replace
the OH– producing more HgI42–.
In which of the following systems would the number of moles of the substances present at
equilibrium NOT be shifted by a change in the volume of the system at constant temperature?
(A) CO(g) + NO(g)
CO2(g) + 1/2 N2(g)
(B) N2(g) + 3 H2(g)
2 NH3(g)
(C) N2(g) + 2 O2(g)
2 NO2(g)
(D) N2O4(g)
2 NO2(g)
(E) NO(g) + O3(g)
NO2(g) + O2(g)
PCl3(g) + Cl2(g)
7.
∆H < 0
Equal number of moles of gas on each side so neither
side is favored (less volume would favor more moles
and more volume would favor fewer moles)
PCl5(g) + energy
Some PCl3 and Cl2 are mixed in a container at 200°C and the system reaches equilibrium according
to the equation above. Which of the following causes an increase in the number of moles of PCl5
present at equilibrium?
I. Decreasing the volume of the container
II. Raising the temperature
III. Adding a mole of He gas at constant volume
(A) I only
(B) II only
(C) I and III only
(D) II and III only
Yes I. b/c less volume would favor fewer moles of gas.
Not II. b/c adding energy is adding a product which would shift left.
Not III. b/c adding an inert gas would have no effect (not in K expression,
doesn’t change P per V ratio of gases involved)
(E) I, II, and III
2
2 SO3(g)
8.
2 SO2(g) + O2(g)
When 0.70 mole of SO3 and 0.80 mole of SO2 are placed in an evacuated 1.00-liter flask, the
reaction represented above occurs. After the reactants and the product reach equilibrium and the
initial temperature is restored, the flask is found to contain 0.20 mole of SO3 .
Based on these results, what is the equilibrium constant expression, Kc , of the reaction?
(substituted values in the expression without calculating the Kc value)
I
C
E
2 SO3(g)
0.70 M
–0.50
0.20 M
2 SO2(g) + O2(g)
0.80 M
0M
+0.50
+0.25
1.30 M
0.25 M
I2(g) + Br2(g)
9.
Kc = (1.30)2(0.25)
(0.20)2
(as a multiple choice answer)
(no calculation)
2 IBr(g)
A sealed 1.0 L flask is charged with 0.35 mol of I2 and 0.35 mol of Br2 .
An equilibrium reaction ensues. When the container contents achieve equilibrium, the flask
contains 0.50 mol of IBr . The value of Keq is __________.
I
C
E
I2(g) + Br2(g)
0.35 M 0.35 M
–0.25 –0.25
0.10 M 0.10 M
2 IBr(g)
0M
+0.50
0.50 M
H2(g) + Br2(g)
Kc = [IBr]2 = (0.50)2 = 0.25 = 25
[I2][H2]
(0.10)2
0.01
2 HBr(g)
10. At a certain temperature, the value of the equilibrium constant, K, for the reaction represented above is
2.0 x 105. What is the value of K for the reverse reaction at the same temperature?
(A) –2.0 x 10–5
(B) 5.0 x 10–6
(C) 2.0 x 10–5
Reverse reactions have K that is 1/K1
So reverse K is 1/(2.0 x 105)
Or, 5.0 x 10-6
(D) 5.0 x 10–5
(E) 5.0 x 10–4
3
Questions 11-12 refer to the following.
PCl5(g)
PCl3(g) + Cl2(g)
PCl5(g) decomposes into PCl3(g) and Cl2(g) according to the equation above. A pure sample of PCl5(g) is
placed into a rigid, evacuated 1.00 L container. The initial pressure of the PCl5(g) is 1.00 atm.
The temperature is held constant until the PCl5(g) reaches equilibrium with its decomposition products.
The figures below show the initial and equilibrium conditions of the system.
PCl5 , PCl3 , and Cl2
Ptotal = 1.40 atm
PCl5
Ptotal = 1.00 atm
Figure 1: Initial
11.
Figure 2: Equilibrium
As the reaction progresses toward equilibrium, the rate of the forward reaction
(A)
(B)
(C)
(D)
increases until it becomes the same as the reverse reaction rate at equilibrium.
stays constant before and after equilibrium is reached.
A is half true, but “decreases”,
decreases to become a constant nonzero rate at equilibrium.
not “increases.”
decreases to become zero at equilibrium.
Greater pressure of reactant initially (Q = 0/1.00 = 0) so forward rate is faster due to greater collision
frequency of reactant particles. The forward rate slows over time as reactant is consumed and there is a
lower collision frequency of reactant particles, but it does not reach zero b/c more reactant particles are
forming from products as the reverse rate increases over time until the rates are equal at equilibrium.
12.
Which of the following statements about Kp , the equilibrium constant for the reaction, is correct?
(A) Kp > 1
(B) Kp < 1
(C) Kp = 1
(D) It cannot be determined whether Kp > 1 , Kp < 1 , or Kp = 1 without additional information.
Initially, there is (PPCl5)in = 1.00 atm , (PPCl3)in = 0 atm , and (PCl2)in = 0 atm ,
but the Change and Equilibrium values are unknown.
PCl5(g)
PCl3(g) + Cl2(g)
I 1.00 atm
C –x
E 1.00 – x
0 atm
+x
x
0 atm
+x
x
At Equilibrium, there Ptotal = 1.40 atm, therefore...
(1.00 – x) + x + x = 1.40 atm
1.00 + x = 1.40
x = 0.40 atm
Kp = (0.40)(0.40) = (0.16)
(1.00 – 0.40)
(0.60)
Numerator is smaller than
denominator so K < 1
(no calculation needed)
4
13. How is the reaction quotient (Q) of a reaction related to the equilibrium constant (Keq) of the reaction?
(A)
Q does not depend on the concentrations or partial pressures of reaction components.
(B)
Q is the same as Keq when a reaction is at equilibrium.
(C)
When Q < K, the reaction will proceed to the left.
(D)
Keq does not change with temperature, whereas Q is temperature dependent.
(E)
K does not depend on the concentrations or partial pressures of reaction components.
2 HF(g)
H2(g) + F2(g)
14. At 25oC, 10.0 mol of HF(g) are combined with 1.0 moles each of H2(g) and F2(g) in a 1.0 L rigid
container. The value of Kc at 25oC is 0.010. Which of the following will result as the reaction occurs?
(A)
More H2(g) and F2(g) will form.
(B)
More HF(g) will form.
(C)
The total pressure will increase.
(D)
The total pressure will remain constant.
K = [H2][F2]
[HF]2
K = 0.10
Q = (1.0)(1.0)
(10.0)2
Q= 1
100
.
Q = 0.010
Q = K , the system is at equilibrium
___15. Which of the following compounds would be more soluble in a nitric acid solution than in water?
(A)
(B)
(C)
(D)
(E)
HCl
Ba(OH)2
Al(OH)3
HF
KNO3
HCl, Ba(OH)2 , and KNO3 are strong
electrolytes already 100% dissolved in water
so they cannot be “more” soluble in acid.
(Strong Acid, Strong Base, Soluble Salt)
Al(OH)3
Al3+ + 3 OH–
Adding HNO3 is adding H+ which reacts
with OH– removing a product so it shifts right
causing more Al(OH)3 to dissolve.
___16. Barium sulfate is LEAST soluble in a 0.01-molar solution of which of the following?
(A) Al2(SO4)3
(B) (NH4)2SO4
(C) Na2SO4
(D) NH3
(E) BaCl2
BaSO4(s)
Ba2+(aq) + SO42–(aq)
Al2(SO4)3 contains the most of a common ion (SO42–)
product which shifts the solubility reaction left
dissolving less solid BaSO4 .
___17. In which of the following aqueous solutions would you expect PbI2 to have the greatest solubility?
(A) pure water
(B) 0.020 M BaI2
(C) 0.020 M Pb(NO3)2
(D) 0.030 M NaI
(E) 0.020 M KI
PbI2(s)
Pb2+(aq) + 2 I2–(aq)
Pure water contains NO common ions which increases the solubility of the solid PbI2 .
(or least it does not decrease it at all).
All the other solutions contain common ions which decrease the solubility of the solid.
(see #14 above)
5
___18. What is the molar solubility in water of Ag2CrO4? (The Ksp for Ag2CrO4 is 8 x 10–12)
(A) 8 x 10–12
(B) 2 x 10–12
(C) (4 x 10–12)1/2
(D) (4 x 10–12)1/3
(E) (2 x 10–12)1/3
2 Ag+ + CrO42–
Ag2CrO4
I
x
0
0
C
–x
+2x
+x
E
0
2x
x
Ksp = [Ag+]2[CrO42–]
8 x 10–12 = (2x)2(x)
8 x 10–12 = 4x3
2 x 10–12 = x3
so… x = 3√(2 x 10–12)
___19. The solubility of MnCO3 is 5 x 10–10 M . What is the solubility product constant, Ksp, for MnCO3?
(A)
(B)
(C)
(D)
(E)
2.5 x 10–21
5.0 x 10–20
2.5 x 10–19
2.2 x 10–5
2.2 x 10–20
I
MnCO3
Mn2+
5 x 10–10 M
0
C –5 x 10–10 M
E
0
CO32–
+
0
+5 x 10–10 M
+5 x 10–10 M
5 x 10–10 M
5 x 10–10 M
Ksp = [Mn2+][CO32–]
Ksp = (5 x 10¯10)2
Ksp = 25 x 10–20 = 2.5 x 10–19
___20. In a saturated solution of Zn(OH) 2 at 25oC the value of [OH–] is 2.0 x 10–6 M. What is the value
of the solubility-product constant, Ksp, for Zn(OH)2 at 25oC?
(A) 4.0 x 10–18
(B) 8.0 x 10–18
(C) 1.6 x 10–17
(D) 4.0 x 10–12
(E) 2.0 x 10–6
[OH–] is 2.0 x 10–6 M at EQUILIBRIUM in a saturated solution.
Zn(OH)2
I
x
C –1.0 x 10–6
E
0
Zn2+
0
+
2 OH–
0
+1.0 x 10–6 +2.0 x 10–6
1.0 x 10–6
2.0 x 10–6 M
Ksp = [Zn2+][OH–]2
Ksp = (1.0 x 10¯6) (2.0 x 10¯6)2
Ksp = (1.0 x 10¯6) (4.0 x 10¯12)
Ksp = 4.0 x 10¯18
If [OH–] = 2.0 x 10–6 M, then [Zn2+] = 1.0 x 10–6 M, b/c the mol ratio is 1:2 for Zn2+ : OH–
6
CO(g) + 2 H2O(g)
Kc = 1.5 x 103
CO2(g) + 2 H2(g)
___21. A 2.0 mol sample of CO(g) and a 2.0 mol sample of H2O(g) are introduced into a previously
evacuated 1.00 L rigid container, and the temperature is held constant as the reaction represented
above reaches equilibrium. Which of the following is true at equilibrium?
(A) [H2O] > [CO] and [CO2] > [H2]
(B) [H2O] > [H2]
(C) [CO2] > [CO]
(D) [CO] = [H2O] = [CO2] = [H2]
CO + 2 H2O
I 2.0 M 2.0 M
C –x
–2x
E
CO2 + 2 H2
0M
0M
+x
+2x
K = [CO2][H2]2
[CO][H2O]2
K = 1.5 x 103
The system will shift right faster to reach equilibrium.
B/c K > 1, at equilibrium there will be more products than reactants.
Due to stoichiometric ratios, [H2O] < [CO] and [H2] > [CO2]
X(g) + 2 Q(g)
Kc = 1.3 x 105 at 50oC
R(g) + Z(g)
___22. A 1.0 mol sample of X(g) and a 1.0 mol sample of Q(g) are introduced into an evacuated, rigid
10.00 L container and allowed to reach equilibrium at 50oC according to the equation above.
At equilibrium, which of the following is true about the concentrations of the gases?
(A) [R] = ½ [Q]
(B) [Q] = ½ [X]
(C) [R] = [Z] > [Q]
(D) [X] = [Q] = [R] = [Z]
X + 2Q
I 0.10 M 0.10 M
C –x
–2x
E
R + Z
0M 0M
+x
+x
K = [R][Z]
[X][Q]2
K = 1.3 x 105
The system will shift right faster to reach equilibrium.
B/c K > 1, at equilibrium there will be more products than reactants, so [Z] > [Q] .
and [R] = [Z] b/c R and Z form at a 1:1 ratio.
Compound
Ksp at 298 K
Ag2SO4
PbSO4
1 x 10–5
1 x 10–8
___23. A 1.0 L solution of AgNO3(aq) and Pb(NO3)2(aq) has Ag+ and Pb2+ concentrations each of 0.0010 M.
A 0.0010 mol sample of K2SO4(s) is added to the solution.
Based on the information in the table above, which of the following will occur?
(Assume that the volume change of the solution is negligible.)
(A) No precipitate will form.
(B) Only Ag2SO4(s) will precipitate.
(C) Only PbSO4(s) will precipitate.
“Will a precipitate form?”
Is Q > K? (shift left faster, making solid)
(D) Both Ag2SO4(s) and PbSO4(s) will precipitate
Ag2SO4(s)
2 Ag+ + SO42–
0.0010 M 0.0010 M
Ksp = [Ag+]2[SO42–]
Ksp = 1 x 10¯5
PbSO4(s)
Pb+ + SO42–
0.0010 M 0.0010 M
Ksp = [Pb2+][SO42–]
Ksp = 1 x 10¯8
Q = (1.0 x 10¯3)2(1.0 x 10¯3)
Q = (1.0 x 10¯3)(1.0 x 10¯3)
Q = (1.0 x 10¯6)(1.0 x 10¯3)
Q = 1.0 x 10¯6
Q = 1.0 x 10¯9
Q < Ksp , shifts right faster
Q > Ksp , shifts left faster
forming solid precipitate
7
Section II Free Response
Calculator Allowed
CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS.
It is to your advantage to do this, because you may earn partial credit if you do and little or no credit if you
do not. Attention should be paid to significant figures.
C(s) + CO2(g)
2 CO(g)
1. Solid carbon and carbon dioxide gas at 1,160 K were placed in a rigid 2.00 L container, and the
reaction represented above occurred. As the reaction proceeded, the total pressure in the container
was monitored. When equilibrium was reached, there was still some C(s) remaining in the
container. Results are recorded in the table below.
Time
Total Pressure of
(hours) Gases in Container
at 1,160 K (atm)
(a)
0.0
5.00
2.0
6.26
4.0
7.09
6.0
7.75
8.0
8.37
10.0
8.37
Write the expression for the equilibrium constant, Kp for the reaction. (1)
Kp = (PCO)2
(PCO2)
(b) Calculate the number of moles of CO2(g) initially placed in the container. (Assume that the
volume of the solid carbon is negligible.) (2)
PV = nRT
(5.00 atm)(2.00 L) = n(0.08206)(1160 K)
n = 0.105 mol CO2
(c)
For the reaction mixture at equilibrium at 1,160 K, the partial pressure of the CO2(g) is
1.63 atm. Calculate
(i)
the partial pressure of CO(g), and (1)
Ptotal = PCO2 + PCO
8.37 atm = PCO + 1.63 atm
PCO = 6.74 atm
(ii)
the value of the equilibrium constant, Kp (2)
Kp =
(6.74)2
1.63
= 27.9
8
(d) If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure
of the gases at equilibrium be greater than, less than, or equal to the final total pressure of the
gases at equilibrium without the catalyst? Justify your answer. (Assume that the volume of the
solid catalyst is negligible.) (1)
With a catalyst, the final total pressure of the gases at equilibrium would be equal to the final total pressure of
the gases at equilibrium without a catalyst because a catalyst can cause a reaction to reach equilibrium faster,
but it does not affect the values of Q or K.
In another experiment involving the same reaction, a rigid 2.00 L container initially contains 10.0 g
of C(s), plus CO(g) and CO2(g), each at a partial pressure of 2.00 atm at 1,160 K.
(e) Predict whether the partial pressure of CO2(g) will increase, decrease, or remain the same as
this system approaches equilibrium. Justify your prediction with a calculation. (2)
C(s) + CO2(g)
2 CO(g)
Kp = 27.9 from part (c)(ii)
Q = (PCO)2
(PCO2)
Q = (2.00)2 = 2.00
(2.00)
Q < K , the reaction will shift right faster to reach equilibrium causing
PCO2 to decrease as it is consumed.
CO2(g) + H2(g)
H2O(g) + CO(g)
2. When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the equation
above. In one experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/L
(a) What is the mole fraction of CO(g) in the equilibrium mixture? (1)
XCO = 0.55 = 0.34
1.60
(b) Using the equilibrium concentrations given above, calculate the value of Kc , the equilibrium
constant for the reaction. (1)
Kc = [H2O][CO] = (0.55)2 = 5.0
[CO2][H2]
(0.30)(0.20)
9
(c) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g)
is converted back to CO2(g) . Calculate the value of Kc at this lower temperature. (2)
30% of CO is changed from the 0.55 M at equilibrium in part (c).
30% of 0.55 M CO is
(0.30) x (0.55) = 0.17 M
Therefore, 0.17 M of CO is changed to CO2 .
CO2(g) + H2(g)
I 0.30 M
0.20 M
C +0.17
+0.17
E 0.47 M
0.37 M
H2O(g) + CO(g)
0.55 M 0.55 M
–0.17
–0.17
0.38 M
0.38 M
Tricky part:
The INITIAL pressures in part (c) are the
previous equilibrium pressures from part (b)
b/c the system was at equilibrium, then the
temperature changed in part (c), so it will shift
(change) to re-establish equilibrium.
Kc = [H2O][CO] = (0.38)2 = 0.83
[CO2][H2]
(0.47)(0.37)
(d) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a
3.0-liter reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter,
of CO(g) at this temperature. (2)
0.50 mol = 0.17 M
3.0 L
Initially, there is 0.17 M H2 & 0.17 M CO2 , but the Change and Equilibrium values are unknown.
CO2(g) + H2(g)
I 0.17 M 0.17 M
C –x
–x
E 0.17 – x 0.17 – x
H2O(g) + CO(g)
0M
0M
+x
+x
x
x
Kc = 5.0
(0.17 – x) is NOT ≈ 0.17
Kc = [H2O][CO]
[CO2][H2]
No need to ignore x as negligible because
avoiding a polynomial is not an issue here.
5.0 =
x2
(0.17 – x)2
(square root both sides)
2.2 =
x
(0.17 – x)
(multiply 0.17 – x to the left side and distribute 2.2  1.7 and 2.2  x)
0.37 – 2.2x = x
0.37 = 3.2x
Not able to ignore x as negligible because
K is NOT <<< 1.
(add +2.2x to both sides)
(divide by 3.2)
x = 0.12 M = [CO]
10
3 H2(g) + N2(g)
2 NH3(g)
3. After 0.40 mol of N2(g) and 1.00 mol of H2(g) are placed into an evacuated 1.0 L container at
500 K, the reaction represented above occurs.
The concentration of N2(g) as a function of time is shown below.
1.00
[H2]
0.80
0.60
0.40
[N2]
0.20
0 0
[NH3]
Time
(a) Write the expression for the equilibrium constant, Kc , for the reaction. (1)
Kc = [NH3]2
[H2]3[N2]
(b) What is [N2] at equilibrium? (1)
0.20 M (from graph)
(c) Determine the equilibrium concentrations of H2(g) and NH3(g) . (2)
3H2(g) + N2(g)
I
C
E
2 NH3(g)
1.00 M 0.40 M
–0.60 –0.20
0.40 M 0.20 M
[H2] = 0.40 M
0M
+0.40
0.40 M
[NH3] = 0.40 M
(d) On the graph above, make a sketch that shows how the concentrations of H2(g) and NH3(g)
change as a function of time. (2)
(see
on graph)
11
(e) Calculate the value of the following equilibrium constant, Kc , at 500 K. (1)
Kc = [NH3]2
[H2]3[N2]
Kc = (0.40)2 = 13
(0.40)3(0.20)
Kc = 13
(f) At 1500 K, the value of Kc for the reaction is 1.20. In an experiment, 0.50 mol H2(g) ,
0.80 mol N2(g) , and 0.10 mol NH3(g) are placed in a 1.0 L container and allowed to reach
equilibrium at 1500 K. Determine whether the equilibrium concentration of H2(g) will be
greater than, equal to, or less than the initial concentration of H2(g) . Justify your answer. (2)
3H2(g) + N2(g)
2 NH3(g)
Kc = 1.20
Kc = [NH3]2
[H2]3[N2]
Given initial values for H2 , N2 , and NH3 , and have no equilibrium values, so…
Q = (0.10)2 = 0.10
(0.50)3(0.80)
Q < Kc
Q < K , the reaction will proceed to the right causing [H2] to be less than the initial.
4. Answer the following questions relating to the solubility of the sulfides of copper and silver.
(a) At 25oC, 8.8 x 10–22 g CuS(s) will dissolve in 100 mL of water.
(i)
Write the equation for the dissociation of CuS(s) in water. (1)
CuS(s)
Cu2+ + S2–
Calculate the solubility, in mol∙L–1, of CuS(s) in water at 25oC. (2)
(ii)
8.8 x 10–22 g x
1 mol = 9.2 x 10–24 mol
95.61 g
0.100 L
= 9.2 x 10–23 M CuS
Calculate the value of the solubility product constant, Ksp, of CuS(s) at 25oC. (1)
(iii)
CuS(s)
I
9.2 x 10–23 M
C –9.2 x 10–23
E
0M
Cu2+
0M
+9.2 x 10–23
9.2 x 10–23 M
+
S2–
Ksp = [Cu2+][S2–]
0M
Ksp = (9.2 x 10–23)2
+9.2 x 10–23
Ksp = 8.5 x 10–45
9.2 x 10–23 M
12
(b) At 25oC, the value of Ksp for Ag2S(s) is 1.6 x 10–49 and the value of Ksp for CuS(s) is 8.5 x 10–45.
(i) Calculate the equilibrium value of [Ag+(aq)] in 1000 L of saturated Ag2S solution to
which 0.0050 mole Na2S(s) has been added. Assume no change in volume. (1)
0.0050 mol Na2S = 5.0 x 10–6 M Na2S , so… [S2–] = 5.0 x 10–6 M
1000 L
Ag2S(s)
2 Ag+
I
x
C
–x
+2x
E
0
2x
+
0M
(b/c Na2S dissociates into Na+ & S2–)
Ksp = 1.6 x 10–49
S2–
5.0 x 10–6 M
Ksp = [Ag+]2[S2–]
1.6 x 10–49 = (2x)2(5.0 x 10–6)
+x
5.0 x 10–6 + x
3.2 x 10–44 = 4x2
8.0 x 10–45 = x2
8.9 x 10–23 = x
[Ag+] = 1.8 x 10–22 M
(ii) If 9.0 mL of 0.0600 M Na2S(aq) is added to 9.0 mL of 0.0500 M AgNO3(aq), will a precipitate
form? Assume that volumes are additive. Show calculations to support your answer. (3)
Ag2S(s)
2 Ag+ + S2–
Ksp = [Ag+]2[S2–]
Ksp = 1.6 x 10–49
[Ag+] = _?_ (after mixing):
M1V1 = M2V2
(0.0500 M AgNO3)(9.0 mL) = M2(18 mL)
[Ag+] = 0.030 M
[S2–] = _?_ (after mixing):
M1V1 = M2V2
(0.0600 M Na2S)(9.0 mL) = M2(18 mL)
–
[S2 ] = 0.030 M
Q = [Ag+]2[S2–]
Q = (0.025)2(0.030) = 1.9 x 10–5
Q > K, therefore a precipitate will form as the
reaction proceeds to the left to reach equilibrium.
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ANSWER KEY
1.
2008 #1
2.
1995
(EQUILIBRIUM)
3.
2003B #1
(edited)
4.
2001 #1
(edited)
14