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Moles of AgNO3 = 0.050 L x 0.100 M = 0.05 moles AgNO3 = 0.05 moles
Moles of HCl = 0.050 L x 0.100 M = 0.05 moles HCl = 0.05 moles Cl(aq)−
Since the reaction runs to completion, and all species are in a 1:1 ratio, 0.05 moles of AgCl are produced (you can prove this to yourself using an ICE table)
Heat lost by reaction = heat gained by solution
Heat gain = m s T
= 100.0 g H2O x 4.18J
g °C− 1 × (23.40 °C– 22.60 °C)
= 330 J = Heat Loss, this is the heat evolved from the formation of 0.050 moles of AgCl
Therefore, the heat produced per mol of AgCl =
1330 16.60
0.050 1000
J kJkJ mol
mol AgCl J
Problem 2. Answer: -1370 kJ mol-1
Heat capacity of calorimeter = C
q=(360 s) (4000 J/s) = 1440 kJ
q = C ∆T = C (50.3 - 25.0) rearrange and solve C = 56.92 kJC-1
mol / kJ 1370mol) (0.100
C) 25.0)(27.4C kJ (56.92Q
1
rxn
Problem 3. Solution: -14.2 J
Absorbing heat is an endothermic process (positive)
Work done by the system on the surroundings is negative by convention, therefore
To determine the final concentrations, the first thing needed are the initial reactant concentrations and an expression for the reaction coefficient Q.
[A2(g)] = 1.00
0.250
mol
L= 4.00 M, [AB(g)] = [B2(g)] =
2.00
0.250
mol
L= 8.00 M,
and
2 2
2 2
(8.00)2.0
(4.00) (8.00)init
init init
ABQ
A B
The direction of the reaction needs to be determined. To do this, we compare Q vs K.
Since Q = 2.0 > K = 0.5, the reaction proceeds to the left.
To determine what final concentrations will be from initial concentrations, a handy tool – the Initial, Change, Equilibrium (ICE) table can be used.
All compounds involved in the reaction are included in an ICE table as follows, with the species that will be consumed on the left side, and the species that will be produced on the right side. Since the reaction proceeds to the left, A2 and B2 will be formed and AB will be consumed.
Concentration (M)
2 AB(g) A2(g) B2(g)
Initial 8.00 M 4.00 M 8.00 M
Change -2x +x +x
Equilibrium 8.00 – 2x 4.00 + x 8.00 + x
As the reaction proceeds, x moles of A2(g) and B2(g) are formed as 2x moles of AB(g) are consumed.
Make sure to consider stoichiometric coefficients appropriately.
The equilibrium values are simply the sum of the initial + change concentrations.
Substitute the equilibrium concentrations into K, and solve for x (remember that K is defined for the reaction in the way that it was initially described):
Solution: Pure solids and pure liquids and solvents are not included in the expression. Products go over Reactants, and coefficients in the equation are written as superscripts.
Problem 19. Solution:
Kc = [CO2(g)]
Remember: Solids and liquids are not included in the expression.
Problem 20. Solution: Equation 2 is equal the double and reverse of equation1 therefore
Kp2 = Kp1−2 = (
1
0.157) 2 = 40.6
Problem 21. Answer: Equilibrium Constant = K1−1/2
=1
√K1
Solution: The second equation if reversed (1/K) and halved (K1/2). Combining these two gives 1/K1/2.
Problem 22. Answer: The reaction equation is:
2 CO(g) + O2(g) ⇌ 2 CO2(g) Kc = 3.3 × 1091
The relationship between Kc and Kp is:
Kp = Kc(RT)∆n gasIn this case there are 3 moles of gas in the reactants and 2 moles of gas in the
Problem 30. Solution: Equal volumes of 0.1 M NaF and 0.1 M HF
Adding an acid and the salt of its conjugate base can form a buffer, this is the case in (b).
Problem 31. Solution: 1 and 3.
A buffer normally consists of a weak acid and its conjugate base in roughly equal amounts. The ratio should be no greater than 0.1 to 10 of weak base to weak acid. In 1, the ratio of acid to base is 3:1. In 3 the ratio is 4:1. In 2, there is a greater amount of strong base than acid, so all of the acetic acid is consumed so it is not a buffer. 4 is not a buffer solution since all of the weak acid is consumed with strong base, 5 is not a buffer because sodium acetate is a weak base, so you have a weak base with a strong base in solution.
Problem 32. Solution: B
This is a buffer solution, therefore:
2[ ] 0.400
log log(1.20 10 ) log 1.74[ ] 0.600
eq
eq
ApH pKa
HA
Problem 33. Solution: This is a buffer solution, therefore:
[ ]log
[ ]
eq
eq
ApH pKa
HA
= 3[ ]
4.75 log0.100
CH COONa
=4.75
Therefore 3[ ]log
0.100
CH COONa
=0 and 3[ ]
0.100
CH COONa
=1
[CH3COONa] = 0.10M
# moles CH3COONa = 0.10 moles
Problem 34. Solution: For this equilibrium [H+] = [In-] = 10-8 = 1 x 10-8
Ka = 1 x 10-6 = ][
]][[
HIn
InH
= ][
)10 x 1( 2-8
HIn
[HIn] = 1 x 10-10, therefore [HIn]/[In- ] = 1 x 10-10/1 x 10-8 =0.01 = 1/100
This is a buffer system as there are equal amounts of conjugate acid and base, but the addition of H+ can change the pH slightly. The original pH of the buffer is approximately equal to pKa. When H+ ions are added from the strong acid HCl, A- is converted into HA. Therefore the addition of 0.01 moles H+ produces 0.01 moles HA and consumes 0.01 moles A-.
nHA= 0.5 + 0.01 = 0.51 mol
nA- = 0.5 – 0.01 = 0.49 mol
[H+]=Ka x nHA/nA- = (1.8 x 10-4) (0.51/0.49) = 1.873 x 10-4
pH = -log (1.873 x 10-4) = 3.727
Problem 37. Solution: C
There are more moles of carbonate buffer in solution c than any of the other solutions
Problem 38. Solution: E
The solution is a buffer with pH above 7. A buffer is resistant to both addition of strong acid and strong base and the concentration of the hydronium ion is not more than the hydroxide ion (pH >7).
A buffer normally consists of a weak acid and its conjugate base in roughly equal amounts. The ratio should be no greater than 0.1 to 10 of weak base to weak acid. In 1, the ratio of acid to base is 3:1. In 3 the ratio is 4:1. In 2, there is a greater amount of strong base than acid.
Problem 41. Solution: C or E
In order to determine which solutions are able to act as buffer solutions, determine what ions will be found in the solution and whether those ions are acidic, basic or spectator ions. If acidic and/or basic ions are found, calculate the amount of ions that are present. For solution A) all ions present (H+, NO3-, and Na+) are spectator ions. No buffer abilities possible. For solution B) all ions present (Na+, OH- and Cl-) are spectator ions. No buffer abilities possible. For solution C) This is a 1:1 ratio of conjugate acid to its conjugate base. This IS a buffer. For solution D) the HCl completely dissociates to form H+ and Cl- ions. NH3 can form an equilibrium where NH3 + H+ ↔ NH4+. However, HCl and its dissociated ions are present in 0.01mol amounts—the same as the amount of NH3. Therefore, all the NH3 is “used up” in reacting with the HCl, so no buffering abilities are possible. For solution E) the NaOH completely dissociates to form Na+ and OH- ions. The 0.004mol of OH- will react with the 0.01mol of HF to form 0.004mol of F- and have 0.006mol left of HF since the OH- is the limiting reagent. This means we have HA and A- both present in a ratio of 3:2. This is a buffer.
Problem 42. Solution: C
B + H2O ↔ BH+ + OH-
Calculate from the given pH, the concentration of OH- ions that dissociate form from the reaction of the base with water.
pOH = 14 – pH = 14 – (8.88) = 5.12
[OH-] = 10-pOH = 10-5.12 = 7.585 x 10-6M
Assume 1 L of solution, therefore the [B] = 0.40mol/L and [BH+] = 0.250mol/L.
Because equal volumes of the acid and weak base are being mixed, all concentrations (M) can be treated as moles (mol). HNO3 is a strong acid so it completely dissociates
HNO3 H+ + NO3-
0.1 0.1 0.1
NH3 will react with the H+ released by the HNO3 to form NH4+. Initially there is 0.3M or 0.3moles of NH3. Upon addition of 0.1moles H+ (from the HNO3), 0.1mol of NH3 will react to form 0.1mol of NH4+, leaving 0.2mol NH3 unreacted.
Therefore,
NH3 + H+ ↔ NH4+
Using the following equation, solve for [H+].
[H+] = Ka x [HA]
[A-]
Kw = Ka x Kb so that Ka = Kw = 1.0 x 10-14 = 5.56 x 10-10
Write out the equations that are occurring in the solution described above. Because both solutes are being added to 1L of water, all concentrations (M) can be treated as moles (mol).
C2H5COONa is a soluble salt so it completely dissociates
C2H5COONa C2H5COO- + Na+
0.1 0.1 0.1
HCl is a strong acid so it completely dissociates
HCl H+ + Cl-
0.01 0.01 0.01
C2H5COO- will react will all the H+ to form C2H5COOH. Initially there is 0.1 mole of C2H5COO-. When 0.01mol of H+ is added, the C2H5COO- reacts leaving 0.09mol. There is also 0.1mol of C2H5COOH to start, but when the C2H5COO- reacts with the H+, it forms 0.01mol more C2H5COOH so that the total amount of C2H5COOH is 0.11mol.
C2H5COOH ↔ C2H5COO- + H+
Before HCl is added 0.1 0.1
After HCl is added 0.11 0.09
Calculate [H+] using the following equation. (n = moles)
[H+] = Ka x nHA = 1.41 x 10-5 x (0.11) = 1.72 x 10-5M
In the reaction above, Sn(aq)2+ is being oxidized to Sn(aq)
4+ (it lost 2 e−) and Fe(aq)3+ is being reduced to
Fe(aq)2+ (gaining 1 e−). Ecell = Ereduction + Eoxidation. Since all E values are given as Ereduction values,
the Ered value for Fe in the reaction is +0.77 V. The E value for Sn in the reaction must be reversed because Sn is undergoing oxidation, therefore Eox = −0.15V.
Calculate the Ecell for all three reactions. In statement I), Ecell= -1.86V (Cd(aq)2+ is being reduced and
Cl2(g) is being oxidized). In statement II), Ecell= -0.29V (Sn(aq)2+ is being oxidized and reduced into
Sn(aq)4+ and Sn respectively). In statement II) Ecell = 0.91V (Sn is being oxidized and Fe3+ is being
reduced). Ecell values that are positive means that the redox reactions will occur spontaneously, while a negative Ecell value means the redox reaction is not spontaneous.
Problem 58. Answer: I, II, and III
Solution: Statement I) is true becauseCu2+ has a more positive Ereduction value than Cr3+, which means that Cu2+ is a better oxidizing agent than Cr3+ (remember that an oxidizing agent oxidizes other substances and becomes reduced in the process).
Problem 59. Answer: -0.26
In the reaction given above, Ag(aq)+ is being reduced to Ag while Ni is being oxidized to Ni(aq)
2+ . The
overall Ecello for the reaction is calculated as
Ecello = Ereduction + Eoxidation
The reduction half of the reaction is given as + 0.80V. In order to calculate the E value for the oxidation half, rearrange the Ecell
o equation to solve for Eoxidation.
Eoxidation = Ecello − Ereduction = (+1.06V)– (+0.80V) = +0.26V. The Eoxidation value is the opposite
sign from the Ereduction value, so in order to solve for the Ereductionn value as asked, the sign on the
Eoxidation value must be reversed. Therefore, Ereduction for Ni(aq)2+ = −0.26V
Increasing the temperature increases the kinetic energy of the molecules in the liquid causing the molecules to move more quickly and increase the vapor pressure.
Solution: To determine which gas has the highest density, consider the equation
Density =molar mass (
gmol
) × pressure(kPa)
R(gas constant) × temperature(K)
If measured at the same pressure and temperature, the gas that will have the highest density will have the largest molar mass. The molar masses of the gases are as follows: F2, 38g/mol; C2H6, 30.08g/mol; H2S, 34.086g/mol; NO, 30.01g/mol and SiH4, 32.13g/mol. Therefore, F2 will have the highest density because it has the largest molar mass.
Problem 85. Answer: D
Solution: PV = nRT 1 1 2 2
1 2
PV PV
T T
2
(1.00 )(600 ) (2.00 )(1200 )
280
atm mL atm mL
K T
T2 = 847C
Problem 86. Solution:
We need to determine what mass is in the gas phase, and the rest must be in the liquid phase. We make the assumption that the liquid takes up a negligible volume of the container.
The pressure of the water vapour will be 233.7mmHg
P=1 atm / 760mmHg * 233.7 mmHg = 0.3075 atm
Now calculate the moles of gas:
n =PV
RT =
0.3075 atm ∗ 0.5 L
0.08206 L atm mol − 1 K − 1 ∗ (273.15 + 70)K= 0.0055 𝑚𝑜𝑙𝑒
mass of gaseous water is:
18g
mol× 0.0055 mol = 0.098 g
So the mass of liquid water is 0.5-0.098 = 0.402 g.