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Question 1

Acid-Base Indicators

Answer 1

• The endpoint of a titration is the point at which the color of the indicator changes.
• Table 17.3

Question 2

Solubility Equillibria: Solubility Product Expression and K_sp

Answer 2

• Quantitative predictions about how much of a given ionic compound will dissolve in water is possible with the solubility product constant, K_sp.
• AgCl(s) <> Ag+(aq) + Cl-(aq); K_sp = [Ag+][Cl-]

Question 3

Calculations involving K_sp and Solubility

Answer 3

• Molar solubility is the number of moles of solute in 1L of a saturated solution (mol/L)
• Solubility is the number of grams of solute in 1L of a saturated solution (g/L)
• To calculate a compound’s molar solubility:
  1. Construct an equilibrium table
  2. Fill in what is known
  3. Determine the unknowns

Question 4

The K_sp of silver bromide is 7.7*10^-13. Calculate the molar solubility.

Answer 4

AgBr(s) <> Ag+(aq) + Br-(aq)
I (M). 0. 0
C (M). +s. +s
E (M). s. s

K_sp = [Ag+][Br-] = 7.710^-13
s^2 = 7.710^-13
s = 8.810^-7M
(8.810^-7mol AgBr/ 1L) * (187.8g/1mol AgBr) = 1.7*10^-4g/L

Question 5

Calculate the solubility of copper(II) hydroxide [Cu(OH)2] in g/L.

Answer 5

Cu(OH)2(s) <> Cu2+(aq) + 2OH-(aq)
0. 0
+s. +2s
s. 2s
K_sp = [Cu2+][OH-]^2
K_sp of Cu(OH)2 = 2.210^-20
Molar mass = 97.57g/mol
(s)(2s)^2 = 2.210^-20 = 4s^3
S = 1.810^-7M
Solubility of Cu(OH)2 = (1.810^-7mol/L) * (97.57g/1mol)
= 1.7*10^-5g/L

Question 6

Predicting Precipitation Reactions

Answer 6

• For the dissociation of an ionic solid in water, the following conditions may exist:
  1. The solution is unsaturated
  2. The solution is saturated
  3. The solution is supersaturated
• The following relationships are useful in making predictions on when a precipitate might form.
  QK_sp; a precipitate forms

Question 7

Example: Predict whether a precipitate will form when each of the following is added to 650mL of 0.080M K2SO4:
A. 250mL of 0.0040M BaCl2
B. 175 mL of 0.15M AgNO3
C. 325mL of 0.25M Sr(NO3)2
(Assume volumes are additive)

Answer 7

A. BaSO4 K_sp = 1.110^-10
[Ba2+] = (250mL0.0040M)/(650mL+250mL) = 0.0011M
[SO42-] = (650mL0.0080M)/(650mL+250mL) = 0.0058M
Q = [Ba2+][SO42-] = (0.0011)(0.0058) = 6.410^-6

Question 8

Factors affecting solubility

Answer 8

Several factors exist that affect the solubility of ionic ocmpounds:

• Common ion effect
• pH
• Formation of complex ion

Question 9

Example: Calculate the molar solubility of silver chloride in a solution that 6.5*10^-3M

Answer 9

AgCl(s) <> Ag+(aq) + Cl-(aq)
I. 6.510^-3. 0
C. +s. +s
E. 6.510^-3 +s. s
K_sp = 1.610-10 = [Ag+][Cl-]
1.610^-10 = (6.510^-3+a)(s)
1.610^-10 - 6.510^-3 * s
S = (1.610^-10)/(6.510^-3) = 2.510^-8
H2O: square root (1.610^10) = 1.310^-5M

Question 10

pH

Answer 10

Mg(OH)2(s) <> Mg2+(aq) + 2OH-(aq)
2H+(aq) + 2OH-(aq) > 2H2O (l)
Overall: Mg(OH)2(s) + 2H+(aq) <> Mg2+(aq) + 2H2O(l)
K_sp = [Mg2+][OH-]^2 = 1.2*10^-11
s(2s)^2 = 4s^3 = 1.2*10^-11
s = 1.4*10^-4M
At equilibrium:
[OH]- = 2(1.4*10^-4M) = 2.8 * 10^-4M
pOH = -log(2.8*10^-4) = 3.55
pH = 14.00-3.55 = 10.45
In a solution with a pH of less than 10.45, the solubility of Mg(OH)2 increases
If the pH of the medium were higher than 10.45, [OH-] would be higher and the solubility of Mg(OH)2, would decrease because of the common ion (OH-) effect.

Question 11

pH

Answer 11

BaF2(s) <> Ba2+(aq) + 2F-(aq)
2H+(aq) + 2F-(aq) > 2HF(aq)
Overall: BaF2(s) + 2H+(aq) <> Ba2+(aq) +2HF(aq)
As the concentration of F- decreases, the concentration of Ba2+ must increase so statistics the equality: K_sp = [Ba2+][HF]^2

Question 12

Example: Which of the following compounds will be more soluble in acidic solution than in water:
A. CuS
B. AgCl
C. PbSO4

Answer 12

A. CuS(s) <> Cu2+(aq) + S2-(aq)
Note: S2- is CB of WA (HS-)
> More soluble
B. AgCl(s) <> Ag+(aq) + Cl-(aq)
> No more or less soluble
C. PbSO4(s) <> Pb2+(aq) + SO42-(aq)
Note: SO42- is CB of WA (HSO4-)
> More soluble