CHM1 19AnswersEoCh Collection

S = 2.05E–5 mol/0.100 L = 2.05E-4 M

K_s = [Ca²⁺][F^–]² = (S)(2S)² = 4 × (2.05E–4)³ = 3.44E–11

4. The equation for the dissolution is

_La(IO3)₃ La³⁺ + 3 IO₃^–

La³⁺ + 3 IO₃^–

If the solubility is S, then the equilibrium concentrations of the ions will be
[La³⁺] = S and [IO₃^–] = 3S. Then K_s = [La³⁺][IO₃^–]³ = S(3S)³ = 27S⁴

27S⁴ = 6.2 × 10^–12, S = ( ( 6.2 ÷ 27) × 10^–12 )^¼ = 6.92 × 10^–4 M

[IO₃^–] = 3S = 2.08 × 10^–5 (M)

5. (a) In pure water, K_s = [Sr²⁺][SO₄^2–] = S²

S = ?K_s = (2.8 × 10^–7)^½ = 5.3 × 10^–4

(b) In 0.10 mol L^–1 Na₂SO₄, we have

K_s = [Sr²⁺][SO₄^2–] = S × (0.10 + S) = 2.8 × 10^–7

Because S is negligible compared to 0.10 M, we make the approximation

K_s = [Sr²⁺][SO₄^2–] ? S × (0.10 <underbar>M</underbar> ) = 2.8 × 10^–7

so S ? (2.8 × 10^–7) / 0.10M = 2.8 × 10^–6 M — which is roughly 100 times smaller than the result from (a)6. As with most real-world problems, this is best approached as a series of smaller problems, making simplifying approximations as appropriate.

Volume of treated water: 1000 L + 10 L = 1010 L

Concentration of OH^– on addition to 1000 L of pure water:
(4 M) × (10 L)/(1010 L) = .040 M

Initial concentration of Cd²⁺ in 1010 L of water:
(1.6E–5 M) x (100/101) ? 1.6E–5 M

The easiest way to tackle this is to start by assuming that a stoichiometric quantity of Cd(OH)₂ is formed — that is, all of the Cd²⁺ gets precipitated.

 

Concentrations	[Cd2+], M	[OH–], M
initial	1.6E–5	0.04
change	–1.6E–5	–3.2E–5
final:	0	.04 – 3.2E–5 ? .04

Now "turn on the equilibrium" — find the concentration of Cd²⁺ that can exist in a 0.04M OH^– solution:

Concentrations	[Cd2+], M	[OH–], M
initial	o	0.04
change	+x	+2x
at equilibrium	x	.04 + 2x ? .04

Substitute these values into the solubility product expression:

Cd(OH)₂(s) = [Cd²⁺] [OH^–]² = 2.5E–14

[Cd²⁺] = (2.5E–14) / (16E–4) = 1.6E–13 M

Note that the effluent will now be very alkaline: pH = 14 + log .04 = 12.6,
so in order to meet environmental standards an equivalent quantity of strong acid must be added to neutralize the water before it is released.

7. (d) 

8. The silver ion concentrations required to precipitate the two salts are found by substituting into the appropriate solubility product expressions:

to precipitate AgCl: [Ag⁺] = 1.8E-10 / .001 = 1.8E-7 M

to precipitate Ag₂CrO₄: [Ag⁺] = (2.0E-12 / .01)^½ = 1.4E–5 M

The first solid to form as the concentration of Ag⁺ increases will be AgCl. Eventually the Ag⁺ concentration reaches 1.4E-5 M and Ag₂CrO₄ begins to precipitate. At this point the concentration of chloride ion in the solution will be 1.3E-5 M which is about 13% of the amount originally present.