Some additional molar solubility problem

Solving Ksp Problems: Part One - s2

Solving Ksp Problems: Part Two - 4s3

Solving Ksp Problems: Part Three - 27s4

Solving Ksp Problems: Part Four - 108s5

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Request: if you came here first without studying any of the other problems (or knowing what a K_sp expression is), I'd like to recommend you go study those problems first.

Problem #1: The K_sp of Sn₃(BO₃)₄ is 6.128 x 10¯¹². Determine the molar solubility.

Solution:

1) Write the dissociation equation:

Sn₃(BO₃)₄(s) ⇌ 3Sn⁴⁺(aq) + 4BO₃³¯(aq)

2) Write the K_sp expression:

K_sp = [Sn⁴⁺]³ [BO₃³¯]⁴

3) Solve for the molar solubility:

6.128 x 10¯¹² = (3s)³ (4s)⁴

6.128 x 10¯¹² = (27s³) (256s⁴)

6.128 x 10¯¹² = 6912s⁷

s = 0.007074 M

Problem #2: Given that the K_sp value for MX, M₂X and MX₃ are the same, determine the solubility order.

(a) MX > M₂X > MX₃
(b) MX₃ > M₂X > MX
(c) M₂X > MX₃ > MX
(d) MX > MX₃ > M₂X

Solution:

1) Since the K_sp values for all three are the same, I will assume a K_sp value to use. Since K_sp values are quite small, I should pick a reasonable value. In other words:

K_sp = 1 x 10¯⁶ is reasonable

K_sp = 1 is not reasonable, because it is too large compared to what actual K_sp values are.

2) What I need to do is calculate each substance's molar solubility:

MX ---> s₁² = 1.00 x 10¯⁶

M₂X ---> 4s₂³ = 1.00 x 10¯⁶

MX₃ ---> 27s₃⁴ = 1.00 x 10¯⁶

If you need to review why the solutions just above are set up the way they are, please go here. The link take you to the explanation for MX. Contained within that file are links to explanations concerning M₂X and MX₃.

3) The answers:

s₁ = 0.00100 M <--- molar solubility for MX

s₂ = 0.00630 M <--- molar solubility for M₂X

s₃ = 0.0139 M <--- molar solubility for MX₃

Answer choice (b) is the correct answer.

4) Sometimes, the answer choices use a 'less than' sign rather than a 'greater than' sign. If the were the case, here are the answer choices:

(a) MX < M₂X < MX₃
(b) MX₃ < M₂X < MX
(c) M₂X < MX₃ < MX
(d) MX < MX₃ < M₂X

If these were the answers, then choice (a) is the correct answer.

Problem #3: How much water is required to dissolve 25.2 mg of CaF₂?

Solution:

1) We need to look up the K_sp of CaF₂:

3.9 x 10¯¹¹

Found here.

2) We need to know the solubility of CaF₂ in mg/L. To do that, first calculate the molar solubility of CaF₂, then convert to g/L and on to mg/L:

CaF₂(s) ⇌ Ca²⁺(aq) + 2F¯(aq)

K_sp = [Ca²⁺] [F¯]²

3.9 x 10¯¹¹ = (s) (2s)²

4s³ = 3.9 x 10¯¹¹

s = 0.000213633 M

(0.000213633 mol/L) (78.074 g/mol) = 0.01668 g/L

(0.01668 g/L) (1000 mg / 1 g) = 16.68 mg/L

The solution to the g/L value is also given in the link given above.

3) A simple ratio and proportion leads us to the answer:

25.2 mg		16.68 mg
–––––––	=	–––––––
x		1 L

x = 1.5 L (answer to two sig figs, based on the K_sp)

Problem #4: Two forms of barium oxalate (called A and B just below) have different K_sp values at 18 °C:

(A)  BaC2O4 ⋅ 2H2O      1.2 x 10¯7

(B)  BaC2O4 ⋅ 1⁄2H2O     2.2 x 10¯7

Describe what will happen if 1.0 g of A and 1.0 g of B are added to 100. mL of water under 1.00 atm of pressure.

Solution:

1) Write the chemical equation and the K_sp expression:

BaC₂O₄(s) ⇌ Ba²⁺(aq) + C₂O₄²¯(aq)

K_sp = [Ba²⁺] [C₂O₄²¯]

2) The 1.0 g of A and the 1.0 g of B are added to the water and the entire system is stirred. What happens?

The instant before anything dissolves and ionizes, both A and B "see" pure water, so each starts dissolving.

A tiny amount of A and a tiny amount of B dissolve, then ionize. This results in a value for [Ba²⁺] and a value for [C₂O₄²¯].

However, let us stipulate that Q (the reaction quotient) that results from these tiny amounts of A and B that dissolve is less than the K_sp.

Both A and B "see" a solution that can accept more dissolved ions. This means a tiny bit more BaC₂O₄ (some from A and some from B) dissolves, then ionizes, thus pushing the Q to a larger value, getting it even closer to the K_sp.

More of A and B dissolve and ionize, until Q equals the K_sp. At this point (where Q = K_sp), an equilibrium is established. This equilibrium is between the undissolved A, the undissolved B, and the ions in solution. The values of [Ba²⁺] and [C₂O₄²¯] will stop increasing and will remain constant.

3) However, which K_sp value has been reached at equilibrium? The answer is the K_sp value for A, the less soluble form of barium oxalate.

Let us take the view of B, the more soluble form. B "sees" a solution that can accept more ions since its K_sp has not been reached. So, some of B dissolves and ionizes.

However, A now "sees" a solution where its K_sp has been exceeded. As a result, some ions reform into the solid state and the K_sp value for A is restored.

4) Let us assume that equal amounts of A and B dissolve to produce the K_sp value of A. One can perform a calculation to determine the tiny amount of A and of B that do dissolve. This exercise is left to the reader.

5) Calcium sulfate also shows this difference in solubilities between two hydrates. Barium iodate shows the difference between the anhydrate and one hydrate (as does calcium iodate). Magnesium carbonate (and strontium iodate) show differences in solubility between two hydrates and the anhydrate.

Solving Ksp Problems: Part One - s2

Solving Ksp Problems: Part Two - 4s3

Solving Ksp Problems: Part Three - 27s4

Solving Ksp Problems: Part Four - 108s5

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