Solving K _{sp}Problems:

Part One - s^{2}Solving K _{sp}Problems:

Part Two - 4s^{3}Solving K _{sp}Problems:

Part Three - 27s^{4}Solving K _{sp}Problems:

Part Four - 108s^{5}Back to Equilibrium Menu

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 Sn_{3}(BO_{3})_{4} is 6.128 x 10¯^{12}. Determine the molar solubility.

**Solution:**

1) Write the dissociation equation:

Sn_{3}(BO_{3})_{4}(s) ⇌ 3Sn^{4+}(aq) + 4BO_{3}^{3}¯(aq)

2) Write the K_{sp} expression:

K_{sp}= [Sn^{4+}]^{3}[BO_{3}^{3}¯]^{4}

3) Solve for the molar solubility:

6.128 x 10¯Note: three tin(IV) ions are produced in solution for every one Sn^{12}= (3s)^{3}(4s)^{4}6.128 x 10¯

^{12}= (27s^{3}) (256s^{4})6.128 x 10¯

^{12}= 6912s^{7}s = 0.007074 M

**Problem #2:** Given that the K_{sp} value for MX, M_{2}X and MX_{3} are the same, determine the solubility order.

(a) MX > M_{2}X > MX_{3}

(b) MX_{3}> M_{2}X > MX

(c) M_{2}X > MX_{3}> MX

(d) MX > MX_{3}> M_{2}X

**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 reasonableK

_{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 ---> s_{1}^{2}= 1.00 x 10¯^{6}M

_{2}X ---> 4s_{2}^{3}= 1.00 x 10¯^{6}MX

_{3}---> 27s_{3}^{4}= 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 M

_{2}X and MX_{3}.

3) The answers:

s_{1}= 0.00100 M <--- molar solubility for MXs

_{2}= 0.00630 M <--- molar solubility for M_{2}Xs

_{3}= 0.0139 M <--- molar solubility for MX_{3}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 < M_{2}X < MX_{3}

(b) MX_{3}< M_{2}X < MX

(c) M_{2}X < MX_{3}< MX

(d) MX < MX_{3}< M_{2}XIf these were the answers, then choice (a) is the correct answer.

**Problem #3:** How much water is required to dissolve 25.2 mg of CaF_{2}?

**Solution:**

1) We need to look up the K_{sp} of CaF_{2}:

3.9 x 10¯^{11}Found here.

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

CaF_{2}(s) ⇌ Ca^{2+}(aq) + 2F¯(aq)K

_{sp}= [Ca^{2+}] [F¯]^{2}3.9 x 10¯

^{11}= (s) (2s)^{2}4s

^{3}= 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) BaC _{2}O_{4}⋅2H_{2}O 1.2 x 10¯^{7}(B) BaC _{2}O_{4}⋅^{1}⁄_{2}H_{2}O 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:

BaC_{2}O_{4}(s) ⇌ Ba^{2+}(aq) + C_{2}O_{4}^{2}¯(aq)K

_{sp}= [Ba^{2+}] [C_{2}O_{4}^{2}¯]

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 [Ba

^{2+}] and a value for [C_{2}O_{4}^{2}¯].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 BaC

_{2}O_{4}(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 [Ba^{2+}] and [C_{2}O_{4}^{2}¯] 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 K _{sp}Problems:

Part One - s^{2}Solving K _{sp}Problems:

Part Two - 4s^{3}Solving K _{sp}Problems:

Part Three - 27s^{4}Solving K _{sp}Problems:

Part Four - 108s^{5}Back to Equilibrium Menu