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})

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