Fifteen Examples

I added other gram-type solubility units such as g/L or g/250mL. No matter what the unit is, it will have to be converted to molar solubility.

The general format of this type of problem is:

"Determine the K_{sp}of ____, given a solubility of ____ g/100mL (or g/L, etc).

When the solubility is given in grams per 100mL, (rather than molar solubility, which is in mol/L), the K_{sp} can still be calculated, albeit with an additional step or two. We have to first convert the g/100mL data to the corresponding mol/L (molar) solubility value. Then the molar solubility value is used for the remainder of the calculation.

Here is how to convert a g/100mL value to molar solubility:

(1) multiply the g/100mL value by 10/10. This converts it to g / 1000 mL or, more commonly, g / L. (Sometimes the data is given in g/L. When that happens, this step is skipped.)(2) divide the g/L value by the molar mass of the substance. This gives moles / L, which is molar solubility.

After the above conversion, the problem becomes calculate the K_{sp} from molar solubility data.

In these types of problem, the 100mL in the unit never sets the number of significant figures. It is part of the unit, not a value in the calculations. Use the amount that dissolves, such as, for example, 2.97 x 10¯^{10} g/100mL in determining significant figures.

**Example #1:** Determine the K_{sp} of nickel sulfide (NiS), given that its solubility is 2.97 x 10¯^{10} g / 100mL.

**Solution:**

1) Convert to g per 1000 mL, then moles per liter:

(2.97 x 10¯^{10}g / 100mL) (10/10) = 2.97 x 10¯^{9}g / 1000mL(2.97 x 10¯

^{9}g / L) / 90.77 g/mol = 3.27 x 10¯^{11}mol/L

2) When NiS dissolves, it dissociates like this:

NiS(s) ⇌ Ni^{2+}(aq) + S^{2}¯(aq)

3) The K_{sp} expression is:

K_{sp}= [Ni^{2+}] [S^{2}¯]

4) There is a 1:1 ratio between NiS and Ni^{2+} and there is a 1:1 ratio between NiS and S^{2}¯. This means that, when 3.27 x 10¯^{11} mole per liter of NiS dissolves, it produces 3.27 x 10¯^{11} mole per liter of Ni^{2+} and 3.27 x 10¯^{11} mole per liter of S^{2}¯ in solution.

5) Putting the values into the K_{sp} expression, we obtain:

K_{sp}= (3.27 x 10¯^{11}) (3.27 x 10¯^{11}) = 1.07 x 10¯^{21}

**Example #2:** Determine the K_{sp} of magnesium fluoride (MgF_{2}), given that its solubility is 1.65 x 10¯^{3} g / 100mL.

**Solution:**

1) Convert to g 1000 mL, then moles /L:

(1.65 x 10¯^{3}g / 100mL) (10/10) = 1.65 x 10¯^{2}g / 1000mL(1.65 x 10¯

^{2}g / L) / 62.30 g/mol = 2.65 x 10¯^{4}mol/L

2) When MgF_{2} dissolves, it dissociates like this:

MgF_{2}(s) ⇌ Mg^{2+}(aq) + 2F¯(aq)

3) The K_{sp} expression is:

K_{sp}= [Mg^{2+}] [F¯]^{2}

4) There is a 1:1 ratio between MgF_{2} and Mg^{2+}, BUT there is a 1:2 ratio between MgF_{2} and F¯. This means that, when 2.65 x 10¯^{4} mole per liter of MgF_{2} dissolves, it produces 2.65 x 10¯^{4} mole per liter of Mg^{2+} and 5.30 x 10¯^{4} mole per liter of F¯ in solution.

5) Putting the values into the K_{sp} expression, we obtain:

K_{sp}= (2.65 x 10¯^{4}) (5.30 x 10¯^{4})^{2}= 7.44 x 10¯^{11}

6) Please note, I __DID__ __NOT__ double the F¯ concentration. I took the MgF_{2} concentration (its molar solubility) and doubled it to get the F¯ concentration. This is because of the 1:2 molar ratio between MgF_{2} and F¯.

**Example #3:** Determine the K_{sp} of manganese(II) iodate [Mn(IO_{3})_{2}], given that its solubility is 1.935 x 10¯^{1} g/100mL.

**Solution:**

1) Convert to g per 1000 mL, then moles per liter:

(1.935 x 10¯^{1}g / 100mL) (10/10) = 1.935 g / 1000mL1.935 g / L / 404.746 g/mol = 4.78 x 10¯

^{3}mol/L

2) When Mn(IO_{3})_{2} dissolves, it dissociates like this:

Mn(IO_{3})_{2}(s) ⇌ Mn^{2+}(aq) + 2IO_{3}¯(aq)

3) The K_{sp} expression is:

K_{sp}= [Mn^{2+}] [IO_{3}¯]^{2}

4) There is a 1:1 ratio between Mn(IO_{3})_{2} and Mn^{2+}, BUT there is a 1:2 ratio between Mn(IO_{3})_{2} and IO_{3}¯. This means that, when 4.78 x 10¯^{3} mole per liter of Mn(IO_{3})_{2} dissolves, it produces 4.78 x 10¯^{3} mole per liter of Mn^{2+} and 9.56 x 10¯^{3} mole per liter of IO_{3}¯ in solution.

5) Putting the values into the K_{sp} expression, we obtain:

K_{sp}= (4.78 x 10¯^{3}) (9.56 x 10¯^{3})^{2}= 4.37 x 10¯^{7}

**Example #4:** Determine the K_{sp} of calcium arsenate [Ca_{3}(AsO_{4})_{2}], given that its solubility is 0.13 g/L.

**Solution:**

1) Since the solubility is already in g/L, we can proceed directly to calcuating the solubility in moles per liter:

0.13 g / L / 398.078 g/mol = 3.2657 x 10¯^{4}mol/L

2) The K_{sp} expression is:

K_{sp}= [Ca^{2+}]^{3}[AsO_{4}^{2}¯]^{2}

3) Putting concentration values into the K_{sp} expression, we obtain:

K_{sp}= (9.7971 x 10¯^{4})^{3}(6.5314 x 10¯^{4})^{2}= 4.01 x 10¯^{18}

I used the Wikipedia entry for calcium arsenate, where the solubility is given as 0.013 g/100mL.

**Example #5:** A saturated solution of CaF_{2} contains 0.00680 g/250mL of solution. Find the K_{sp} of CaF_{2}.

**Solution:**

1) Calculate the grams of CaF_{2} in one liter:

(0.00680 g/250mL) (4) = 0.0272 g/LComment: there are four 250mL amounts in 1 liter

2) Convert g/L to mol/L:

0.0272 g/L divided by 78.074 g/mol = 3.48387 x 10¯^{4}mol/L (a few guard digits)

3) Set up K_{sp} expression and solve for it:

K_{sp}= [Ca^{2+}] [F¯]^{2}K

_{sp}= (3.48387 x 10¯^{4}) (6.96774 x 10¯^{4})^{2}K

_{sp}= 1.69 x 10¯^{10}

**Example #6:** 300. mL of a saturated solution of Cu(IO_{4})_{2} contains 0.380 grams of dissolved salt. Determine the K_{sp}.

**Solution:**

1) Calculate the grams of Cu(IO_{4})_{2} in one liter:

0.380 g / 0.300 L = x / 1.00 Lx = 1.267 g/L

2) Convert g/L to mol/L

1.267 g/L / 445.338 g/mol = 0.002845 mol/L

3) Plug into K_{sp} expression for Cu(IO_{4})_{2}:

K_{sp}= [0.002845] [0.005690]^{2}K

_{sp}= 9.21 x 10¯^{8}

Given the solubility in g / 100 mL, calculate the K_{sp}

7) PbBr_{2}; 1.04 x 10¯^{8} g/100mL

8) ZnS; 5.28 x 10¯^{12} g/100mL

9) Cd_{3}(PO_{4})_{2}; 6.25 x 10¯^{6} g/100mL

Answers only. No detailed solutions

**Example #10:** A saturated solution of BaSO_{4} contains 0.000245 g/100mL Calculate its K_{sp}.

**Solution:**

1) Calculate the grams of BaSO_{4} in one liter:

(0.000245 g/100mL) (10 100mL/L) = 0.00245 g/L

2) Convert g/L to mol/L:

0.00245 g/L / 233.391 g/mol = 0.0000104974 mol/L <--- keep a few guard digits

3) BaSO_{4} dissociates as follows:

BaSO_{4}(s) ⇌ Ba^{2+}(aq) + SO_{4}^{2}¯(aq)

4) Write the K_{sp} expression for BaSO_{4} and calculate the K_{sp}:

K_{sp}= [Ba^{2+}] [SO_{4}^{2}¯]K

_{sp}= (0.0000104974) (0.0000104974)K

_{sp}= 1.10 x 10¯^{10}

**Example #11:** The solubility of Ba_{3}(PO_{4})_{2} is 7.571 x 10¯^{4} g/100mL of water at 25 °C. Determine the K_{sp} of Ba_{3}(PO_{4})_{2} at this temperature.

**Solution:**

1) Grams in one liter:

(7.571 x 10¯^{4}g/100mL) (10 100mL/L) = 7.571 x 10¯^{3}g/L

2) Convert g/L to mol/L:

7.751 x 10¯^{3}g/L / 601.93 g/mol = 1.25779 x 10¯^{5}mol/L <--- keep a few guard digits

3) Write the K_{sp} expression for Ba_{3}(PO_{4})_{2} and calculate the K_{sp}:

K_{sp}= [Ba^{2+}]^{3}[PO_{4}^{3}¯]^{2}<--- I'll skip writing the chemical equation.[Ba

^{2+}] = (1.25779 x 10¯^{5}) (3) = 0.0000377337 M

[PO_{4}^{3}¯] = (1.25779 x 10¯^{5}) (2) = 0.0000251558 MK

_{sp}= (0.0000377337)^{3}(0.0000251558)^{2}= 3.40 x 10¯^{23}

This source is where I got the K_{sp} value and then I back-calculated to get the g/100mL value to start the problem. Pretty smart, huh?

**Example #12:** A saturated solution of mercury(I) bromide has a solubility of 0.039 mg per liter at 25 °C. Determine the K_{sp}.

**Solution:**

1) Convert mg/L to moles/L

0.039 mg 1 g 1 mol ––––––– x ––––––– x ––––––– = 6.952 x 10¯ ^{8}mol/L1 L 1000 mg 560.988 g

2) Write the dissociation equation and the K_{sp} expression:

Hg_{2}Br_{2}(s) ⇌ Hg_{2}^{2+}(aq) + 2Br¯K

_{sp}= [Hg_{2}^{2+}] [Br¯]^{2}

3) Calculate the K_{sp}:

K_{sp}= (6.952 x 10¯^{8}) (1.3904 x 10¯^{7})^{2}= 1.3 x 10¯^{21}

**Example #13:** The solubility of CaSO_{4} (MM = 136.04 g/mol) is 0.955 g/L. Calculate the K_{sp} of CaSO_{4}.

**Solution:**

1) Write the chemical equation for the dissolving of calcium sulfate and write the K_{sp} expression.

CaSO_{4}(s) ⇌ Ca^{2+}(aq) + SO_{4}^{2+}(aq)K

_{sp}= [Ca^{2+}] [SO_{4}^{2+}]

2) Convert g/L to moles/L:

(0.955 g/L) (1.00 mol / 136.04 g) = 0.00701999 mol/L

3) Determine the K_{sp}:

K_{sp}= (0.00701999) (0.00701999) = 4.93 x 10¯^{5}

**Example #14:** Calculate the mass of BaSO_{4} that can dissolve in 100.0 mL of solution. The K_{sp} value for BaSO_{4} is 1.5 x 10¯^{9}. The MM for barium sulfate is 233.391 g/mol.

**Solution:**

1) Write the chemical equation for the dissolving of barium sulfate. Then, write the K_{sp} expression for barium sulfate.

BaSO_{4}(s) ⇌ Ba^{2+}(aq) + SO_{4}^{2}¯(aq)K

_{sp}= [Ba^{2+}] [SO_{4}^{2}¯]

2) Solve the K_{sp} expression for the molar solubility of BaSO_{4}:

1.5 x 10¯^{9}= (s) (s)s = 0.00003873 M

3) Convert molar solubility to grams per liter:

(0.00003873 mol/L) (233.391 g/mol) = 0.0090392 g/L

4) Convert to g/100mL. There are ten 100mL portions in one liter:

(0.00904 g/L) (1.00 L / 10 100mL) = 0.000904 g/100mL

**Example #15:** The solubility of an ionic compound M_{2}X_{3} (molar mass = 284 g/mol) is 5.0 x 10^{-17} g/L. What is K_{sp} for the compound?

**Solution:**

5.0 x 10¯^{17}g/L / 284 g/mol = 1.76 x 10¯^{19}mol/LM

_{2}X_{3}(s) ⇌ 2M^{3+}(aq) + 3X^{2}¯(aq)K

_{sp}= [M^{3+}]^{2}[X^{2-}]^{3}[M

^{3+}] = (2) (1.76 x 10¯^{19}mol/L)

[X^{2-}] = (3) (1.76 x 10¯^{19}mol/L)K

_{sp}= (3.52 x 10¯^{19})^{2}(5.28 x 10¯^{19})^{3}= 1.82 x 10¯^{92}