Sometimes, the solubility is given in grams per 100 mL, rather than molar solubility (which is in mol/L). The K_{sp} can still be calculated from these data, but indirectly. First, we have to convert the g/100mL data to mol/L (molar) solubility data. Then the molar solubility value is used to perform the actual 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 grams per 1000 mL or, better yet, grams per liter. (Sometimes the data is given in g/L. When that happens, this step is skipped.)

2) divide the grams per liter value by the molar mass of the substance. This gives moles per liter, which is molar solubility.

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

Here are all the problems. The general format is this: "Determine the K_{sp} of ____, given the solubility of ___.

1) nickel sulfide (NiS), 2.97 x 10¯ ^{10}g/100mL6) Cu(IO _{4})_{2}, 0.380 g/300. mL2) magnesium fluoride (MgF _{2}), 1.65 x 10¯^{3}g/100mL7) PbBr _{2}, 1.04 x 10¯^{8}g/100mL3) manganese(II) iodate [Mn(IO _{3})_{2}], 1.935 x 10¯^{1}g/100mL8) ZnS, 5.28 x 10¯ ^{12}g/100mL4) calcium arsenate [Ca _{3}(AsO_{4})_{2}], 0.043 g/L9) Cd _{3}(PO_{4})_{2}, 6.25 x 10¯^{6}g/100mL5) CaF _{2}, 0.00680 g/250.0 mL

Note that one problem uses 250mL and another uses 300mL. In each case, a conversion is made to the equivalent g/L value and from there to the mol/L value.

In these types of problem, the 100mL in the unit never sets the number of significant figures. Use the amount that dissolves, such as, for example, 2.97 x 10¯^{10} g/100mL has three sig. figs.

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

**Solution:**

Convert to grams per 1000 mL, then moles per liter:

2.97 x 10¯^{10}grams / 100mL x (10/10) = 2.97 x 10¯^{9}grams / 1000mL2.97 x 10¯

^{9}grams / L divided by 90.77 g/mol = 3.27 x 10¯^{11}mol/L

When NiS dissolves, it dissociates like this:

NiS (s) ⇌ Ni^{2+}(aq) + S^{2}¯ (aq)

The K_{sp} expression is:

K_{sp}= [Ni^{2+}] [S^{2}¯]

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.

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} grams per 100mL.

**Solution:**

Convert to grams per 1000 mL, then moles per liter:

1.65 x 10¯^{3}grams / 100mL x (10/10) = 1.65 x 10¯^{2}grams / 1000mL1.65 x 10¯

^{2}grams / L divided by 62.30 g/mol = 2.65 x 10¯^{4}mol/L

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

MgF_{2}(s) ⇌ Mg^{2+}(aq) + 2 F¯ (aq)

The K_{sp} expression is:

K_{sp}= [Mg^{2+}] [F¯]^{2}

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.

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}

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 f 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} grams per 100mL.

**Solution:**

Convert to grams per 1000 mL, then moles per liter:

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

^{3}mol/L

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

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

The K_{sp} expression is:

K_{sp}= [Mn^{2+}] [IO_{3}¯]^{2}

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.

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:**

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

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

The K_{sp} expression is:

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

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}

In checking this problem, I found an online listing for the K_{sp} of calcium arsenate. The value given was 6.8 x 10¯^{19}

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 gram of CaF_{2} in 250.0 mL of solution. Find the K_{sp} of CaF_{2}.

**Solution:**

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

0.00680 g/250mL x 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:** A 300. mL 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 grams per 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