### Calculating the Ksp from the gram per 100 mL solubility

Sometimes, the solubility is given in grams per 100 mL, rather than molar solubility (which is in mol/L). The Ksp can still be calculated from these data, 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 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 Ksp from molar solubility data.

Here are all the problems. The general format is this: "Determine the Ksp of ____, given a solubility of ____.

 1) nickel sulfide (NiS), 2.97 x 10¯10 g/100mL 6) Cu(IO4)2, 0.380 g/300mL 2) magnesium fluoride (MgF2), 1.65 x 10¯3 g/100mL 7) PbBr2, 1.04 x 10¯8 g/100mL 3) manganese(II) iodate [Mn(IO3)2], 1.935 x 10¯1 g/100mL 8) ZnS, 5.28 x 10¯12 g/100mL 4) calcium arsenate [Ca3(AsO4)2], 0.043 g/L 9) Cd3(PO4)2, 6.25 x 10¯6 g/100mL 5) CaF2, 0.00680 g/250mL 10) BaSO4, 0.000245 g/100mL

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 Ksp 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 / 1000mL

2.97 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) ⇌ Ni2+(aq) + S2¯(aq)

The Ksp expression is:

Ksp = [Ni2+] [S2¯]

There is a 1:1 ratio between NiS and Ni2+ and there is a 1:1 ratio between NiS and S2¯. 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 Ni2+ and 3.27 x 10¯11 mole per liter of S2¯ in solution.

Putting the values into the Ksp expression, we obtain:

Ksp = (3.27 x 10¯11) (3.27 x 10¯11) = 1.07 x 10¯21

Example #2: Determine the Ksp of magnesium fluoride (MgF2), 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 / 1000mL

1.65 x 10¯2 grams / L divided by 62.30 g/mol = 2.65 x 10¯4 mol/L

When MgF2 dissolves, it dissociates like this:

MgF2(s) ⇌ Mg2+(aq) + 2F¯(aq)

The Ksp expression is:

Ksp = [Mg2+] [F¯]2

There is a 1:1 ratio between MgF2 and Mg2+, BUT there is a 1:2 ratio between MgF2 and F¯. This means that, when 2.65 x 10¯4 mole per liter of MgF2 dissolves, it produces 2.65 x 10¯4 mole per liter of Mg2+ and 5.30 x 10¯4 mole per liter of F¯ in solution.

Putting the values into the Ksp expression, we obtain:

Ksp = (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 MgF2 concentration (its molar solubility) and doubled it to get the F¯ concentration. This is because of the 1:2 molar ratio between MgF2 and F¯.

Example #3: Determine the Ksp of manganese(II) iodate [Mn(IO3)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 / 1000mL

1.935 grams / L divided by 404.746 g/mol = 4.78 x 10¯3 mol/L

When Mn(IO3)2 dissolves, it dissociates like this:

Mn(IO3)2(s) ⇌ Mn2+(aq) + 2IO3¯(aq)

The Ksp expression is:

Ksp = [Mn2+] [IO3¯]2

There is a 1:1 ratio between Mn(IO3)2 and Mn2+, BUT there is a 1:2 ratio between Mn(IO3)2 and IO3¯. This means that, when 4.78 x 10¯3 mole per liter of Mn(IO3)2 dissolves, it produces 4.78 x 10¯3 mole per liter of Mn2+ and 9.56 x 10¯3 mole per liter of IO3¯ in solution.

Putting the values into the Ksp expression, we obtain:

Ksp = (4.78 x 10¯3) (9.56 x 10¯3)2 = 4.37 x 10¯7

Example #4: Determine the Ksp of calcium arsenate [Ca3(AsO4)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 Ksp expression is:

Ksp = [Ca2+]3[AsO42¯]2

Putting concentration values into the Ksp expression, we obtain:

Ksp = (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 Ksp 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 CaF2 contains 0.00680 g/250mL of solution. Find the Ksp of CaF2.

Solution:

1) Calculate the grams of CaF2 in one liter:

0.00680 g/250mL x 4 = 0.0272 g/L

Comment: 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 Ksp expression and solve for it:

Ksp = [Ca2+] [F¯]2

Ksp = (3.48387 x 10¯4) (6.96774 x 10¯4)2

Ksp = 1.69 x 10¯10

Example #6: 300. mL of a saturated solution of Cu(IO4)2 contains 0.380 grams of dissolved salt. Determine the Ksp.

Solution:

1) Calculate the grams of Cu(IO4)2 in one liter:

0.380 g / 0.300 L = x / 1.00 L

x = 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 Ksp expression for Cu(IO4)2:

Ksp = [0.002845] [0.005690]2

Ksp = 9.21 x 10¯8

Given the solubility in grams per 100 mL, calculate the Ksp

7) PbBr2; 1.04 x 10¯8 g/100mL

8) ZnS; 5.28 x 10¯12 g/100mL

9) Cd3(PO4)2; 6.25 x 10¯6 g/100mL

Example #10: A saturated solution of BaSO4 contains 0.000245 g/100mL Calculate its Ksp.

Solution:

1) Calculate the grams of BaSO4 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) Write the Ksp expression for BaSO4 and calculate the Ksp:

Ksp = [Ba2+] [SO42¯]

Ksp = (0.0000104974) (0.0000104974)

Ksp = 1.10 x 10¯10

4) By the way, BaSO4 dissociates like this:

BaSO4(s) ⇌ Ba2+(aq) + SO42¯(aq)

Example #11: The solubility of Ba3(PO4)2 is 7.571 x 10¯4 g/100mL of water at 25 °C. Determine the Ksp of Ba3(PO4)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 Ksp expression for Ba3(PO4)2 and calculate the Ksp:

Ksp = [Ba2+]3 [PO43¯]2 <--- I'll skip writing the chemical equation.

[Ba2+] = (1.25779 x 10¯5) (3) = 0.0000377337 M
[PO43¯] = (1.25779 x 10¯5) (2) = 0.0000251558 M

Ksp = (0.0000377337)3 (0.0000251558)2 = 3.40 x 10¯23

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