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

The generic problem is:

Calculate the molar solubility (in mol/L) of a saturated solution of the substance.

I'm going to assume you've gone through the other files first, so the presentations here will be a bit abbreviated.

**Example #1:** The K_{sp} of Sn(OH)_{4} has been determined to be 1 x 10¯^{57}. Calculate its molar solubility.

**Solution:**

1) Write the dissociation equation for Sn(OH)_{4} and its K_{sp} expression:

Sn(OH)_{4}⇌ Sn^{4+}+ 4OH¯

K_{sp}= [Sn^{4+}] [OH¯]^{4}

2) Substitute into the K_{sp} expression and solve:

1 x 10¯^{57}= (s) (4s)^{4}256s

^{5}= 1 x 10¯^{57}s = 1 x 10¯

^{12}M (to one significant figure)The value for s is the molar solubility of the tin(IV) ion. Since every Sn(IV) ion comes from a dissolved Sn(OH)

_{4}formula unit, the value for s is the molar solubility of Sn(OH)_{4}.

**Example #2:** Given that the K_{sp} of Ti(OH)_{4} is 7 x 10¯^{53}, determine its solubility in grams per liter of saturated solution.

**Solution:**

1) Write the dissociation equation for Ti(OH)_{4} and its K_{sp} expression:

Ti(OH)_{4}⇌ Ti^{4+}+ 4OH¯

K_{sp}= [Ti^{4+}] [OH¯]^{4}

2) Substitute into the K_{sp} expression and solve:

7 x 10¯^{53}= (s) (4s)^{4}256s

^{5}= 7 x 10¯^{53}s = 1.222845571 x 10¯

^{11}MThe value for s is the molar solubility of the titanium(IV) ion. Since every Ti(IV) ion comes from a dissolved Ti(OH)

_{4}formula unit, the value for s is the molar solubility of Ti(OH)_{4}.

3) Determine concentration in grams per liter:

(1.222845571 x 10¯^{11}mol/L) (115.8946 g/mol) = 1 x 10¯^{9}g/L (to one sig fig)

**Example #3:** The log K_{sp} for thorium(IV) hydroxide has been determined to be -50.52 ± 0.08. What is the molar solubility for Th(OH)_{4}?

**Solution:**

1) Write the dissociation equation for Th(OH)_{4} and its K_{sp} expression:

Th(OH)_{4}⇌ Th^{4+}+ 4OH¯

K_{sp}= [Th^{4+}] [OH¯]^{4}

2) Substitute into the K_{sp} expression and solve:

K_{sp}= 10¯^{50.52}= 3.0 x 10¯^{51}3.0 x 10¯

^{51}= (s) (4s)^{4}256s

^{5}= 3.0 x 10¯^{51}s = 2.6 x 10¯

^{11}M

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