Problems 1 - 10

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**Problem #1:** The solubility product of Mg(OH)_{2} is 1.2 x 10¯^{11}. What minimum OH¯ concentration must be attained (for example, by adding NaOH) to decrease the Mg^{2+} concentration in a solution of Mg(NO_{3})_{2} to less than 1.1 x 10¯^{10} M?

**Solution:**

K_{sp} expression:

K_{sp}= [Mg^{2+}] [OH¯]^{2}

We set [Mg^{2+}] = 1.1 x 10¯^{10} and [OH¯] = s. Substituting into the K_{sp} expression:

1.2 x 10¯^{11}= (1.1 x 10¯^{10}) (s)^{2}x = 0.33 M

Any sodium hydroxide solution greater than 0.33 M will reduce the [Mg^{2+}] to less than 1.1 x 10¯^{10} M.

**Problem #2:** Calculate the pH at which zinc hydroxide just starts to precipitate from a 0.00857 M solution of zinc nitrate. K_{sp} for zinc hydroxide = 3.0 x 10^{-17}

**Solution:**

1) K_{sp} expression:

K_{sp}= [Zn^{2+}] [OH¯]^{2}

2) Substitute and solve for [OH¯]:

3.0 x 10^{-17}= (0.00857) (s)^{2}x = 5.91657 x 10

^{-8}M (I kept a few guard digits.)

3) Compute the pH:

pOH = 7.228pH = 6.772

Note how zinc hydroxide would precipitate even when the solution is slightly acidic.

**Problem #3:** Calculate the number of moles of Ag_{2}CrO_{4} that will dissolve in 1.00 L of 0.010 M K_{2}CrO_{4} solution. K_{sp} for Ag_{2}CrO_{4} = 9.0 x 10^{-12}.

**Solution:**

1) Concentration of dichromate ion from potassium chromate:

0.010 M

2) Calculate solubility of Ag^{+}:

K_{sp}= [Ag^{+}]^{2}[CrO_{4}^{2}¯]9.0 x 10

^{-12}= (s)^{2}(0.010)x = 3.0 x 10

^{-5}M

Since there is a 2:1 ratio between the moles of aqueous silver ion and the moles of silver chromate that dissolved, 1.5 x 10^{-5} M is the molar solubility of Ag_{2}CrO_{4} in 0.010 M K_{2}CrO_{4} solution.

Since we were asked for the moles of silver chromate that would disolve in 1.00 L, the final answer is:

1.5 x 10^{-5}mol

**Problem #4:** What is the maximum concentration of Mg^{2+} ion that can remain dissolved in a solution that contains 0.7147 M NH_{3} and 0.2073 M NH_{4}Cl? (K_{sp} for Mg(OH)_{2} is 1.2 x 10¯^{11}; K_{b} for NH_{3} is 1.77 x 10¯^{5})

**Solution:**

1) Use the acid base data supplied to calculate [OH¯]:

K_{b}= ([NH_{4}^{+}] [OH¯]) / [NH_{3}]1.77 x 10¯

^{5}= [(0.2073) (x)] / 0.7147x = 6.10 x 10¯

^{5}M

2) Use the K_{sp} expression to calculate the [Mg^{2+}]:

K_{sp}= [Mg^{2+}] [OH¯]^{2}1.2 x 10¯

^{11}= (s) (6.10 x 10¯^{5})^{2}s = 3.2 x 10¯

^{3}M

**Problem #5:** 1.1 x 10^{-4} g of Cr(OH)_{3} is added to 120. L of water at 25 °C. Will it all dissolve? (K_{sp} = 6.7 x 10^{-31})

**Solution:**

1) Solve K_{sp} expression for the molar solubility:

K_{sp}= [Cr^{3+}] [OH¯]^{3}6.7 x 10

^{-31}= (s) (3s)^{3}x = 1.255 x 10

^{-8}M

2) Convert to grams per liter:

1.255 x 10^{-8}mol/L times 103.0 g/mol = 1.29 x 10^{-6}g/L

3) Check the problem's data:

1.1 x 10^{-4}g / 120. L = 9.17 x 10^{-7}g/LAll the Cr(OH)

_{3}dissolves.

**Problem #5 - Part 2:** 4.0 x 10^{-4} g of NaOH is added. Will a precipitate form?

**Solution:**

1) Convert g/L to mol/L:

Cr(OH)_{3}9.17 x 10^{-7}g/L divided by 103 g/mol = 8.90 x 10^{-9}MNaOH

4.0 x 10^{-4}g / 120 L = 3.33 x 10^{-6}g/L3.33 x 10

^{-6}g/L divided by 40.0 g/mol = 8.33 x 10^{-8}M

2) Calculate a Q_{sp}:

x = (8.90 x 10^{-9}) (8.33 x 10^{-8})^{3}x = 5.15 x 10

^{-30}Cr(OH)

_{3}precipitates.Q is a 'reaction quotient.' You might have to look that up.

**Problem #5 - Part 3:** Calculate the molar solubility of Cr(OH)_{3} in a solution buffered at pH = 11.00

**Solution:**

1) Determine the [OH¯]:

pOH = 14.00 - 11.00 = 3.00[OH¯] = 10¯

^{pOH}= 10¯^{3.00}= 1.0 x 10¯^{3}M

2) Determine the molar solubility:

6.7 x 10^{-31}= (s) (1.0 x 10¯^{3})^{3}x = 6.7 x 10

^{-22}M

**Problem #6:** The K_{sp} of AgBr is 5.4 x 10¯^{13} at 25 °C. Calculate the molar solubility of AgBr in 0.050 M AgNO_{3}(aq) at 25 °C.

**Solution:**

K_{sp}= [Ag^{+}] [Br¯]The solution already contains 0.050 M Ag

^{+}from the dissociation of the AgNO_{3}.The solution will also contain additional Ag

^{+}, due to the dissociation of the AgBr. This value is small compared to 0.050 M, it can be ignored.Substitute into K

_{sp}equation:5.4 x 10¯

^{13}= (0.050) (s)s = (5.4 x 10¯

^{13}) / 0.050s = 1.08 x 10¯

^{11}MTo two significant digits, the bromide concentration is 1.1 x 10¯

^{11}mol/LIf you decided that the additional amount of silver ion CANNOT be ignored, you would have this:

5.4 x 10¯

^{13}= (0.050 + s) (s)The produces a quadratic:

s

^{2}+ 0.05s - 5.4 x 10¯^{13}= 0Substituting into a quadratic solver gives:

1.0799999997667 x 10¯

^{11}In other words, the ignoring was a perfectly valid thing to do.