Will a precipitate form?

**Example #1:** A sample of tap water is found to be 0.0250 M in Ca^{2+}. If 105 mg of Na_{2}SO_{4} is added to 100.0 mL of the tap water, will any CaSO_{4} precipitate? (The K_{sp} of CaSO_{4} is 7.10 x 10¯^{5}. The molar mass of Na_{2}SO_{4} is 142.04 g/mol.)

**Solution:**

1) Precipitation will occur if the K_{sp} is exceeded by the reaction quotient, commonly symbolized by a capital Q.

Q = [Ca^{2+}]_{o}[SO_{4}^{2}¯]_{o}The subscripted 'o' identifer indicates that we are using the starting concentrations, not the concentrations at equilibrium.

2) Determine the molar concentration of the sulfate ion:

105 mg = 0.105 g0.105 g / 142.04 g/mol = 0.00073923 mol

0.00073923 mol / 0.100 L = 0.0073923 M

3) Calculate the value of Q:

Q = (0.0250) (0.00073923) = 1.85 x 10¯^{4}

4) Conclusion:

Q > K_{sp}precipitation occurs

5) Comments:

The precipitation will occur until the concentrations of Ca^{2+}and SO_{4}^{2}¯, when multiplied together, equal the K_{sp}.If Q had been found to be less than K

_{sp}, no precipitation occurs.If Q = K

_{sp}, then the system is holding the maximum amount of CaSO_{4}that can be dissolved. No precipitation will occur.

**Example #2:** Does precipitation occur when 20.0 mL of 4.0 x 10¯^{4} M CaCl_{2} is mixed with 60.0 mL of 3.0 x 10¯^{4} M Na_{2}CO_{3}? (The K_{sp} for CaCO_{3} is 2.8 x 10¯^{9}.)

**Solution:**

1) Calculate the new molarities after the solutions have been mixed::

Mixing 20.0 mL of 4.0 x 10¯^{4}M CaCl_{2}with 60.0 mL dilutes the 4.0 x 10¯^{4}M Ca^{2+}by a ratio of 20 /80 = 1.0 x 10¯^{4}M Ca^{2+}Mixing 60.0 mL of 3.0 x 10¯

^{4}M Na_{2}CO_{3}with 20.0 mL dilutes the 3.0 x 10¯^{4}M CO_{3}^{2}¯ by a ratio of 60 / 80 = 2.25 x 10¯^{4}M CO_{3}^{2}¯Comment: the above was not written by the ChemTeam. I decided to include it because it is different from how the ChemTeam would approach the solution. I would have used M

_{1}V_{1}= M_{2}V_{2}.

2) Find the ion product (another term for reaction quotient):

Q = [Ca^{2+}]_{o}[CO_{3}^{2}¯]_{o}Q = (1.0 x 10¯

^{4}) (2.25 x 10¯^{4}] = 2.25 x 10¯^{8}

3) Since Q > K_{sp}, we conclude precipitation will take place.

**Example #3:** 0.96 g Na_{2}CO_{3} is combined with 0.20 g BaBr_{2} in a 10.0 L solution. Will a precipitate form? (K_{sp} = 2.8 x 10¯^{9})

**Solution:**

1) Calculate molarities using MV = grams / molar mass:

sodium carbonate(x) (10.0 L) = 0.96 g / 105.988 g/molx = 0.000905763 M

barium bromide

(x) (10.0 L) = 0.20 g / 297.138 g/molx = 0.0000673088 M

2) Calculate Q:

Q = [Ba^{2+}]_{o}[CO_{3}^{2}¯]_{o}Q = (0.000905763) (0.0000673088) = 6.1 x 10¯

^{8}

Compare Q to K_{sp}:

Q > K_{sp}A precipitate will form.

**Example #4:** Will a precipitate of PbCl_{2} form when the following solutions are mixed? 100. mL of 0.010 M Pb(NO_{3})_{2} and 100. mL of 0.0020 M NaCl. The K_{sp} for PbCl_{2} is 1.6 x 10¯^{5}.

**Solution:**

1) Calculate new molarities:

Pb^{2+}: (0.010 mol/L) (0.10 L) = 0.0010 mol

Cl¯: (0.0020 mol/L) (0.10 L) = 0.00020 molPb

^{2+}: 0.0010 mol / 0.20 L = 0.0050 M

Cl¯: 0.0002 mol / 0.20 L = 0.0010 M

2) Calculate Q and compare it to K_{sp}

Q = [Pb^{2+}]_{o}[Cl¯]$\text{}{\text{}}_{\mathrm{o}}^{2}$Q = (0.005) (0.001)

^{2}= 5 x 10¯^{9}Q < K

_{sp}A precipitate will not form.

**Example #5:** Show whether a precipitate will form when 0.10 mL of 0.0011 M AgNO_{3} is mixed into 125. mL of 0.00102M HCl? (K_{sp} for AgCl is 1.77 x 10¯^{10})

**Solution:**

1) New molarities after solutions are mixed:

AgNO_{3}---> (0.0011 mol/L) (0.10 mL) = (x) (125.1 mL); x = 8.793 x 10¯^{7}MHCl ---> (0.00102 mol/L) (125 mL) = (y) (125.1 mL); y = 0.001019 M

2) Calculate Q:

Q = [Ag^{+}]_{o}[Cl¯]_{o}Q = (8.793 x 10¯

^{7}) (0.001019) = 8.96 x 10¯^{10}

3) Compare Q to K_{sp}:

Q > K_{sp}A precipitate will form

**Example #6:** A 200.0 mL solution of 4.00 x 10¯^{3} M BaCl_{2} is added to a 600.0 mL solution of 8.00 x 10¯^{3} M K_{2}SO_{4}. Assuming that the volumes are additive, will BaSO_{4} (K_{sp} = 1.08 x 10¯^{10}) precipitate from this solution?

**Solution:**

1) Determine original molarities afer addition of solutions, but before any precipitation might take place.

M_{1}V_{1}= M_{2}V_{2}(0.004 M) (200 mL) = (M

_{2}) (800 mL)M

_{2}= 0.00100 M (for BaCl_{2})(0.008 M) (600 mL) = (M

_{2}) (800 mL)M

_{2}= 0.00300 M (for K_{2}SO_{4})

2) Determine Q:

Q = [Ba^{2+}]_{o}[SO_{4}^{2}¯]_{o}Q = (0.001) (0.003) = 3 x 10¯

^{6}

3) Compare Q to K_{sp}:

Q = 3 x 10¯^{6}and K_{sp}= 1.08 x 10¯^{10}Q > K

_{sp}Precipitation will take place.

**Example #7:** Will a precipitate of Ca(OH)_{2} (K_{sp} = 5.02 x 10¯^{6}) form if 2.00 mL of 0.200 M NaOH is added to 1.00 x 10^{3} mL of 0.100 M CaCl_{2}? (Assume the volumes are additive.)

**Solution:**

1) Calculate new molarities for the two reactants:

M_{1}V_{1}= M_{2}V_{2}(0.2 M) (2 mL) = (M

_{2}) (1002 mL)M

_{2}= 0.003992 M <--- this is the [NaOH] just before precipitation, if any(0.1 M) (1000 mL) = (M

_{2}) (1002 mL)M

_{2}= 0.09980 M <--- this is the [CaCl_{2}] before precipitation, if any

2) Calculate Q:

Q = [Ca^{2+}]_{o}$\text{[OH\xaf]}{\text{}}_{\mathrm{o}}^{2}$Q = (0.09980) (0.003992)

^{2}= 1.6 x 10¯^{6}Q < K

_{sp}no precipitation

**Example #8:** Lead(II) chromate has a K_{sp} of 1.8 x 10¯^{16}. Exactly 4.0 mL of 0.0040 M lead(II) nitrate is mixed with 2.0 mL of 0.00020 M sodium chromate.

(a) Write the precipitation reaction (net ionic) also known as the equation for the solid precipitate dissociating.

(b) Write the K_{sp}expression for this solid precipitate dissociating.

(c) Will a precipitate form? Show calculations to support your answer.

(d) What would be the effect on the solubility equilibrium system if concentrated potassium chromate solution is added?

**Solution:**

(a)

PbCrO_{4}(s) ⇌ Pb^{2+}(aq) + CrO_{4}^{2}¯(aq)

(b)

K_{sp}= [Pb^{2+}] [CrO_{4}^{2}¯]

(c)

Q = [Pb^{2+}]_{o}[CrO_{4}^{2}¯]_{o}Q = (0.00267) (0.0000667) = 1.8 x 10¯

^{7}Q > K

_{sp}therefore precipitation occurs.Note: the reader is left to determine the concentrations after mixing (but before precipitation starts). 6.0 mL is the value for V

_{2}.

(d)

An increase in a product will cause the position of the equilibrium to shift left, to the reactant side.The result will be an increase in PbCrO

_{4}(s), a decrease in [Pb^{2+}], and an increase in [CrO_{4}^{2}¯] (because of the addition of more chromate).The K

_{sp}value will not change.

**Example #9:** A solution is prepared by dissolving 0.957 mol of SrF_{2} in 1000. mL of hot water. (a) Find Q (the ion product). (b) Upon cooling the solution to 25 °C, will a precipitate form? Support your conclusion. (c) If precipitation occurs, what mass percentage of the original amount of SrF_{2} remains in solution? (Assume that a negligable change in volume occurs during the operations. K_{sp} = 2.5 x 10¯^{9})

**Solution to (a):**

[Sr^{2+}]_{o}= 0.957, [F¯]_{o}= 1.914Q = [Sr

^{2+}]_{o}$\text{[F\xaf]}{\text{}}_{\mathrm{o}}^{2}$Q = (0.957) (1.914)

^{2}= 3.505

**Solution to (b):**

Q > K_{sp}As the solution cools to 25 °C, most of the SrF

_{2}will precipitate. This reduces the value of Q. Precipitation ceases when Q = K_{sp}.

**Solution to (c):**

1) Determine molar solubility of SrF_{2}:

SrF_{2}(s) ⇌ Sr^{2+}(aq) + 2F¯(aq)K

_{sp}= [Sr^{2+}] [F¯]^{2}2.5 x 10¯

^{9}= (s) (2s)^{2}s = 0.000855 M

2) Determine grams in 1.00 L of solution. Determine grams in 0.957 mol:

(0.000855 mol/L) (125.62 g/mol) = 0.1074 g(0.957 mol) (125.62 g/mol) = 120.22 g

3) Mass percent that remains in solution:

(0.1074 g / 120.22 g) (100) = 0.0893%

**Example #10:** 50.0 mL of 4.5 x 10¯^{6} M Hg_{2}(NO_{3})_{2} and 25.0 mL of 5.0 x 10¯^{6} M NaCl are mixed. K_{sp} for Hg_{2}Cl_{2} is 1.3 x 10¯^{18}. Determine whether precipitation will occur and justify your answer.

**Solution:**

[$\text{Hg}{\text{}}_{2}^{\mathrm{2+}}$]_{o}= 4.5 x 10¯^{6}diluted 50 to 75 mL = (4.5 x 10¯^{6}) x 50 /75 = 3.00 x 10¯^{6}[Cl¯]

_{o}= 5.0 x 10¯^{6}M diluted 25 to 75 = (5.0 x 10¯^{6}) x 25 / 75 = 1.667 x 10¯^{6}Hg

_{2}Cl_{2}(s) ⇌ $\text{Hg}{\text{}}_{2}^{\mathrm{2+}}$(aq) + 2Cl¯(aq)Q = [$\text{Hg}{\text{}}_{2}^{\mathrm{2+}}$]

_{o}[Cl¯]_{o}Q = (3.00 x 10¯

^{6}) (1.667 x 10¯^{6})^{2}= 8.33 x 10¯^{18}Q > K

_{sp}(or K_{sp}< Q)There is a precipitate because Q (the reactant quotient, which is the state of the reaction at a certain time) is greater than the K

_{sp}.