The Reaction Quotient:
Will a precipitate form?

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Example #1: A sample of tap water is found to be 0.0250 M in Ca²⁺. If 105 mg of Na₂SO₄ is added to 100.0 mL of the tap water, will any CaSO₄ precipitate? (The K_sp of CaSO₄ is 7.10 x 10¯⁵. The molar mass of Na₂SO₄ 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 = [Ca²⁺]_o [SO₄²¯]_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 g

0.105 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 Ca²⁺ and SO₄²¯, 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 CaSO₄ that can be dissolved. No precipitation will occur.

Example #2: Does precipitation occur when 20.0 mL of 4.0 x 10¯⁴ M CaCl₂ is mixed with 60.0 mL of 3.0 x 10¯⁴ M Na₂CO₃? (The K_sp for CaCO₃ 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 CaCl₂ with 60.0 mL dilutes the 4.0 x 10¯⁴ M Ca²⁺ by a ratio of 20 /80 = 1.0 x 10¯⁴ M Ca²⁺

Mixing 60.0 mL of 3.0 x 10¯⁴ M Na₂CO₃ with 20.0 mL dilutes the 3.0 x 10¯⁴ M CO₃²¯ by a ratio of 60 / 80 = 2.25 x 10¯⁴ M CO₃²¯

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 M₁V₁ = M₂V₂.

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

Q = [Ca²⁺]_o [CO₃²¯]_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 Na₂CO₃ is combined with 0.20 g BaBr₂ 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/mol

x = 0.000905763 M

barium bromide

(x) (10.0 L) = 0.20 g / 297.138 g/mol

x = 0.0000673088 M

2) Calculate Q:

Q = [Ba²⁺]_o [CO₃²¯]_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 PbCl₂ form when the following solutions are mixed? 100. mL of 0.010 M Pb(NO₃)₂ and 100. mL of 0.0020 M NaCl. The K_sp for PbCl₂ is 1.6 x 10¯⁵.

Solution:

1) Calculate new molarities:

Pb²⁺: (0.010 mol/L) (0.10 L) = 0.0010 mol
Cl¯: (0.0020 mol/L) (0.10 L) = 0.00020 mol

Pb²⁺: 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 = [Pb²⁺]_o [Cl¯]${}_{\mathrm{o}}^2$

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 AgNO₃ 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:

AgNO₃ ---> (0.0011 mol/L) (0.10 mL) = (x) (125.1 mL); x = 8.793 x 10¯⁷ M

HCl ---> (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 BaCl₂ is added to a 600.0 mL solution of 8.00 x 10¯³ M K₂SO₄. Assuming that the volumes are additive, will BaSO₄ (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.

M₁V₁ = M₂V₂

(0.004 M) (200 mL) = (M₂) (800 mL)

M₂ = 0.00100 M (for BaCl₂)

(0.008 M) (600 mL) = (M₂) (800 mL)

M₂ = 0.00300 M (for K₂SO₄)

2) Determine Q:

Q = [Ba²⁺]_o [SO₄²¯]_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 10³ mL of 0.100 M CaCl₂? (Assume the volumes are additive.)

Solution:

1) Calculate new molarities for the two reactants:

M₁V₁ = M₂V₂

(0.2 M) (2 mL) = (M₂) (1002 mL)

M₂ = 0.003992 M <--- this is the [NaOH] just before precipitation, if any

(0.1 M) (1000 mL) = (M₂) (1002 mL)

M₂ = 0.09980 M <--- this is the [CaCl₂] before precipitation, if any

2) Calculate Q:

Q = [Ca²⁺]_o $\text{[OH¯]}{}_{\mathrm{o}}^2$

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)

PbCrO₄(s) ⇌ Pb²⁺(aq) + CrO₄²¯(aq)

(b)

K_sp = [Pb²⁺] [CrO₄²¯]

(c)

Q = [Pb²⁺]_o [CrO₄²¯]_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 V₂.

(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 PbCrO₄(s), a decrease in [Pb²⁺], and an increase in [CrO₄²¯] (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 SrF₂ 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 SrF₂ remains in solution? (Assume that a negligable change in volume occurs during the operations. K_sp = 2.5 x 10¯⁹)

Solution to (a):

[Sr²⁺]_o = 0.957, [F¯]_o = 1.914

Q = [Sr²⁺]_o $\text{[F¯]}{}_{\mathrm{o}}^2$

Q = (0.957) (1.914)² = 3.505

Solution to (b):

Q > K_sp

As the solution cools to 25 °C, most of the SrF₂ will precipitate. This reduces the value of Q. Precipitation ceases when Q = K_sp.

Solution to (c):

1) Determine molar solubility of SrF₂:

SrF₂(s) ⇌ Sr²⁺(aq) + 2F¯(aq)

K_sp = [Sr²⁺] [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 Hg₂(NO₃)₂ and 25.0 mL of 5.0 x 10¯⁶ M NaCl are mixed. K_sp for Hg₂Cl₂ is 1.3 x 10¯¹⁸. Determine whether precipitation will occur and justify your answer.

Solution:

[$\text{Hg}{}_2^{\mathrm{2+}}$]_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

Hg₂Cl₂(s) ⇌ $\text{Hg}{}_2^{\mathrm{2+}}$(aq) + 2Cl¯(aq)

Q = [$\text{Hg}{}_2^{\mathrm{2+}}$]_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.

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