### The Reaction Quotient:Will a precipitate form?

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 Ksp 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 Ksp 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 > Ksp

precipitation occurs

The precipitation will occur until the concentrations of Ca2+ and SO42¯, when multiplied together, equal the Ksp.

If Q had been found to be less than Ksp, no precipitation occurs.

If Q = Ksp, 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 Ksp 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 > Ksp, 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? (Ksp = 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 Ksp:

Q > Ksp

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 Ksp 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 Ksp

Q = [Pb2+]o [Cl¯]$\text{}{\text{}}_{o}^{2}$

Q = (0.005) (0.001)2 = 5 x 10¯9

Q < Ksp

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? (Ksp 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 Ksp:

Q > Ksp

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 (Ksp = 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 Ksp:

Q = 3 x 10¯6 and Ksp = 1.08 x 10¯10

Q > Ksp

Precipitation will take place.

Example #7: Will a precipitate of Ca(OH)2 (Ksp = 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 $\text{[OH¯]}{\text{}}_{o}^{2}$

Q = (0.09980) (0.003992)2 = 1.6 x 10¯6

Q < Ksp

no precipitation

Example #8: Lead(II) chromate has a Ksp 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 Ksp expression for this solid precipitate dissociating.
(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)

Ksp = [Pb2+] [CrO42¯]

(c)

Q = [Pb2+]o [CrO42¯]o

Q = (0.00267) (0.0000667) = 1.8 x 10¯7

Q > Ksp 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 Ksp 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. Ksp = 2.5 x 10¯9)

Solution to (a):

[Sr2+]o = 0.957, [F¯]o = 1.914

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

Q = (0.957) (1.914)2 = 3.505

Solution to (b):

Q > Ksp

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

Solution to (c):

1) Determine molar solubility of SrF2:

SrF2(s) ⇌ Sr2+(aq) + 2F¯(aq)

Ksp = [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. Ksp for Hg2Cl2 is 1.3 x 10¯18. Determine whether precipitation will occur and justify your answer.

Solution:

[$\text{Hg}{\text{}}_{2}^{2+}$]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) ⇌ $\text{Hg}{\text{}}_{2}^{2+}$(aq) + 2Cl¯(aq)

Q = [$\text{Hg}{\text{}}_{2}^{2+}$]o [Cl¯]o

Q = (3.00 x 10¯6) (1.667 x 10¯6)2 = 8.33 x 10¯18

Q > Ksp (or Ksp < 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 Ksp.