Unit 7 - Equilibrium

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Unit 7 - Equilibrium

7.1 Introduction to Equilibrium

7.2 Direction of Reversible Reactions

7.3 Reaction Quotient and Equilibrium Constant 7.4 Calculating the Equilibrium Constant

7.5 Magnitude of the Equilibrium Constant 7.6 Properties of the Equilibrium Constant 7.7 Calculating Equilibrium Concentrations 7.8 Representations of Equilibrium

7.9 Introduction to Le Châtelier’s Principle

7.10 Reaction Quotient & Le Châtelier’s Principle 7.11 Introduction to Solubility Equilibria

7.12 Common-Ion Effect

7.13 pH and Solubility

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7.12 Common-Ion Effect

Common Ion Effect

Up to now, all the dissolved ions have come from the original solid but sometimes other substances will be present, that may also be contributing ions to the mixture - may have ions in common with the original solid.

e.g. add solution of NaCl _(aq) to PbCl _2(s) ⇋ Pb ²⁺ _(aq) + 2 Cl ^— _(aq) Quite often we do not need to do a calculation. Realising that the extra Cl ^— _(aq) will push the equilibrium to the left, meaning that less PbCl _2(s) will dissolve and K _sp will be lowered is enough.

Alternatively, two solutions may be mixed, and only then does the possibility of forming a solid exist as between them they both have ions that belong to a potential precipitation reaction

e.g. mix solutions of Pb(NO ₃ ) _(aq) and KI _(aq) to initiate

PbI _2(s) ⇋ Pb ²⁺ _(aq) + 2 I ^— _(aq)

This is an opportunity to use another version of the Reaction Quotient, Q, to see how a mixture of chemicals would compare with the equilibrium mixture described by K _sp .

If Q < K _sp then ions present are not enough to form precipitate - they remain disssolved

If Q = K then solution is saturated - so precipitate will form

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Example 1: Will a precipitate form when 23 mL of 0.020 M Na ₂ CO ₃ is added to 12 mL of 0.010 M MgCl ₂ ?

MgCO 3(s) ⇋ Mg ⁺ _(aq) + CO ₃ ²⁻ _(aq) K _sp = 4.0 x 10 ^-5 Na ₂ CO ₃ : n = M x L n = 0.020 x 0.023 = 4.6 x 10 ^-4 moles Na ₂ CO ₃

= 4.6 x 10 ^-4 moles CO ₃ ^2-

M = n / L = 4.6 x 10 ^-4 / 0.035 = 1.31 x 10 ^-2 M so [CO ₃ ^2- ] = 1.31 x 10 ^-2 M MgCl ₂ : n = M x L n = 0.010 x 0.012 = 1.2 x 10 ^-4 moles MgCl ₂

= 1.2 x 10 ^-4 moles Mg ²⁺

M = n / L = 1.2 x 10 ^-4 / 0.035 = 3.43 x 10 ^-3 M so [Mg ²⁺ ] = 3.43 x 10 ^-3 M

Q = [Mg ²⁺ ] [CO ₃ ^2- ] = (1.31 x 10 ^-2 ) (3.43 x 10 ^-3 ) = 4.49 x 10 ^-5

Since Q > K _sp , the solution is supersaturated and a precipitate of MgCO _3(s) will form

Example 2: A solution contains 0.020 M Cl ^— ions and 0.020 M Br ^— ions. To separate the Cl ^— ions from the Br ^— ions, solid AgNO ₃ is slowly added to the solution without

changing the volume.

What concentration of Ag ⁺ ions (in mol/L) is needed to precipitate as much AgBr as possible without precipitating AgCl?

AgBr: AgBr _(s) ⇋ Ag ⁺ _(aq) + Br ^— _(aq) K _sp = 7.7 x 10 ^-13

[Br ^— ] = 0.020 K _sp = [Ag ⁺ ][Br ^— ] so [Ag ⁺ ] = 7.7 x 10 ^-13 / 0.020

= 3.9 x 10 ^-11 M

Thus, [Ag ⁺ ] > 3.9 x 10 ^-11 M is required to start the precipitation of AgBr.

AgCl: AgCl _(s) ⇋ Ag ⁺ _(aq) + Cl ^— _(aq) K _sp = 1.6 x 10 ^-10

[Cl ^— ] = 0.020 K _sp = [Ag ⁺ ][Cl ^— ] so [Ag ⁺ ] = 1.6 x 10 ^-10 / 0.020

= 8.0 x 10 ^-9 M

Thus, [Ag ⁺ ] > 8.0 x 10 ^-9 M is required to start the precipitation of AgCl.

To precipitate the Br ^— ions as AgBr without precipitating the Cl ^— ions as AgCl, then,

[Ag ⁺ ] must be greater than 3.9 x 10 ^-11 M and lower than 8.0 x 10 ^-9 M.

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7.12 Practice Problems

1. CdF _2(s) ⇋ ^Cd ²⁺ _(aq) + 2 F ⁻ _(aq)

A saturated aqueous solution of CdF ₂ is prepared. The equilibrium in the solution is represented above. In the solution, [Cd ²⁺ ] _eq = 0.0585M and [F ⁻ ] _eq = 0.117M.

Some 0.90M NaF is added to the saturated solution. Which of the following identifies the molar solubility of CdF ₂ in pure water and explains the effect that the addition of NaF has on this solubility?

A The molar solubility of CdF ₂ in pure water is 0.0585 M, and adding NaF decreases this solubility because the equilibrium shifts to favor the

precipitation of some CdF ₂ .

B The molar solubility of CdF ₂ in pure water is 0.0585M, and adding NaF has no effect on the solubility because only changes in temperature can

increase or decrease the molar solubility of an ionic solid.

C The molar solubility of CdF ₂ in pure water is 0.117M, and adding NaF decreases this solubility because the equilibrium shifts to favor the

precipitation of some CdF ₂ .

D The molar solubility of CdF ₂ in pure water is 0.176M, and adding NaF

increases this solubility because the Na ⁺ ions displace the Cd ²⁺ ions, causing the equilibrium to shift to favor the products.

2. CuBr _(s) ⇋ Cu ⁺ _(aq) + Br ^— _(aq)

Shown above is the chemical equation for the dissolution of the slightly soluble salt CuBr _(s) . Its K _sp value in pure water was experimentally determined.

CuBr was found to be much less soluble in a 0.001M NaBr solution than in pure water.

Which of the following correctly explains the decrease in solubility of CuBr in 0.001 M NaBr ?

A The concentration of water is much less in the NaBr solution than in pure water, reducing the rate of dissolution of the salt.

B The presence of Na ⁺ ions in the solution inhibits the dissolution of salts.

C The presence of additional Br ⁻ already in the solution decreases the value of the K _sp for CuBr, causing the solubility to decrease.

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3. AgCl _(s) ⇋ Ag _(aq) + Cl _(aq) K _sp = 1.8 × 10

Shown above is information about the dissolution of AgCl _(s) in water at 298K. In a

chemistry lab a student wants to determine the value of S, the molar solubility of AgCl, by measuring [Ag ⁺ ] in a saturated solution prepared by mixing excess AgCl and distilled water.

How would the results of the experiment be altered if the student mixed excess AgCl

with tap water (in which [Cl ⁻ ]=0.010M) instead of distilled water and the student did not account for the Cl ⁻ in the tap water?

A The value obtained for K _sp would be too small because Cl ⁻ _(aq) ions would be attracted to the Ag ⁺ ions in the AgCl crystals, thus preventing water molecules

from reaching the crystals.

B The value obtained for K _sp would be too small because less AgCl _(s) would dissolve because of the common ion effect due to the Cl ⁻ _(aq) already in the water.

C The value obtained for K _sp would be too large because more AgCl _(s) would dissolve because of the attractions between Ag ⁺ ions in the AgCl crystals and the Cl ⁻ _(aq) ions in the water.

D The results of the experiment would not be altered because 0.010M is such a small concentration of Cl ⁻ _(aq) ions and thus has no effect on the dissolution of

AgCl _(s) .

4. Answer the following questions about the solubility of AgCl _(s) . The value of K _sp for AgC _l(s) is 1.8 × 10 ⁻¹⁰ .

a) Calculate the value of [Ag ⁺ ] in a saturated solution of AgCl in distilled water.

1 point is earned for the correct value with support work (units not necessary).

Let x = equilibrium concentration of Ag ⁺ (and of Cl ⁻ ).

Then K _sp = 1.8 x 10 ⁻¹⁰ = x ² so x = 1.3 x 10 ⁻⁵ M b) The concentration of Cl ⁻ _(aq) in seawater is 0.54 M.

i) Calculate the molar solubility of AgCl _(s) in seawater.

1 point is earned for the correct value with support work (units not necessary).

1.8 x 10 ^-10 = [Ag ⁺ ] x (0.54) so [Ag ⁺ ] = 3.3 x 10 ^-10 M

ii) Explain why AgCl _(s) is less soluble in seawater than in distilled water.

An increased [Cl ^- ] will decrease the solubility of AgCl _(s) since the K _sp is a product of the [Ag ⁺ ] and [Cl ^- ]. (This is an example of the common ion effect.)

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For the next question, parts a) to d) were done in 7.11.

5. Several reactions are carried out using AgBr, a cream-colored silver salt for which the value of the solubility-product constant, K _sp , is 5.0 x 10 ⁻¹³ at 298 K.

e) A student mixes 10.0 mL of 1.5 x 10 ⁻⁴ M AgNO ₃ with 2.0 mL of 5.0 x 10 ⁻⁴ M NaBr and stirs the resulting mixture. What will the student observe? Justify your answer with calculations.

1 point is earned for calculation of concentration of ions.

[Ag ⁺ ] = 10 / 12 x 1.5 x 10 ^-4 = 1.3 x 10 ^-4 M [Br ^— ] = 2 / 12 x 5.0 x 10 ^-4 = 8.3 x 10 ^-5 M

1 point is earned for calculation of Q and conclusion based on comparison between Q and K _sp .

Q = [Ag ⁺ ][Br ⁻ ] = (1.3 x 10 ^-4 M)(8.3 x 10 ^-5 M) = 1.1 x 10 ^-8 1 point is earned for indicating the precipitation of AgBr.

Q > K since 1.1 x 10 ^-8 > 5.0 x 10 ^-13 , so a precipitate will form since direction will be from right to left.

f) The color of another salt of silver, AgI _(s) , is yellow. A student adds a solution of NaI to a test tube containing a small amount of solid, cream-colored AgBr.

After stirring the contents of the test tube, the student observes that the solid in the test tube changes color from cream to yellow.

i) Write the chemical equation for the reaction that occurred in the test tube.

1 point is earned for the correct equation.

AgBr _(s) + I ⁻ _(aq) → AgI _(s) + Br ⁻ _(aq)

OR AgBr _(s) + NaI _(aq) → AgI _(s) + NaBr _(aq)

ii) Which salt has the greater value of K _sp : AgBr or AgI ? Justify your answer.

1 point is earned for the correct choice with justification.

AgBr has the greater value of K _sp . The precipitate will consist of the less soluble salt when both I ⁻ _(aq) and Br ⁻ _(aq) are present. Because the color of the precipitate in the test tube turns yellow, it must be AgI _(s) that precipitates;

therefore K _sp for AgBr must be greater than K _sp for AgI.

OR K for the displacement reaction is K / K .

Unit 7 - Equilibrium

Unit 7 - Equilibrium

7.1 Introduction to Equilibrium

7.2 Direction of Reversible Reactions

7.3 Reaction Quotient and Equilibrium Constant 7.4 Calculating the Equilibrium Constant

7.5 Magnitude of the Equilibrium Constant 7.6 Properties of the Equilibrium Constant 7.7 Calculating Equilibrium Concentrations 7.8 Representations of Equilibrium

7.9 Introduction to Le Châtelier’s Principle

7.10 Reaction Quotient & Le Châtelier’s Principle 7.11 Introduction to Solubility Equilibria

7.12 Common-Ion Effect

7.13 pH and Solubility

7.12 Common-Ion Effect

Common Ion Effect

Up to now, all the dissolved ions have come from the original solid but sometimes other substances will be present, that may also be contributing ions to the mixture - may have ions in common with the original solid.

e.g. add solution of NaCl (aq) to PbCl 2(s) ⇋ Pb 2+ (aq) + 2 Cl — (aq) Quite often we do not need to do a calculation. Realising that the extra Cl — (aq) will push the equilibrium to the left, meaning that less PbCl 2(s) will dissolve and K sp will be lowered is enough.

Alternatively, two solutions may be mixed, and only then does the possibility of forming a solid exist as between them they both have ions that belong to a potential precipitation reaction

e.g. mix solutions of Pb(NO 3 ) (aq) and KI (aq) to initiate

PbI 2(s) ⇋ Pb 2+ (aq) + 2 I — (aq)

This is an opportunity to use another version of the Reaction Quotient, Q, to see how a mixture of chemicals would compare with the equilibrium mixture described by K sp .

If Q < K sp then ions present are not enough to form precipitate - they remain disssolved

If Q = K then solution is saturated - so precipitate will form

Example 1: Will a precipitate form when 23 mL of 0.020 M Na 2 CO 3 is added to 12 mL of 0.010 M MgCl 2 ?

MgCO 3(s) ⇋ Mg + (aq) + CO 3 2− (aq) K sp = 4.0 x 10 -5 Na 2 CO 3 : n = M x L n = 0.020 x 0.023 = 4.6 x 10 -4 moles Na 2 CO 3

= 4.6 x 10 -4 moles CO 3 2-

M = n / L = 4.6 x 10 -4 / 0.035 = 1.31 x 10 -2 M so [CO 3 2- ] = 1.31 x 10 -2 M MgCl 2 : n = M x L n = 0.010 x 0.012 = 1.2 x 10 -4 moles MgCl 2

= 1.2 x 10 -4 moles Mg 2+

M = n / L = 1.2 x 10 -4 / 0.035 = 3.43 x 10 -3 M so [Mg 2+ ] = 3.43 x 10 -3 M

Q = [Mg 2+ ] [CO 3 2- ] = (1.31 x 10 -2 ) (3.43 x 10 -3 ) = 4.49 x 10 -5

Since Q > K sp , the solution is supersaturated and a precipitate of MgCO 3(s) will form

Example 2: A solution contains 0.020 M Cl — ions and 0.020 M Br — ions. To separate the Cl — ions from the Br — ions, solid AgNO 3 is slowly added to the solution without

changing the volume.

What concentration of Ag + ions (in mol/L) is needed to precipitate as much AgBr as possible without precipitating AgCl?

AgBr: AgBr (s) ⇋ Ag + (aq) + Br — (aq) K sp = 7.7 x 10 -13

[Br — ] = 0.020 K sp = [Ag + ][Br — ] so [Ag + ] = 7.7 x 10 -13 / 0.020

= 3.9 x 10 -11 M

Thus, [Ag + ] > 3.9 x 10 -11 M is required to start the precipitation of AgBr.

AgCl: AgCl (s) ⇋ Ag + (aq) + Cl — (aq) K sp = 1.6 x 10 -10

[Cl — ] = 0.020 K sp = [Ag + ][Cl — ] so [Ag + ] = 1.6 x 10 -10 / 0.020

= 8.0 x 10 -9 M

Thus, [Ag + ] > 8.0 x 10 -9 M is required to start the precipitation of AgCl.

To precipitate the Br — ions as AgBr without precipitating the Cl — ions as AgCl, then,

[Ag + ] must be greater than 3.9 x 10 -11 M and lower than 8.0 x 10 -9 M.

7.12 Practice Problems

1. CdF 2(s) ⇋ Cd 2+ (aq) + 2 F − (aq)

A saturated aqueous solution of CdF 2 is prepared. The equilibrium in the solution is represented above. In the solution, [Cd 2+ ] eq = 0.0585M and [F − ] eq = 0.117M.

Some 0.90M NaF is added to the saturated solution. Which of the following identifies the molar solubility of CdF 2 in pure water and explains the effect that the addition of NaF has on this solubility?

A The molar solubility of CdF 2 in pure water is 0.0585 M, and adding NaF decreases this solubility because the equilibrium shifts to favor the

precipitation of some CdF 2 .

B The molar solubility of CdF 2 in pure water is 0.0585M, and adding NaF has no effect on the solubility because only changes in temperature can

increase or decrease the molar solubility of an ionic solid.

C The molar solubility of CdF 2 in pure water is 0.117M, and adding NaF decreases this solubility because the equilibrium shifts to favor the

precipitation of some CdF 2 .

D The molar solubility of CdF 2 in pure water is 0.176M, and adding NaF

increases this solubility because the Na + ions displace the Cd 2+ ions, causing the equilibrium to shift to favor the products.

2. CuBr (s) ⇋ Cu + (aq) + Br — (aq)

Shown above is the chemical equation for the dissolution of the slightly soluble salt CuBr (s) . Its K sp value in pure water was experimentally determined.

CuBr was found to be much less soluble in a 0.001M NaBr solution than in pure water.

Which of the following correctly explains the decrease in solubility of CuBr in 0.001 M NaBr ?

A The concentration of water is much less in the NaBr solution than in pure water, reducing the rate of dissolution of the salt.

B The presence of Na + ions in the solution inhibits the dissolution of salts.

C The presence of additional Br − already in the solution decreases the value of the K sp for CuBr, causing the solubility to decrease.

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3. AgCl (s) ⇋ Ag (aq) + Cl (aq) K sp = 1.8 × 10

Shown above is information about the dissolution of AgCl (s) in water at 298K. In a

chemistry lab a student wants to determine the value of S, the molar solubility of AgCl, by measuring [Ag + ] in a saturated solution prepared by mixing excess AgCl and distilled water.

How would the results of the experiment be altered if the student mixed excess AgCl

with tap water (in which [Cl − ]=0.010M) instead of distilled water and the student did not account for the Cl − in the tap water?

A The value obtained for K sp would be too small because Cl − (aq) ions would be attracted to the Ag + ions in the AgCl crystals, thus preventing water molecules

from reaching the crystals.

B The value obtained for K sp would be too small because less AgCl (s) would dissolve because of the common ion effect due to the Cl − (aq) already in the water.

C The value obtained for K sp would be too large because more AgCl (s) would dissolve because of the attractions between Ag + ions in the AgCl crystals and the Cl − (aq) ions in the water.

D The results of the experiment would not be altered because 0.010M is such a small concentration of Cl − (aq) ions and thus has no effect on the dissolution of

AgCl (s) .

4. Answer the following questions about the solubility of AgCl (s) . The value of K sp for AgC l(s) is 1.8 × 10 −10 .

a) Calculate the value of [Ag + ] in a saturated solution of AgCl in distilled water.

1 point is earned for the correct value with support work (units not necessary).

Let x = equilibrium concentration of Ag + (and of Cl − ).

Then K sp = 1.8 x 10 −10 = x 2 so x = 1.3 x 10 −5 M b) The concentration of Cl − (aq) in seawater is 0.54 M.

i) Calculate the molar solubility of AgCl (s) in seawater.

1 point is earned for the correct value with support work (units not necessary).

e.g. add solution of NaCl _(aq) to PbCl _2(s) ⇋ Pb ²⁺ _(aq) + 2 Cl ^— _(aq) Quite often we do not need to do a calculation. Realising that the extra Cl ^— _(aq) will push the equilibrium to the left, meaning that less PbCl _2(s) will dissolve and K _sp will be lowered is enough.

e.g. mix solutions of Pb(NO ₃ ) _(aq) and KI _(aq) to initiate

PbI _2(s) ⇋ Pb ²⁺ _(aq) + 2 I ^— _(aq)

This is an opportunity to use another version of the Reaction Quotient, Q, to see how a mixture of chemicals would compare with the equilibrium mixture described by K _sp .

If Q < K _sp then ions present are not enough to form precipitate - they remain disssolved

Example 1: Will a precipitate form when 23 mL of 0.020 M Na ₂ CO ₃ is added to 12 mL of 0.010 M MgCl ₂ ?

MgCO 3(s) ⇋ Mg ⁺ _(aq) + CO ₃ ²⁻ _(aq) K _sp = 4.0 x 10 ^-5 Na ₂ CO ₃ : n = M x L n = 0.020 x 0.023 = 4.6 x 10 ^-4 moles Na ₂ CO ₃

= 4.6 x 10 ^-4 moles CO ₃ ^2-

M = n / L = 4.6 x 10 ^-4 / 0.035 = 1.31 x 10 ^-2 M so [CO ₃ ^2- ] = 1.31 x 10 ^-2 M MgCl ₂ : n = M x L n = 0.010 x 0.012 = 1.2 x 10 ^-4 moles MgCl ₂

= 1.2 x 10 ^-4 moles Mg ²⁺

M = n / L = 1.2 x 10 ^-4 / 0.035 = 3.43 x 10 ^-3 M so [Mg ²⁺ ] = 3.43 x 10 ^-3 M

Q = [Mg ²⁺ ] [CO ₃ ^2- ] = (1.31 x 10 ^-2 ) (3.43 x 10 ^-3 ) = 4.49 x 10 ^-5

Since Q > K _sp , the solution is supersaturated and a precipitate of MgCO _3(s) will form

Example 2: A solution contains 0.020 M Cl ^— ions and 0.020 M Br ^— ions. To separate the Cl ^— ions from the Br ^— ions, solid AgNO ₃ is slowly added to the solution without

What concentration of Ag ⁺ ions (in mol/L) is needed to precipitate as much AgBr as possible without precipitating AgCl?

AgBr: AgBr _(s) ⇋ Ag ⁺ _(aq) + Br ^— _(aq) K _sp = 7.7 x 10 ^-13

[Br ^— ] = 0.020 K _sp = [Ag ⁺ ][Br ^— ] so [Ag ⁺ ] = 7.7 x 10 ^-13 / 0.020

= 3.9 x 10 ^-11 M

Thus, [Ag ⁺ ] > 3.9 x 10 ^-11 M is required to start the precipitation of AgBr.

AgCl: AgCl _(s) ⇋ Ag ⁺ _(aq) + Cl ^— _(aq) K _sp = 1.6 x 10 ^-10

[Cl ^— ] = 0.020 K _sp = [Ag ⁺ ][Cl ^— ] so [Ag ⁺ ] = 1.6 x 10 ^-10 / 0.020

= 8.0 x 10 ^-9 M

Thus, [Ag ⁺ ] > 8.0 x 10 ^-9 M is required to start the precipitation of AgCl.

To precipitate the Br ^— ions as AgBr without precipitating the Cl ^— ions as AgCl, then,

[Ag ⁺ ] must be greater than 3.9 x 10 ^-11 M and lower than 8.0 x 10 ^-9 M.

1. CdF _2(s) ⇋ ^Cd ²⁺ _(aq) + 2 F ⁻ _(aq)

A saturated aqueous solution of CdF ₂ is prepared. The equilibrium in the solution is represented above. In the solution, [Cd ²⁺ ] _eq = 0.0585M and [F ⁻ ] _eq = 0.117M.

Some 0.90M NaF is added to the saturated solution. Which of the following identifies the molar solubility of CdF ₂ in pure water and explains the effect that the addition of NaF has on this solubility?

A The molar solubility of CdF ₂ in pure water is 0.0585 M, and adding NaF decreases this solubility because the equilibrium shifts to favor the

precipitation of some CdF ₂ .

B The molar solubility of CdF ₂ in pure water is 0.0585M, and adding NaF has no effect on the solubility because only changes in temperature can

C The molar solubility of CdF ₂ in pure water is 0.117M, and adding NaF decreases this solubility because the equilibrium shifts to favor the

precipitation of some CdF ₂ .

D The molar solubility of CdF ₂ in pure water is 0.176M, and adding NaF

increases this solubility because the Na ⁺ ions displace the Cd ²⁺ ions, causing the equilibrium to shift to favor the products.

2. CuBr _(s) ⇋ Cu ⁺ _(aq) + Br ^— _(aq)

Shown above is the chemical equation for the dissolution of the slightly soluble salt CuBr _(s) . Its K _sp value in pure water was experimentally determined.

B The presence of Na ⁺ ions in the solution inhibits the dissolution of salts.

C The presence of additional Br ⁻ already in the solution decreases the value of the K _sp for CuBr, causing the solubility to decrease.

3. AgCl _(s) ⇋ Ag _(aq) + Cl _(aq) K _sp = 1.8 × 10

Shown above is information about the dissolution of AgCl _(s) in water at 298K. In a

chemistry lab a student wants to determine the value of S, the molar solubility of AgCl, by measuring [Ag ⁺ ] in a saturated solution prepared by mixing excess AgCl and distilled water.

with tap water (in which [Cl ⁻ ]=0.010M) instead of distilled water and the student did not account for the Cl ⁻ in the tap water?

A The value obtained for K _sp would be too small because Cl ⁻ _(aq) ions would be attracted to the Ag ⁺ ions in the AgCl crystals, thus preventing water molecules

B The value obtained for K _sp would be too small because less AgCl _(s) would dissolve because of the common ion effect due to the Cl ⁻ _(aq) already in the water.

C The value obtained for K _sp would be too large because more AgCl _(s) would dissolve because of the attractions between Ag ⁺ ions in the AgCl crystals and the Cl ⁻ _(aq) ions in the water.

D The results of the experiment would not be altered because 0.010M is such a small concentration of Cl ⁻ _(aq) ions and thus has no effect on the dissolution of

AgCl _(s) .

4. Answer the following questions about the solubility of AgCl _(s) . The value of K _sp for AgC _l(s) is 1.8 × 10 ⁻¹⁰ .

a) Calculate the value of [Ag ⁺ ] in a saturated solution of AgCl in distilled water.

Let x = equilibrium concentration of Ag ⁺ (and of Cl ⁻ ).

Then K _sp = 1.8 x 10 ⁻¹⁰ = x ² so x = 1.3 x 10 ⁻⁵ M b) The concentration of Cl ⁻ _(aq) in seawater is 0.54 M.

i) Calculate the molar solubility of AgCl _(s) in seawater.

1.8 x 10 ^-10 = [Ag ⁺ ] x (0.54) so [Ag ⁺ ] = 3.3 x 10 ^-10 M

ii) Explain why AgCl _(s) is less soluble in seawater than in distilled water.

An increased [Cl ^- ] will decrease the solubility of AgCl _(s) since the K _sp is a product of the [Ag ⁺ ] and [Cl ^- ]. (This is an example of the common ion effect.)

5. Several reactions are carried out using AgBr, a cream-colored silver salt for which the value of the solubility-product constant, K _sp , is 5.0 x 10 ⁻¹³ at 298 K.

e) A student mixes 10.0 mL of 1.5 x 10 ⁻⁴ M AgNO ₃ with 2.0 mL of 5.0 x 10 ⁻⁴ M NaBr and stirs the resulting mixture. What will the student observe? Justify your answer with calculations.

[Ag ⁺ ] = 10 / 12 x 1.5 x 10 ^-4 = 1.3 x 10 ^-4 M [Br ^— ] = 2 / 12 x 5.0 x 10 ^-4 = 8.3 x 10 ^-5 M

1 point is earned for calculation of Q and conclusion based on comparison between Q and K _sp .

Q = [Ag ⁺ ][Br ⁻ ] = (1.3 x 10 ^-4 M)(8.3 x 10 ^-5 M) = 1.1 x 10 ^-8 1 point is earned for indicating the precipitation of AgBr.

Q > K since 1.1 x 10 ^-8 > 5.0 x 10 ^-13 , so a precipitate will form since direction will be from right to left.

f) The color of another salt of silver, AgI _(s) , is yellow. A student adds a solution of NaI to a test tube containing a small amount of solid, cream-colored AgBr.

AgBr _(s) + I ⁻ _(aq) → AgI _(s) + Br ⁻ _(aq)

OR AgBr _(s) + NaI _(aq) → AgI _(s) + NaBr _(aq)

ii) Which salt has the greater value of K _sp : AgBr or AgI ? Justify your answer.

AgBr has the greater value of K _sp . The precipitate will consist of the less soluble salt when both I ⁻ _(aq) and Br ⁻ _(aq) are present. Because the color of the precipitate in the test tube turns yellow, it must be AgI _(s) that precipitates;

therefore K _sp for AgBr must be greater than K _sp for AgI.