Calculate the solubility of silver bromide, AgBr, in moles per liter, in pure water. Compare this value with the molar solubility of AgBr in 225 mL of water to which 0.15 g of NaBr has been added.

Interpretation:

The solubility of silver bromide,AgBr in mole per liter unit, in pure water has to be calculated and also this value with the molar solubility of AgBr in 225 ml of water containing 0.15 gNaBr has to be compared.

Concept introduction:

The solubility of a salt is defined as the maximum amount of salt that can be dissolved in definite amount of solvent. It is expressed in moles per liter or grams per liter. Solubility in terms of moles per liter is called molar solubility and is defined as the number of moles of solute (salt) dissolved in per liter of solution.

Solubility product constant Ksp is an equilibrium constant and is defined as the product of the equilibrium concentration of the ions of the salt raised to the power of their coefficients in the balanced chemical equation.

For example, general salt AxBy when dissolved in water dissociates as,

  AxBy(s)⇌xAy+(aq)+yB−x(aq)

The expression for Ksp of a salt is,

Ksp=[Ay+]x[B−x]y (1)

The ICE table (1) for salt AxBy, which relates the equilibrium concentration of ions in the solution is given as follows,

EquationAxBy⇌xAy++yB−xInitial(M)00Change(M)+xs+ysEquilibrium(M)xsys

From the table,

[Ay+]=xs[B−x]=ys

Substitute xs for [Ay+] and ys for [B−x] in equation (1).

Ksp=(xs)x(ys)y=xxyy(s)x+y

Rearrange for s.

s=(Kspxxyy)1/(x+y)

Here,

• x is the coefficient of the cation A+y.
• y is the coefficient of the anion B−x.
• s is the molar solubility.

The solubility of a salt decreases if in the solution one of the ion common to the dissolved ion of salt is already present before the dissolution of salt due to common ion effect. This can be explained on the basis of Le-Chatelier’s principle. According to which reaction will be more on the left side rather than right if one the ion from product side is already present before equilibrium.

For example, if anion B−x is already present in the solution before equilibrium, therefore a modified ICE table (2) is used to give the concentration relationships between ions.

EquationAxBy(s)=xAy+(aq)+yB−x(aq)Initial(M)0s'Change(M)+xs+ysEquilibrium(M)xss'+ys

Here,

• s' is the initial concentration of the anion B−x (common ion coming from strong electrolyte) present in the solution before the dissociation of a weak salt.

 From the ICE table (2),

 [Ay+]=xs[B−x]=s'+ys

 Substitute xs for [Ay+] and s'+ys for [B−x] in equation (1).

Ksp=(xs)x(s'+ys)y

The value of s is very small in comparison to the value of s'. Therefore it can be neglected and the expression of Ksp changes to,

Ksp=(xs)x(s')y (2)

• The molar solubility of the salt AgBr in pure water and in water containing 0.15 gNaBr is calculated below.

Given:

Refer to the Appendix J in the textbook for the value of Ksp.

The value of solubility product constant,Ksp of AgBr is 5.4×10−13.

The volume of the solution is 225 ml.

Amount of NaBr added to the solution is 0.15 g.

Calculation of molar solubility of AgBr in pure water:

The balanced chemical reaction for the dissolution of AgBr in water is,

  AgBr(s)⇌ Ag+(aq) + Br−(aq)

The ICE table(2) is as follows,

EquationAgBr(s)⇌Ag+(aq)+Br−(aq)Initial (M)00Change (M)+s+sEquilibrium (M) s s

The Ksp expression for AgBr is,

Ksp = [Ag+][Br−1] (3)

From the table,

[Ag+]=s[Br−1]=s

Substitute s for [Ag+] and [Br−1] in equation (3).

Ksp = (s)(s)

Here,

• Ksp is solubility product constant
• s is the molar solubility of the salt AgBr.

Ksp = s2

Rearrange for s.

s= Ksp2

Substitute 5.4×10−13 for Ksp.

s=5.4×10−132= 7.35×10−7 mol⋅L−1

• Calculation of molar solubility of AgBr in 225 ml of water containing 0.15 gNaBr.

NaBr is a strong salt and undergoes complete dissociation to give sodium ions and bromide ions in the solution and therefore the concentration of both the sodium and bromide ions is equal to the initial concentration of NaBr.

NaBr(s)→ Na+(aq) +Br−(aq)

The concentration of sodium bromide (NaBr) is calculated as follows,

[NaBr]=wMV  (4)

Here,

• w is the given mass of sodium bromide.
• M is the gram molecular mass of the sodium bromide.
• V  is the volume of the solution in liter.

The volume of the solution is,

V =(225 ml)(1.00 L1000 ml)=0.225 L

Gram molecular mass of sodium bromide is 102.9 g⋅mol.

Substitute 0.15 g for w, 102.9 g⋅mol for M and 0.225 L for V  in equation (4).

[NaBr]=0