Question

Free Response Show your calculation approach. Clarity of approach and relevant equations is necessary for partial credit b. Now let's explore the reaction if we take into account non-ideal behavior. Calculate the concentration of Ag" in solution if the solid if the AgBr is dissolved in a solution with an ionic strength of 0.1 (-0.1). Size of Ag' and Br ions (a) are 250 and 300 pm respectively. 1. Solid silver bromide (AgBr) will disassociate slightly in water to produce Ag' and Br. AgBroAgfag) + Br(aq) Kup - 1.3 x 10 a. Calculate the equilibrium concentration of Ag' if the solution if the solid AgBr is dissolved in pure water. Assume that the solution is behaving ideally (y = 1)

Answer 1

ANSWER - Question 1a: Ideal solution

Reaction:

$AgBr\, (s) \rightleftharpoons Ag^{+}\, (aq)+Br^{-}\, (aq)$

If the solution has an ideal behavior:

$Ksp=[Ag^{+}][Br^{-}]=1.3x10^{-5}$

At the equilibrium of dissociation of solid AgBr , the [Ag⁺] = [Br^-] :

	AgBr	<-->	Ag+	+	Br-
Initial	--		0		0
Final	--		X		X

$[Ag^{+}][Br^{-}]=X^{2}=1.3x10^{-5}$

$X=\sqrt{1.3x10^{-5}}=\mathbf{3.61x10^{-3}\, M}$

Then the equilibrium concentration of Ag⁺ is

$X=[Ag^{+}]=\mathbf{3.61x10^{-3}\, M}$

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ANSWER - Question 1b: non-ideal solution

Reaction:

$AgBr\, (s) \rightleftharpoons Ag^{+}\, (aq)+Br^{-}\, (aq)$

If the solution has a non-ideal behavior:

$Ksp= A_{Ag^{+}}\times A_{Br^{-}}=1.3x10^{-5}$

where A = activity and is defiened as

$A_{x}=\gamma _{x}[X]$

γ = activity coefficient of X

replacing into K'sp definition

$Ksp= \gamma A_{Ag^{+}}[Ag^{+}]\times \gamma _{Br^{-}}[Br^{-}]$

$Ksp= \gamma A_{Ag^{+}} \gamma _{Br^{-}} [Ag^{+}][Br^{-}]$

$[Ag^{+}][Br^{-}]=\frac{Ksp}{ \gamma A_{Ag^{+}} \gamma _{Br^{-}} }=K'sp$

to find the [Ag⁺] we need to calculate the values of γ for Ag⁺ and Br^-. The activity coefficients are calculated with the Debye-Huckel equation

$-log\left ( \gamma \right )=\frac{0.51z^{2}\sqrt{\mu }}{1+3.3\alpha \sqrt{\mu }}$

• z = charge of ion
• α = effective hydrated radius of ion in nm
• μ = ionic strength of solution

For Ag⁺ (α= 250 pm = 0.25 nm) and Br^- (α= 300 pm = 0.30 nm)

$-log\left ( \gamma_{Ag^{+}} \right )=\frac{0.51(1)^{2}\sqrt{0.1 }}{1+3.3(0.25) \sqrt{0.1 }}=0.128$

$\gamma_{Ag^{+}} =10^{-p\gamma_{Ag^{+}}}=10^{0.128}=\mathbf{0.745}$

$-log\left ( \gamma_{Br^{-}} \right )=\frac{0.51(1)^{2}\sqrt{0.1 }}{1+3.3(0.35) \sqrt{0.1 }}=0.123$

$\gamma_{Br^{-}} =10^{-p\gamma_{Br^{-}}}=10^{0.122}=\mathbf{0.754}$​​​​​​​

then

$K'sp=\frac{Ksp}{ \gamma A_{Ag^{+}} \gamma _{Br^{-}} }=\frac{1.3x10^{-5}}{0.745\times 0.754}=\mathbf{2.31x10^{-5}}$

Now, we can calculate the equilibrium concentration of Ag⁺. At this point the [Ag⁺] = [Br^-​​​​​​​] :

$[Ag^{+}][Br^{-}]=[Ag^{+}]^{2}=2.31x10^{-5}$

$[Ag^{+}]=\sqrt{2.31x10^{-5}}=\mathbf{4.81x10^{-3}\, M}$

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ANSWER - Question 1c:

• In real equilibrium we consider the activity of ions instead of concentrations because ratio of concentrations is not constant under all conditions and the ionic strength of solution is not considered. The activity of ions is related to its activity coefficient which measures the deviation from ideal behavior. According to Debye-Huckel equation

  $-log\left ( \gamma \right )=\frac{0.51z^{2}\sqrt{\mu }}{1+3.3\alpha \sqrt{\mu }}$

  • z = charge of ion
  • α = effective hydrated radius of ion in nm
  • μ = ionic strength of solution

This parameter takes into account the ionic strength of solution and the fact that ions in solution (hydrated ions) have a greater ionic radius than the reported value:

lonic radius (172 pm) Ionic radius (185 pm) Effective hydrated radius (250 pm) Effective hydrated radius (300 pm)

Taking in consideration the molecules of water closely associated to ion, we can see that for hydrated ions (non-ideal behavior) the concentration must be higher in order to obtain the necessary activity to reach the equilibrium of solubility