Solubility Equilibria Of Sparingly Soluble Salts

OSTWALD THEORY OF INDICATORS

An indicator generally a weak organic acid or weak organic bases is a substance which is used to determine the end point in a titration. They change their colours within a certain pH range generally the colour change is due to shifting of indicator equilibrium, eg, for phenolphthalein (HPh).

HPh $\rightleftharpoons$ H⁺ + Ph^-

Colourless pink

and shifting of this equilibrium from left to right produces pink colours

$K_{In}=\frac{\left[H^{+}\right]\left[P{h}^{-}\right]}{\left[HPh\right]}andpH=p{K}_{In}+\frac{\left[P{h}^{-}\right]}{\left[HPh\right]}$

At equilibrium point [Ph^-] = [HPh]

$\therefore$ pH = pK_In. also K_In = [H⁺]

Where K_In is the ionization constant of the indicator

SOLUBILITY PRODUCT

Consider a binary electrolyte A_xB_y with solubility S mole/litre.

\begin{gathered} {A_x}{B_y} \rightleftharpoons x{A^{ + y}} + y{B^{ - x}} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xS\,\,\,\,\,\,\,\,\,\,yS \hfill \\ \end{gathered}

$K_{eq}=\frac{{[x^{+y}]}^x{[B^{-x}]}^y}{[A_x{B}_y]}$

$K_{eq}[A_x{B}_y]={[x^{+y}]}^x{[B^{-x}]}^y$

$K_{sp}={[x^{+y}]}^x{[B^{-x}]}^y$= x^xy^yS^(x+y)

K_sp is called solubility product.

Solubility product may be defines as "the product of the concentration of ions in a saturated solution of an electrolyte at a given temperature".

From solubility product, we may conclude that;

Case I: If, [ionic product] < K_sp

Then, the solution is unsaturated i.e. more solute go into the solution.

Case II: If, [ionic product] = K_­sp

The solution is just saturated i.e. no more solute can be dissolved.

Case III: If, [ionic product] > K_sp

The solution is supersaturated i.e. precipitation takes place.

Simultaneous solubility

Solubility of two sparingly soluble salt having common ion dissolved in same solution is called simultaneous solubility. Let S₁ and S₂ be the simultaneous solubility of AgBr and AgSCN respectively.

\begin{align}  AgBr(s)A{{g}^{+}}(aq)+B{{r}^{-}}(aq)\,;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{K}_{sp}}=[A{{g}^{+}}]\,[B{{r}^{-}}]=({{S}_{1}}+{{S}_{2}}){{S}_{1}}\,\,\,\,\,.......\,(1) \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({{S}_{1}}+{{S}_{2}})\,\,\,\,\,\,\,\,\,{{S}_{1}} \\ \end{align}

\begin{align}  AgSCN(s)A{{g}^{+}}(aq)+SC{{N}^{-}}(aq)\,;\,\,\,\,\,\,\,\,{{K}_{sp}}=[A{{g}^{+}}]\,[SC{{N}^{-}}]=({{S}_{1}}+{{S}_{2}}){{S}_{2}}\,\,\,\,\,.......\,(2) \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({{S}_{1}}+{{S}_{2}})\,\,\,\,\,\,\,\,\,{{S}_{2}} \\ \end{align}

Thus simultaneous solubility of AgBr and AgSCN can be calculated by the above two expression.

Illustration 1. The following solutions were mixed: 500 ml of 0.01M AgNO₃ and 500 ml. of solution that was both 0.01M in NaCl and 0.01M NaBr. Calculate [Ag⁺], [Cl^–] and [Br^–]

K_sp(AgCl) = 1.0 × 10^–10

K­_sp (AgBr) = 5 × 10^–13

Fractional Precipitation

It is the technique used to separate two or more ions from a solution by adding a selective reagent that precipitates first one ion and then the second.

Considering a solution of 0.1 M Ba²⁺ and 0.1 M Sr²⁺. If K₂Cr₂O₄ is added to this solution as a precipitating agent. K_sp of BaCrO₄ is 1.2 × 10^–10

and K_sp of SrCrO₄ is 3.5 × 10^–5

Concentration of $Cr{O}_4^{2-}$ required to precipitated BaCrO₄ $=\frac{K_{sp}}{[B{a}^{2+}]}=\frac{1.2\times {10}^{-10}}{0.1}=1.2\times {10}^{-9}M$

Concentration of $Cr{O}_4^{2-}$ required to ppt. SrCrO₄ = $\frac{3.8\times {10}^{-5}}{0.1}=3.5\times {10}^{-4}$

Since the required concentration of $Cr{O}_4^{2-}$ is low so BaCrO₄ will ppt. first.

On addition of chromate ions to the solution, BaCrO₄ starts precipitating in the solution when concentration reaches 1.2 × 10^–9 M. On adding further chromate ions to the solution when concentration reaches upto 3.5 × 10^–4, then SrCrO₄ also start precipitating in the solution.

[Ba²⁺] left when SrCrO₄ starts precipitating $=\frac{K_{sp}BaCr{O}_4}{[Cr{O}_4^{2-}]}=\frac{1.2\times {10}^{-10}}{3.5\times {10}^{-4}}$

= 3.4 × 10^–7 M

$\%(B{a}^{+})left=\frac{3.4\times {10}^{-7}}{0.1}\times 100$ = 0.00034%

Calculation of solubilities of salts

We shall now discuss the solubilities of different types of salts under various conditions.

(i) Solubilities of AgCl (salt of a strong acid and strong base) in water

AgCl would dissolve in water as,

AgCl (s) $\to$ Ag⁺ (aq) + Cl^– (aq)

At saturation point,

AgCl (s) $\rightleftharpoons$ Ag⁺(aq) + Cl^– (aq)

If the solubility of the salt is x moles / L.

$\therefore$ [Ag⁺] = xM, [Cl^–] = xM

$\Rightarrow$ x² = K_sp

or x = $\sqrt{K_{sp}}$

(ii) Solubility of AgCl in a solution that is having 0.1M in AgNO₃

AgCl would dissolve and finally reach saturation.

AgCl(s) $\rightleftharpoons$ Ag⁺(aq) + Cl^– (aq)

The K_sp of AgCl is approximately 10^–10 . If AgCl were to be dissolved in water (pure), its solubility would have been 10^–5M (previous section). In the presence of 0.1M AgNO₃ its solubility will decreases due to common ion effect. This means that [Ag⁺] from AgCl would be less than 10^–5 M. Hence, we can ignore the contribution of Ag⁺ from AgCl.

If the solubility of AgCl is x' moles / L in the presence of 0.1M AgNO₃, then

[Ag⁺] = 0.1 M, [Cl^–] = x' M

$\therefore$ x' = $\frac{K_{sp}}{0.1}$ = 10^–9 moles / L

(iii) Solubility of CH₃COOAg (salt of weak acid and strong base) in water

CH₃COOAg dissolves and reaches saturation. Since it is a salt of weak acid and strong base, it would hydrolyse. If the solubility of the salt is x moles/ l then

CH₃COOAg (s) $\rightleftharpoons$ CH₃COO^– (aq) + Ag⁺ (aq)

At eqb: x –y x

CH₃COO^– (aq) + H₂O $\rightleftharpoons$ CH₃COOH (aq) + H⁺ (aq)

At eqb: x –y y y

Where y is the amount of CH₃COO^– ion that is hydrolysed.

$\therefore$ (x –y) x = K_sp

$\frac{y^2}{\left(x-y\right)}$= $\frac{K_w}{K_a}$

Knowing the values of K_sp and K_a, solubility of the salt can be calculated.

(iv) Solubility of CH₃COOAg (salt of a weak acid and strong base) in an acid buffer of pH = 4 (assuming that the buffer does not have any common ion by CH₃COOAg):

CH₃COOAg would dissolve and reach equilibrium. It would then be hydrolysed. If the solubility of the salt is x' M in this solution, then

At eq; CH₃COOAg (s) $\rightleftharpoons$ CH₃COO^– (aq) + Ag⁺ (aq)

x' –y' x'

CH₃COO^– (aq) + H⁺ $\rightleftharpoons$ CH₃COOH (aq)

x' – y' 10^–4 y'

Since the solution is a buffer, the pH will be maintained.

$\therefore$( x' – y') x ' = K_sp

$\frac{y^{\prime }}{{10}^{-4}\times \left(x^{\prime }-y^{\prime }\right)}$= $\frac{1}{K_a}$

Since in presence of basic buffer, the degree of hydrolysis will be suppressed by already existing ^–OH ions, therefore the approximated formula which can be used is

x'² = ${K^{\prime \prime }}_{sp}$(CH₃COOAg) (neglecting y'')

Knowing K_sp and K_a, the solubility can be calculated.

(v) Solubility of CH₃COOAg in an buffer solution of pH = 9

Following the same logic as give in the earlier section,

CH₃COOAg (s) $\rightleftharpoons$ CH₃COO^– (aq) + Ag⁺ (aq)

At eqb: x'' – y'' x''

CH₃COO^–(aq) + H₂O $\rightleftharpoons$ CH₃COOH + OH^–

At eqb: x'' – y'' y'' 10^–5

Where x'' M is the solubility of the salt and y'' the extent to which it is hydrolysed.

$\therefore$( x'' – y'') x'' = K_sp

$\frac{y^{\prime \prime }\times {10}^{-5}}{(x^{\prime \prime }-y^{\prime \prime })}$= $\frac{K_w}{K_a}$

Knowing, K_sp and K_a, the solubility can be calculated.

(vi) Solubility of AgCl in an aqueous solution containing NH₃

Let the amount of NH₃ initially be 'a' M. If the solubility of the salt is x moles/ l, then

AgCl (s) $\rightleftharpoons$ Ag⁺ (aq) + Cl^– (aq)

At eq: x –y x

Ag⁺(aq) + 2NH₃ (aq) $\rightleftharpoons$ Ag(NH₃)₂⁺(aq)

x–y a–2y y

Where y is the amount of Ag⁺ which has reacted with NH₃.

$\therefore$( x –y) x = K_sp

$\frac{y}{\left(x-y\right){\left(a-2y\right)}^2}$= K_f (formation constant of Ag(NH₃)₂⁺)

Knowing K_sp and K_f, the solubility can be calculated.

Illustration 2. The solubility of BaSO₄ in water is 2.3 x 10^–4 gm/100 mL. Calculate the percentage loss in weight when 0.2 gm of BaSO₄ is washed with (a) 1lt of water (b) 1lt of 0.01N Na₂SO₄.

Solution: (a) Solubility is in general expressed in gm/lt,

so solubility of BaSO₄ = 2.3 x10^–3 g /lt

Loss in weight of BaSO₄ = amount of BaSO₄ soluble

$\therefore$ %loss = $\frac{2.3\times {10}^{-3}}{0.2}$ x 100 = 1.15%

(b) Now 0.01 N N Na₂SO₄ $\equiv$ 0.01 N $\equiv$ ions

$\equiv$ 0.005 M $S{O}_4^{2-}$ ions

Now presence of $S{O}_4^{2-}$ prior to washing BaSO₄ will suppress the solubility of BaSO₄ (due to common ion effect). The suppression will be governed by K_sp value of BaSO₄. So first calculate K_sp of BaSO₄.

Solubility of BaSO₄ in fresh water = 2 x 10^–3 g/lt

$\equiv$ $\frac{2.3\times {10}^{-3}}{233.4}$ = Mol/lt = 9.85 x 10^–6 M

K_sp = [Ba²⁺] $\left[S{O}_4^{2-}\right]$ = (9.85 × 10^–6)² = 9.71 x 10^–11

Now let x be solubility in mol/lt

$\Rightarrow$ [Ba²⁺] in solution = x mol/lt and $\left[S{O}_4^{2-}\right]$ in solution

= (x + 0.005) mol/lt

Ionic product = [Ba²⁺] $\left[S{O}_4^{2-}\right]$ = (x) (x + 0.005)

K_sp = Ionic Product at equilibrium (saturation)

$\Rightarrow$ 9.71 x 10^–11 = (x) (x + 0.005) $\Rightarrow$ x² + 0.005 x –9.71 x 10^–11 = 0

$\Rightarrow$ x = $\frac{9.71\times {10}^{-11}}{0.005}$ = 1.94 x 10^–8 mol/lt = 1.94 × 10^–8 x 233.4 g/lt

$\Rightarrow$ 4.53 x 10^–6 gm of BaSO₄ are washed away

$\Rightarrow$ percentage loss = $\frac{4.53\times {10}^{-6}\times 100}{0.2}$ = 2.26 x 10^–3%