91Ó°ÊÓ

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

(a)

Consider the reaction given below,

${\text{2H}}_{\text{2}}{\text{S(g)+3O}}_{\text{2}}\text{(g)}⇌{\text{2H}}_{\text{2}}{\text{O(g)+2SO}}_{\text{2}}\text{(g)}$

To solve for K,

we can use the equation below.

${\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}$

First, we have to solve for using the Gibb's free energy constants found in Appendix

B.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(g)}\right){\text{+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{SO}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{S(g)}\right){\text{+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\right]\\ \text{=([2(-228.6)+2(-300.2)]-[2(-33)+3(0)])kJ/mol}\\ {\text{Δ³Ò}}^{\text{°}}\text{=-991.6kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{Δ³¦}}^{\text{°}}}{\text{KI}}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{-991.6°ì´³/³¾´Ç±ô×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—298°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=400.23}\\ {\text{K=e}}^{\text{400.23}}\\ {\text{°­=6.6×10}}^{\text{173}}\end{array}$

Therefore, the required value is ${\text{6.6×10}}^{\text{173}}$.

(b)

Consider the reaction given below,

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(l)}⇌{\text{H}}_{\text{2}}{\text{O(l)+SO}}_{\text{3}}\text{(g)}$

To solve for K,

we can use the equation below.

${\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}$

First, we have to solve for ${\text{Δ³Ò}}^{\text{°}}$using the Gibb's free energy constants found in Appendix

B.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(l)}\right){\text{+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{SO}}_{\text{3}}\text{(g)}\right)\right]\text{-}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(l)}\right)\right]\\ \text{=([1(-237.192)+1(-371)]-[1(-690.059)])kJ/mol}\\ {\text{Δ³Ò}}^{\text{°}}\text{=81.867kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{Δ°ä}}^{\text{°}}}{\text{KI}}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{81.867°ì´³/³¾´Ç±ô×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—298°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=-33.04}\\ {\text{K=e}}^{\text{-33.04}}\\ {\text{°­=4.5×10}}^{\text{-15}}\end{array}$

Therefore, the required value is ${\text{4.5×10}}^{\text{-15}}$.

(c)

Consider the reaction given below,

$\text{HCN(aq)+NaOH(aq)}⇌{\text{NaCN(aq)+H}}_{\text{2}}\text{O(l)}$

${\text{HCN(aq)+Na}}^{\text{+}}{\text{(aq)+OH}}^{\text{-}}\text{(aq)}⇌{\text{Na}}^{\text{+}}{\text{(aq)+CN}}^{\text{-}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}$

${\text{Na}}^{\text{+}}$is just a spectator ion.

${\text{HCN(aq)+OH}}^{\text{-}}\text{(aq)}⇌{\text{+CN}}^{\text{-}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}$

To solve for K, we can use the equation below.

${\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}$

First, we have to solve for using the Gibb's free energy constants found in Appendix

B.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{CN}}^{\text{-}}\text{(aq)}\right){\text{+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(l)}\right)\right]\text{-}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}{\text{(HCN(aq))+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{OH}}^{\text{-}}\text{(aq)}\right)\right]\\ \text{=([1(166)+1(-237.192)]-[1(112)+1(-157.30)])kJ/mol}\\ {\text{Δ³Ò}}^{\text{°}}\text{=-25.892kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{}{\text{G}}\text{×}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{-25.892°ì´³/³¾´Ç±ô×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—298°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=10.45}\\ {\text{K=e}}^{\text{10.45}}\\ {\text{°­=3.5×10}}^{\text{4}}\end{array}$

Therefore, the required value is ${\text{3.5×10}}^{\text{4}}$.