The solubility product of $\mathrm{Al}(\mathrm{OH})_3$ is $2.7 \times 10^{-11}$. Calculate its solubility in $\mathrm{gL}^{-1}$ and also find out pH of this solution. (Atomic mass of $\mathrm{Al}=27 \mathrm{u}$ )
Let S be the solubility of Al(OH)$$_3$$.
Concentration of species at $t=0$
Concentration of various species at equilibrium
$\begin{aligned} K_{\mathrm{sp}} & =\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3=(\mathrm{S})(3 \mathrm{~S})^3=27 \mathrm{~S}^4 \\ S^4 & =\frac{K_{\mathrm{sp}}}{27}=\frac{2.7 \times 10^{-11}}{27}=1 \times 10^{-12} \\ S & =1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\end{aligned}$
(i) Solubility of $\mathrm{Al}(\mathrm{OH})_3$
Molar mass of $\mathrm{Al}(\mathrm{OH})_3$ is 78 g . Therefore,
Solubility of $\mathrm{Al}(\mathrm{OH})_3$ in $\mathrm{g}^{-1}=1 \times 10^{-3} \times 78 \mathrm{~g} \mathrm{~L}^{-1}=78 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}$
$=7.8 \times 10^{-2} \mathrm{~g} \mathrm{~L}^{-1}$
$\begin{aligned} \text{(ii) pH of the solution}\quad \mathrm{S} & =1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \\ {\left[\mathrm{OH}^{-}\right] } & =3 \mathrm{~S}=3 \times 1 \times 10^{-3}=3 \times 10^{-3} \\ \mathrm{pOH} & =3-\log 3 \\ \mathrm{pH} & =14-\mathrm{pOH}=11+\log 3=11.4771\end{aligned}$
Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution.
$\left(\mathrm{K}_{\text {sp }}\right.$ of $\mathrm{PbCl}_2=3.2 \times 10^{-8}$, atomic mass of $\left.\mathrm{Pb}=207 \mathrm{u}\right)$
Suppose, solubility of PbCl$$_2$$ in water is s mol L$$^{-1}$$
$\begin{aligned} \mathrm{PbCl}_2(\mathrm{~s}) & \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+\underset{2 \mathrm{~s}}{2 \mathrm{Cl}^{-}}(\mathrm{aq}) \\ \mathrm{s}_{\mathrm{sp}} & =\left[\mathrm{Pb}^{2 \uparrow}\right] \cdot\left[\mathrm{Cl}^{-}\right]^2 \\ K_{\mathrm{sp}} & =[\mathrm{s}][2 \mathrm{~s}]^2=4 s^3 \\ 3.2 \times 10^{-8} & =4 \mathrm{~s}^3 \\ s^3 & =\frac{3.2 \times 10^{-8}}{4}=0.8 \times 10^{-8} \\ s^3 & =8.0 \times 10^{-9}\end{aligned}$
$$\begin{aligned} \text{Solubility of}\quad \mathrm{PbCl}_2, \mathrm{~s} & =2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \\ \text{Solubility of}\quad \mathrm{PbCl}_2 \text { in } \mathrm{gL}^{-1} & =278 \times 2 \times 10^{-3}=0.556 \mathrm{~g} \mathrm{~L}^{-1} \end{aligned}$$
$\left(\because\right.$ Molar mass of $\left.\mathrm{PbCl}_2=207+(2 \times 35.5)=278\right)$
0.556 g of $\mathrm{PbCl}_2$ dissolve in 1 L of water.
$\therefore \quad 0.1 \mathrm{~g}$ of $\mathrm{PbCl}_2$ will dissolve in $=\frac{1 \times 0.1}{0.556} \mathrm{~L}$ of water $$=0.1798 \mathrm{~L}$$
To make a saturated solution, dissolution of $0.1 \mathrm{g~PbCl}_2$ in $0.1798 \mathrm{~L} \approx 0.2 \mathrm{~L}$ of water will be required.
A reaction between ammonia and boron trifluoride is given below.
$$: \mathrm{NH}_3+\mathrm{BF}_3 \longrightarrow \mathrm{H}_3 \mathrm{~N}: \mathrm{BF}_3$$
Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of $B$ and $N$ in the reactants?
Although $\mathrm{BF}_3$ does not have a proton but acts as Lewis acid as it is an electron deficient compound. It reacts with $\mathrm{NH}_3$ by accepting the lone pair of electrons from $\mathrm{NH}_3$ and complete its octet. The reaction can be represented by
$$\mathrm{BF}_3+: \mathrm{NH}_3 \longrightarrow \mathrm{BF}_3 \leftarrow \mathrm{NH}_3$$
Lewis electronic theory of acids and bases can explain it. Boron in $\mathrm{BF}_3$ is $s p^2$ hybridised where N in $\mathrm{NH}_3$ is $s p^3$ hybridised.
$$ \begin{aligned} & \text { Following data is given for the reaction } \\ & \mathrm{CaCO}_3(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \\ & \Delta_f H^{\ominus}[\mathrm{CaO}(s)]=-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_f H^{\ominus}\left[\mathrm{CO}_2(g)\right]=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_f H^{\ominus}\left[\mathrm{CaCO}_3(s)\right]=-1206.9 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$
Predict the effect of temperature on the equilibrium constant of the above reaction.
Given that,
$\begin{aligned} \Delta_f H^{\ominus}[\mathrm{CaO}(s)] & =-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta_f H^{\ominus}\left[\mathrm{CO}_2(g)\right] & =-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta_f H^{\ominus}\left[\mathrm{CaCO}_3(s)\right] & =-1206.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
In the reaction,
$\begin{gathered}\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(g) \\ \Delta_f H^{\ominus}=\Delta_f H^{\ominus}[\mathrm{CaO}(\mathrm{s})]+\Delta_f H^{\ominus}\left[\mathrm{CO}_2(g)\right]-\Delta_f H^{\ominus}\left[\mathrm{CaCO}_3(s)\right] \\ \therefore\quad\Delta_f H^{\ominus}=-635.1+(-393.5)-(-1206.9)=178.3 \mathrm{kJmol}^{-1}\end{gathered}$
Because $\Delta H$ value is positive, so the reaction is endothermic. Hence, according to Le-Chatelier's principle, reaction will proceed in forward direction on increasing temperature. Thus, the value of equilibrium constant for the reaction increases.
Match the following equilibria with the corresponding condition.
| A. | Liquid $$\rightleftharpoons$$ Vapour | 1. | Saturated solution |
|---|---|---|---|
| B. | Solid $$\rightleftharpoons$$ Liquid | 2. | Boiling point |
| C. | Solid $$\rightleftharpoons$$ Vapour | 3. | Sublimation point |
| D. | Solute (s) $$\rightleftharpoons$$ Solute (solution) | 4. | Melting point |
| 5. | Unsaturated solution |
A. $\rightarrow(2)$
B. $\rightarrow(4)$
C. $\rightarrow(3)$
D. $\rightarrow$ (1)
A. Liquid $\rightleftharpoons$ Vapour equilibrium exists at the boiling point.
B. Solid $\rightleftharpoons$ Liquid equilibrium exists at the melting point.
C. Solid $\rightleftharpoons$ Vapour equilibrium exists at the sublimation point.
D. Solute $(s) \rightleftharpoons$ Solute (solution) equilibrium exists at saturated solution.