NCERT Solutions for Chapter: Equilibrium, Exercise 1: Short Answer Type

$\mathrm{pH}$ of a solution of a strong acid is $5.0$. What will be the ${\mathrm{p}}^{\mathrm{H}}$ of the solution obtained after diluting the given solution a $100$ times?

A sparingly soluble salt gets precipitated only when the product of the concentration of its ions in the solution $\left({\mathrm{Q}}_{\mathrm{sp}}\right)$ becomes greater than its solubility product. If the solubility of ${\mathrm{BaSO}}_4$ in water is $8 \times {10}^{-4} \mathrm{mol} {\mathrm{dm}}^{-3}$. Calculate its solubility in $0.01 \mathrm{mol} {\mathrm{dm}}^{–3}$ of ${\mathrm{H}}_2{\mathrm{SO}}_4$. 

$\mathrm{pH}$ of $0.08 \mathrm{mol} {\mathrm{dm}}^{–3} \mathrm{HOCl}$ solution is $2.85$. Calculate its ionisation constant. 

Calculate the $\mathrm{pH}$ of a solution formed by mixing equal volumes of two solutions $\mathrm{A}$ and $\mathrm{B}$ of a strong acid having $\mathrm{pH}=6$ and $\mathrm{pH}=4$ respectively.  

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 the $\mathrm{pH}$ of this solution. (Atomic mass of $\mathrm{Al}=27 \mathrm{u})$. 

Calculate the volume of water required to dissolve $0.1 \mathrm{g}$ lead (II) chloride to get a saturated solution.$({\mathrm{K}}_{\mathrm{sp}} \mathrm{of} {\mathrm{PbCl}}_2=3.2 \times {10}^{-8}$, the atomic mass of $\mathrm{Pb}=207 \mathrm{u})$. 

A reaction between ammonia and boron trifluoride is given below:

$:{\mathrm{NH}}_3 + {\mathrm{BF}}_3\to {\mathrm{H}}_3\mathrm{N}:{\mathrm{BF}}_3$

Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of $\mathrm{B}$ and $\mathrm{N}$ in the reactants? 

The following data is given for the reaction: ${\mathrm{CaCO}}_3 \left(\mathrm{s}\right) \to \mathrm{CaO} \left(\mathrm{s}\right) + {\mathrm{CO}}_2 \left(\mathrm{g}\right)$

$$\begin{gathered} {∆}_{\mathrm{f}} {\mathrm{H}}^{⊝}\left[\mathrm{CaO}\right(\mathrm{s}\left)\right] =–635.1 \mathrm{kJ} {\mathrm{mol}}^{–1} \\ {∆}_{\mathrm{f}} {\mathrm{H}}^{⊝}\left[{\mathrm{CO}}_2\right(\mathrm{g}\left)\right] = –393.5 \mathrm{kJ} {\mathrm{mol}}^{–1} \\ {∆}_{\mathrm{f}} {\mathrm{H}}^{⊝}\left[{\mathrm{CaCO}}_3\right(\mathrm{s}\left)\right] = –1206.9 \mathrm{kJ} {\mathrm{mol}}^{–1} \end{gathered}$$

Predict the effect of temperature on the equilibrium constant of the above reaction.