Embibe Experts Solutions for Chapter: Equilibrium, Exercise 4: Exercise-4

The following equilibrium exists in a saturated solution of ${\mathrm{NH}}_4\mathrm{Cl}.$

 ${\mathrm{NH}}_4{\mathrm{Cl}}_{\left(\mathrm{S}\right)}\rightleftharpoons {\mathrm{NH}}_4^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}$ ${\mathrm{\Delta H}}_{25°\mathrm{C}}=3.5 \mathrm{kcal} {\mathrm{mol}}^{-1}$
A change that will shift the equilibrium to the right is

A week acid, $\mathrm{HA},$ has a ${\mathrm{K}}_{\mathrm{a}}$  of  $1.00\times {10}^{-5}$. If  $0.100 \mathrm{mole}$ of this acid is dissolving in $1 \mathrm{L}$ of water, the percentage of acid dissociated at equilibrium is closed to

A solution of a substance is titrated against a strong base (or acid), volume $\mathrm{V}$ of the strong base (or acid) is plotted against $\mathrm{pH}$ of the solution (as shown in the figure). The substance could be:

Select the correct graph(s) for the corresponding acid-base titration:

$100 \mathrm{mL}$ of a clear saturated solution of ${\mathrm{Ag}}_2{\mathrm{SO}}_4$ is added to $250 \mathrm{mL}$ of a clear saturated solution of ${\mathrm{PbCrO}}_4$. Then select the incorrect option(s). Given ${\mathrm{K}}_{\mathrm{sp}}$ values for ${\mathrm{Ag}}_2{\mathrm{SO}}_4,{\mathrm{Ag}}_2{\mathrm{CrO}}_4,{\mathrm{PbCrO}}_4$ and ${\mathrm{PbSO}}_4$are $1.4\times {10}^{-5},2.4\times {10}^{-12},2.8\times {10}^{-13}$ and $1.6\times {10}^{-8}$ respectively.

Calculate $\left[{\mathrm{Ag}}^{+}\right],\left[{\mathrm{CO}}_3^{2-}\right]$ and $\left[{\mathrm{CrO}}_4^{2-}\right]$ in a solution saturated with respect to both ${\mathrm{Ag}}_2{\mathrm{CO}}_3 \& {\mathrm{Ag}}_2{\mathrm{CrO}}_4$
Given: ${\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{Ag}}_2{\mathrm{CO}}_3\right)=4\times {10}^{-12}; {\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)=2.4\times {10}^{-12}\& \frac{1}{4\sqrt[3]{5}}=0.146$.

A sample of $\mathrm{AgCl}$ was treated with $5 \mathrm{mL}$ of $1.5 \mathrm{M} {\mathrm{Na}}_2{\mathrm{CO}}_3$ solution to give ${\mathrm{Ag}}_2{\mathrm{CO}}_3$. The remaining solution contained $0.0026 \mathrm{g}$  ${\mathrm{Cl}}^{-}$ per litre Then $:\left({\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{Ag}}_2{\mathrm{CO}}_3\right)=8.2\times {10}^{-12}\right)$.

The melting points of most of the solid substances increases with an increase of pressure acting on them. However, ice melts at a temperature lower than its usual melting point, when the pressure increases. This is because: