The aqueous solution of sugar does not conduct electricity. However, when sodium chloride is added to water, it conducts electricity. How will you explain this statement on the basis of ionisation and how is it affected by concentration of sodium chloride?
Explanation for the given statement on the basis of ionisation and effect upon the concentration of sodium chloride is given below
(i) Sugar being a non-electrolyte does not ionise in water whereas NaCl ionises completely in water and produces $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ion which help in the conduction of electricity.
(ii) When concentration of NaCl is increased, more $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$ ions will be produced. Hence, conductance or conductivity of the solution increases.
$\mathrm{BF}_3$ does not have proton but still acts as an acid and reacts with $\ddot{\mathrm{N}} \mathrm{H}_3$. Why is it so? What type of bond is formed between the two?
$\mathrm{BF}_3$ is an electron deficient compound and hence acts as Lewis acid. $\ddot{\mathrm{N}} \mathrm{H}_3$ has one lone pair which it can donate to $\mathrm{BF}_3$ and form a coordinate bond. Hence, $\mathrm{NH}_3$ acts as a Lewis base.
$\mathrm{H}_3 \mathrm{N}: \longrightarrow \mathrm{BF}_3$
Ionisation constant of a weak base MOH , is given expression
$$\mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{M}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{MOH}]}$$
Values of ionisation constant of some weak bases at a particular temperature are given below
$\begin{array}{ccccc}\text { Base } & \text { Dimethylamine } & \text { Urea } & \text { Pyridine } & \text { Ammonia } \\ \mathrm{K}_{\mathrm{b}} & 5.4 \times 10^{-4} & 1.3 \times 10^{-14} & \begin{array}{l}1.77 \times 10^{-9}\end{array} & 1.77 \times 10^{-5}\end{array}$
Arrange the bases in decreasing order of the extent of their ionisation at equilibrium. Which of the above base is the strongest?
Given that, ionisation constant of a weak base MOH
$$K_b=\left[M^{+}\right]\left[\mathrm{OH}^{-}\right][\mathrm{MOH}] .$$
Larger the ionisation constant $\left(K_b\right)$ of a base, greater is its ionisation and stronger the base. Hence, dimethyl amine is the strongest base.
$K_b\underset{5.4 \times 10^{-4}}{\text { Dimethyl amine }}>\underset{1.77 \times 10^{-5}}{\text { ammonia }}>\underset{1.77 \times 10^{-9}}{\text { pyridine }}>\underset{1.3 \times 10^{-14}}{\text { urea }}$
Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?
$$\mathrm{OH}^{-}, \mathrm{RO}^{-} \mathrm{CH}_3 \mathrm{COO}^{-}, \mathrm{Cl}^{-}$$
Conjugate acid of the given bases are $\mathrm{H}_2 \mathrm{O}, \mathrm{ROH}, \mathrm{CH}_3 \mathrm{COOH}$ and HCl . Order of their acidio strength is
$$\mathrm{HCl}>\mathrm{CH}_3 \mathrm{COOH}>\mathrm{H}_2 \mathrm{O}>\mathrm{ROH}$$
Hence, order of basic strength of their conjugate bases is
$$\mathrm{Cl}^{-}<\mathrm{CH}_3 \mathrm{COO}^{-}<\mathrm{OH}^{-}<\mathrm{RO}^{-}$$
Arrange the following in increasing order of pH .
$$\left.\mathrm{KNO}_3(\mathrm{aq}), \mathrm{CH}_3 \mathrm{COONa}_{(\mathrm{aq}}\right) \mathrm{NH}_4 \mathrm{Cl}(\mathrm{aq}), \mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4(\mathrm{aq})$$
(i) $\mathrm{KNO}_3$ is a salt of strong acid $\left(\mathrm{HNO}_3\right)$ strong base $(\mathrm{KOH})$, hence its aqueous solution is neutral; $\mathrm{pH}=7$.
(ii) $\mathrm{CH}_3 \mathrm{COONa}$ is a salt of weak acid $\left(\mathrm{CH}_3 \mathrm{COOH}\right)$ and strong base $(\mathrm{NaOH})$, hence, its aqueous solution is basic; $\mathrm{pH}>7$.
(iii) $\mathrm{NH}_4 \mathrm{Cl}$ is a salt of strong acid $(\mathrm{HCl})$ and weak base $\left(\mathrm{NH}_4 \mathrm{OH}\right)$ hence, its aqueous solution is acidic; $\mathrm{pH}<7$.
(iv) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4$ is a salt of weak acid, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}$ and weak base, $\mathrm{NH}_4 \mathrm{OH}$. But $\mathrm{NH}_4 \mathrm{OH}$ is slightly stronger than $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}$. Hence, pH is slightly $>7$.
Therefore, increasing order of pH of the given salts is,
$$\mathrm{NH}_4 \mathrm{Cl}<\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4>\mathrm{KNO}_3<\mathrm{CH}_3 \mathrm{COONa}$$