$\mathrm{BF}_3$ 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,
$$\mathrm{NH}_4 \mathrm{Cl}<\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4>\mathrm{KNO}_3<\mathrm{CH}_3 \mathrm{COONa}$$
The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction $2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})$ is $1 \times 10^{-4}$. At a given time, the composition of reaction mixture is $[\mathrm{HI}]=2 \times 10^{-5} \mathrm{~mol}$, $\left[\mathrm{H}_2\right]=1 \times 10^{-5} \mathrm{~mol}$ and $\left[\mathrm{I}_2\right]=1 \times 10^{-5} \mathrm{~mol}$. In which direction will the reaction proceed?
Given that,
$$\begin{aligned} {[\mathrm{HI}] } & =2 \times 10^{-5} \mathrm{~mol} \\ {\left[\mathrm{H}_2\right] } & =1 \times 10^{-5} \mathrm{~mol} \\ {\left[\mathrm{I}_2\right] } & =1 \times 10^{-5} \mathrm{~mol} \end{aligned}$$
At a given time, the reaction quotient $Q$ for the reaction will be given by the expression
$$\begin{aligned} Q & =\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2} \\ & =\frac{1 \times 10^{-5} \times 1 \times 10^{-5}}{\left(2 \times 10^{-5}\right)^2}=\frac{1}{4} \\ & =0.25=2.5 \times 10^{-1} \end{aligned}$$
As the value of reaction quotient is greater than the value of $K_c$, i.e., $1 \times 10^{-4}$ the reaction will proceed in the reverse reaction.