Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{By the use of Henderson's equation}} \cr & pH = p{K_a} + {\text{lo}}{{\text{g}}_{10}}\frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{acid}}} \right]}} \cr & {\text{When,}}\,\,\,\,{\text{ [salt] = }}\,\,{\text{[acid]}} \cr & \therefore \,\,pH = p{K_a} \cr & \because \,\,p{K_a} = 3.58,\,{\text{thus at this state}} \cr & pH = 3.58 \cr} $$
So, acetoacetic acid $$\left( {p{K_a} = 3.58} \right)$$    is best to use.

$$\eqalign{ & pOH = p{K_b} + {\text{log}}\,\frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{base}}} \right]}} \cr & = - {\text{log}}\,{K_b} + {\text{log}}\frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{base}}} \right]}} \cr & = - {\text{log}}\,1.8 \times {10^{ - 5}} + {\text{log}}\frac{{0.20}}{{0.30}} \cr & = 5 - 0.25 + \left( { - 0.176} \right) \cr & = 4.75 - 0.176 \cr & = 4.57 \cr & \therefore \,\,pH = 14 - 4.57 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9.43 \cr} $$

Nucleophiles are Lewis bases while electrophiles are Lewis acids.

Conjugate acid-base pair differ by only one proton.
$$O{H^ - } \to {H^ + } + {O^ - }$$    Conjugate base of $$O{H^ - }$$ is $${O^{2 - }}$$

Electron rich species are called Lewis base. Among the given, $$B{F_3}$$  is an electron deficient species, so have a capacity of electron accepting instead of donating. That’s why it is least likely to act as a Lewis base. It is a Lewis acid.

NOTE :
$${\text{For basic buffer }}pH{\text{ is more than 7}}{\text{. }}$$
\[\underset{\begin{smallmatrix} \text{Initial moles} \\ \text{Moles}\,\text{after mixing} \end{smallmatrix}}{\mathop{{}}}\,\underset{\begin{smallmatrix} 0.1 \\ \text{0}\text{.02} \end{smallmatrix}}{\mathop{C{{H}_{3}}N{{H}_{2}}}}\,+\underset{\begin{smallmatrix} 0.08 \\ 0 \end{smallmatrix}}{\mathop{HCl}}\,\to \underset{\begin{smallmatrix} 0 \\ 0.08 \end{smallmatrix}}{\mathop{C{{H}_{3}}NH_{3}^{+}C{{l}^{-}}}}\,\]
$${\text{As it is a basic buffer solution}}{\text{.}}$$
$$pOH = p{K_b} + \log \frac{{0.08}}{{0.02}} = $$       $$ - \log 5 \times {10^{ - 4}} + \log 4$$
$$\eqalign{ & = 3.30 + 0.602 \cr & = 3.902 \cr} $$
$$pH = 14 - 3.902 = 10.09\,;$$     $$\left[ {{H^ + }} \right] = 7.99 \times {10^{ - 11}} \approx 8 \times {10^{ - 11}}M$$

$$\eqalign{ & {\text{Total volume}} = 100\,mL \cr & {\text{[acid]}} = 10\,mL \times \frac{{1.0}}{{100}} = 0.1 \cr & {\text{[salt]}} = 20\,mL \times \frac{{0.5}}{{100}} = 0.1 \cr & pH{\text{ of acidic buffer}} = p{K_a} + \log \frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{acid}}} \right]}} \cr & = 4.76 + \log \frac{{0.1}}{{0.1}} \cr & = 4.76 \cr} $$

$$C{H_3}COOH$$   is weak acid while $$NaOH$$  is strong base, so one equivalent of $$NaOH$$  cannot be neutralised with one equivalent of $$C{H_3}COOH.$$   Hence, one equivalent of each does not have $$pH$$ value 7. As the $$NaOH$$  is a strong base, the solution will be basic having a $$pH$$ more than 7.

$$C{O_2} + {H_2}O \rightleftharpoons {H_2}C{O_3} \rightleftharpoons {H^ + } + HCO_3^ - $$
If $$C{O_2}$$  escapes, the equilibrium will shift to $$LHS$$  and $$\left[ {{H^ + }} \right]$$  concentration will decrease

Solutions which resist the change in the value of $$pH$$  when small amount of acid or base is added to them are known as buffers.