Equilibrium MCQs for NEET — Chemistry Questions with Answers

(d) K_p is a characteristic constant for a given reaction and changes only with temperature. 

(d) CH₃COONH₄ is a salt of weak acid and weak base and

           K_acid $\approx$ K_base

    CH₃COOH      NH₄OH

(b) In all these compounds the MnS is most soluble because its solubility product is maximum.

(d) The species formed after adding a proton to the base is known as conjugate acid of the base and the species formed after losing a proton is known as conjugate base of acid. So,

              N${\mathrm{H}}_2^{-}$  + H⁺ $\to$NH₃

             Base                Conjugate acid

Equilibrium constant does not depend on catalyst. So, equilibrium constant remains same i.e. 4 X 10^-4.

On increasing the temperature, degree of dissociation of B increases. This shows that reaction is endothermic. On both sides, number of gaseous molecules are same. So, equilibrium depends on temperature only and rate of backward reaction decreases with increase in temperature.

Use the formula of salt hydrolysis of SA and WB.

$$\begin{gathered} \mathrm{pH}=-\frac{1}{2}\log {\mathrm{K}}_{\mathrm{w}}-\frac{1}{2}\log \mathrm{C}+\frac{1}{2}\log {\mathrm{K}}_{\mathrm{b}} \\ {\mathrm{K}}_{\mathrm{a}}\times {\mathrm{K}}_{\mathrm{b}}={10}^{-14} \\ {10}^{-6}\times {\mathrm{K}}_{\mathrm{b}}={10}^{-14} \\ {\mathrm{K}}_{\mathrm{b}}={10}^{-8} \\ \mathrm{pH}=-\frac{1}{2}\log 14-\frac{1}{2}\log 0.1+\frac{1}{2}\log {10}^{-8} \\ =7 + 0.5-4=3.5 \end{gathered}$$

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}} + \log \frac{\left[\mathrm{salt}\right] + \left[\mathrm{base}\right]}{\left[\mathrm{acid}\right]-\left[\mathrm{base}\right]} \\ =4.8 + \log \frac{0.15+0.05}{0.15-0.05} \\ =4.8 + \log \frac{0.20}{0.10} \\ =4.8 + 0.3 \\ =5.1 \end{gathered}$$

 ${\mathrm{PO}}_4^{3-} \& {\mathrm{C}}_2{\mathrm{O}}_4^{2-}$

$$\begin{gathered} \mathrm{\alpha }=\frac{3.6\times {10}^{-7}}{100}=3.6\times {10}^{-9} \\ \left[{\mathrm{H}}^{+}\right] = 2 \times {10}^{-7} \\ \mathrm{for} \mathrm{pure} {\mathrm{H}}_2\mathrm{O}, \left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right] \\ {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right] \left[{\mathrm{OH}}^{-}\right] = 2 \times {10}^7\times 2\times {10}^{-7} \\ {\mathrm{K}}_{\mathrm{w}}=4 \times {10}^{-14} \\ \mathrm{pH} = -\log \left[{\mathrm{H}}^{+}\right] = -\log 2\times {10}^{-7} \\ \mathrm{pH} = 6.7 \end{gathered}$$