Equilibrium MCQs for NEET — Chemistry Questions with Answers

$$\begin{gathered} {\mathrm{N}}_1{\mathrm{V}}_1 = {\mathrm{N}}_2{\mathrm{V}}_2 \\ {10}^{-4}\times 1={\mathrm{N}}_2\times \frac{10}{3} \\ {\mathrm{N}}_2=3 \times {10}^{-5} \\ \left[{\mathrm{H}}^{+}\right] = 3 \times {10}^{-5} \\ \mathrm{pH} = -\log \left[{\mathrm{H}}^{+}\right] = 5- \log 3 \\ \mathrm{pH} = 5-0.48 = 4.52 \end{gathered}$$

$$\begin{gathered} \left|\begin{array}{c}{\mathrm{N}}_1{\mathrm{V}}_1-{\mathrm{N}}_2{\mathrm{V}}_2\\ \left(\mathrm{HCl}\right) \left(\mathrm{NaOH}\right)\end{array}\right|=\mathrm{NV} \\ \left|\begin{array}{c}0.4\times 50-0.2\times 50\end{array}\right|=\mathrm{N}\times 100 \\ \mathrm{N}={10}^{-1} \end{gathered}$$

Because milliequivalents of HCl is greater than NaOH. Hence, solution is acidic.

N=[H⁺]=10^-1

pH = -log[H⁺]=-log10^-1 

pH=1

pH = 10.48,    pOH = 3.52

     [OH^-] = antilog of -3.52

     [OH^-] = 3 X 10^-4

     [OH^-] = $\sqrt{{\mathrm{K}}_{\mathrm{b}}\times \mathrm{C}}$

     (3 X 10^-4)² = K_b X 0.1

     K_b = 9 X 10^-7

CH₃COONa is completely dissociated while ionisation of CH₃COOH is supressed due to common ion effect.

Due to common ion effect, ionisation of NH₄OH is supressed and decreases [OH^-] concentration. Hence, pH decreases.

As the total Cationic or Anionic charge increases, the degree of hydrolysis decreases. Hence, NH₄Cl is more hydrolysed.

Here, a of CH₃COOH is not negligible due to very dilute solution. Here, a is approx 0.9.

[H⁺] = C. a =10^-6 X 0.9 =9 X10^-7

pH=-log[H⁺]= 7- log9

pH=-7-0.954=6.046

${\mathrm{CH}}_3\mathrm{COOH} + \mathrm{NaOH} \stackrel{ }{\to }{\mathrm{CH}}_3\mathrm{COONa} + {\mathrm{H}}_2\mathrm{O}$

At t=0    60ml 1M   20ml 1M         0          0

At t='t'   40ml 1M        0         20ml 1M     -

Now it is the acidic buffer solution.

$$\begin{gathered} \mathrm{pH}=-\mathrm{logKa} + \log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]} \\ \mathrm{pH}=-\log 1.8 \times {10}^{-5}\log \frac{20\times 1}{40\times 1} \\ \mathrm{pH}=4.75 - 0.3 \\ \mathrm{pH}=4.45 \end{gathered}$$

In NH₄OH, Ag⁺ form complex. Hence, Solubility of AgCl is maximum in NH₄OH.

Use, ${\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}(\mathrm{RT}{)}^{∆{\mathrm{n}}_{\mathrm{s}}}$