Chemistry MCQs for NEET — Practice Questions with Answers

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

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

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

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

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

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