${\mathrm{K}}_{{\mathrm{a}}_1},$ ${\mathrm{K}}_{{\mathrm{a}}_2}$ $\mathrm{and}$ ${\mathrm{K}}_{{\mathrm{a}}_3}$ are the respective ionisation constants for the following reactions.
\(\mathrm{H}_2 \mathrm{~S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-}\)
\(\mathrm{HS}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-}\)
\(\mathrm{H}_2 \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}\)
The correct relationship between ${\mathrm{K}}_{{\mathrm{a}}_1},$ ${\mathrm{K}}_{{\mathrm{a}}_2}$ $\mathrm{and}$ ${\mathrm{K}}_{{\mathrm{a}}_3}$ is:
1. \(\mathrm{K}_{\mathrm{a}_3}=\mathrm{K}_{\mathrm{a}_1} \times \mathrm{K}_{\mathrm{a}_2} \)
2. \(\mathrm{K}_{\mathrm{a}_3}=\mathrm{K}_{\mathrm{a}_1}+\mathrm{K}_{\mathrm{a}_2} \)
3. \(K_{a_3}=K_{a_1}-K_{a_2} \)
4. \(\mathrm{K}_{\mathrm{a}_3}=\mathrm{K}_{\mathrm{a}_1} / \mathrm{K}_{\mathrm{a}_2}\)

The concentration of hydrogen ion in a sample of soft drink is $3.8$ $\times {10}^{-3}$ $\mathrm{M}$. The  pH of the soft drink will be:

0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. The pH of the solution will be

The ionization constant of chloroacetic acid is 1.35 × 10^–3. The pH of a 0.1 M acid solution will be:

The ionisation constant of an acid, K_a , is the measure of the strength of an acid. The K_a values of acetic acid, hypochlorous acid and formic acid are $1.74\times {10}^{-5}, 3.0\times {10}^{-8}$ and $1.8\times {10}^{-4}$ , respectively. The  correct order of pH value of 0.1 mol dm^-3 solutions of these acids is:

1. Acetic acid > Hypochlorous acid > Formic acid

2. Hypochlorous acid < Acetic acid > Formic acid

3. Formic acid > Hypochlorous acid > Acetic acid

4. Formic acid < Acetic acid < Hypochlorous acid

PCl₅, PCl₃, and Cl₂ are at equilibrium at 500 K in a closed container and their concentrations are 0.8×10^–3 mol L^–1 , 1.2×10^–3 mol L^–1 and 1.2×10^–3 mol L^–1, respectively.
The value of  K_c  for the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) will be:
1. 1.8 × 10³ mol L^–1
2. 1.8 × 10³
3. 1.8 × 10^–3 mol L^–1
4. 0.55 × 10⁴

Given below are two statements: 

Match the following species with the corresponding conjugate acid:

Codes

Match the standard free energy of the reaction with the corresponding equilibrium constant.

Codes:

A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K.
After equilibrium was attained, a concentration of PCl₅ 
was found to be 0.5 × 10^–1 mol L^–1. If the value of
 K_c is 8.3 × 10^–3 mol L^–1, the concentrations of
PCl₃ and Cl₂ at equilibrium would be:

PCl_5 (g)  $\leftrightharpoons$PCl_3 (g) + Cl_2(g)