# ${\mathrm{K}}_{{\mathrm{a}}_{1}},$ ${\mathrm{K}}_{{\mathrm{a}}_{2}}$ $\mathrm{and}$ ${\mathrm{K}}_{{\mathrm{a}}_{3}}$ 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. $$\mathrm{K}_{\mathrm{a}_3}=\mathrm{K}_{\mathrm{a}_1} / \mathrm{K}_{\mathrm{a}_2}$$

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

 1 3.14 2 2.42 3 11.58 4 6
Subtopic:  pH calculation |
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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

 1 12.7 2 1.3 3 3.14 4 11.7
Subtopic:  pH calculation |
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The ionization constant of chloroacetic acid is 1.35 × 10–3. The pH of a 0.1 M acid solution will be:

 1 1.94 2 6.14 3 3.23 4 5.64

Subtopic:  Salt Hydrolysis & Titration |
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The ionisation constant of an acid, Ka , is the measure of the strength of an acid. The Ka values of acetic acid, hypochlorous acid and formic acid are  and $1.8×{10}^{-4}$ , 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

Subtopic:  pH calculation |
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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  Kc  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

Subtopic:  Kp, Kc & Factors Affecting them |
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Given below are two statements:

 Assertion (A): A solution containing a mixture of acetic acid and sodium acetate maintains a constant value of pH on the addition of small amounts of acid or alkali. Reason (R): A solution containing a mixture of acetic acid and sodium acetate acts as a buffer solution.

 1 Both (A) and (R) are True and (R) is the correct explanation of (A). 2 Both (A) and (R) are True but (R) is not the correct explanation of (A). 3 (A) is True but (R) is False. 4 (A) is False but (R) is True.

Subtopic:  Buffer |
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Match the following species with the corresponding conjugate acid:

 Species Conjugate acid A.  $N{H}_{3}$ 1. $C{O}_{3}^{2-}$ B. $HC{O}_{3}^{-}$ 2. $N{H}_{4}^{+}$ C. ${H}_{2}O$ 3.${H}_{3}{O}^{+}$ D. $HS{O}_{4}^{-}$ 4. ${H}_{2}C{O}_{3}$ 5. ${H}_{2}S{O}_{4}$

Codes

 A B C D 1. 2 5 1 5 2. 2 4 3 5 3. 5 4 3 2 4. 4 5 3 2
Subtopic:  Acids & Bases - Definitions & Classification |
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Match the standard free energy of the reaction with the corresponding equilibrium constant.

 A. $∆{\mathrm{G}}^{⊝}>0$ 1. K>1 B. $∆{\mathrm{G}}^{⊝}<0$ 2. K=1 C. $∆{\mathrm{G}}^{⊝}=0$ 3. K=0 4. K<1

Codes:

 A B C 1. 4 1 2 2. 1 2 3 3. 2 4 3 4. 4 1 3
Subtopic:  Kp, Kc & Factors Affecting them |
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A sample of pure PCl5 was introduced into an evacuated vessel at 473 K.
After equilibrium was attained, a concentration of PCl
5
was found to be 0.5 × 10
–1 mol L–1. If the value of

Kc is 8.3 × 10–3 mol L–1, the concentrations of
PCl
3 and Cl2 at equilibrium would be:

PCl5 (g)  PCl3 (g) + Cl2(g)

 1  $\left[{\mathrm{PCl}}_{3}\right]$ $=$ $0.02$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ $,$  $\left[{\mathrm{Cl}}_{2}\right]$ $=$ $0.04$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ 2 $\left[{\mathrm{PCl}}_{3}\right]=\left[{\mathrm{Cl}}_{2}\right]$ $=$ $0.02$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ 3 $\left[{\mathrm{PCl}}_{3}\right]$ $=$ $0.04$ $\mathrm{mol}$  ${\mathrm{L}}^{-1},$ $\left[{\mathrm{Cl}}_{2}\right]$ $=0.02$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ 4 $\left[{\mathrm{PCl}}_{3}\right]=$ $\left[{\mathrm{Cl}}_{2}\right]$ $=$ $0.04$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$

Subtopic:  Kp, Kc & Factors Affecting them |
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