Equilibrium Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The solubility of \( \ce{Ag2CrO4} \) is \( 1.3 \times 10^{-4} \, \text{M} \). What is its \( K_{sp} \)?

For \( \ce{Ag2CrO4 <=> 2Ag+ + CrO4^{2-}} \), \( [\ce{Ag+}] = 2S = 2.6 \times 10^{-4} \), \( [\ce{CrO4^{2-}}] = S = 1.3 \times 10^{-4} \). \( K_{sp} = (2.6 \times 10^{-4})^2 (1.3 \times 10^{-4}) = 8.79 \times 10^{-12} \).

\( 1.3 \times 10^{-4} \)
\( 6.76 \times 10^{-8} \)
\( 4.0 \times 10^{-12} \)
\( 8.79 \times 10^{-12} \)
4

What is the conjugate base of \( \ce{HSO4-} \)?

\( \ce{HSO4-} \) loses a proton (\( \ce{H+} \)) to form \( \ce{SO4^{2-}} \), its conjugate base.

\( \ce{H2SO4} \)
\( \ce{H3O+} \)
\( \ce{SO4^{2-}} \)
\( \ce{OH-} \)
3

The \( K_a \) of \( \ce{HSO4-} \) is \( 1.2 \times 10^{-2} \). What is the \( K_b \) of \( \ce{SO4^{2-}} \) at 298 K?

\( K_w = K_a \times K_b = 1.0 \times 10^{-14} \), \( K_b = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-2}} = 8.33 \times 10^{-13} \).

\( 1.2 \times 10^{-2} \)
\( 1.0 \times 10^{-14} \)
\( 6.0 \times 10^{-13} \)
\( 8.33 \times 10^{-13} \)
4

What is the effect of adding an inert gas at constant volume to the equilibrium \( \ce{2A(g) <=> B(g) + C(g)} \)?

Adding an inert gas at constant volume does not change the partial pressures or concentrations of the reactants and products, so the equilibrium position remains unaffected.

No effect
Shifts to the left
Shifts to the right
Increases \( K_c \)
1

For \( \ce{X2(g) <=> 2X(g)} \), \( K_c = 0.25 \) at 500 K. If 0.4 mol \( \ce{X2} \) is in a 2 L vessel, what is the degree of dissociation at equilibrium?

Initial: \( [\ce{X2}] = \frac{0.4}{2} = 0.2 \, \text{M} \). Let \( \alpha \) be the degree of dissociation, \( [\ce{X2}] = 0.2(1 - \alpha) \), \( [\ce{X}] = 0.2 \times 2\alpha = 0.4\alpha \). \( K_c = \frac{[\ce{X}]^2}{[\ce{X2}]} = \frac{(0.4\alpha)^2}{0.2(1 - \alpha)} = \frac{0.16\alpha^2}{0.2(1 - \alpha)} = 0.25 \), \( 0.8\alpha^2 = 0.25(1 - \alpha) \), \( 0.8\alpha^2 + 0.25\alpha - 0.25 = 0 \), \( \alpha \approx 0.5 \).

0.25
0.75
0.33
0.5
4

For \( \ce{CO2(g) <=> CO2(aq)} \), the equilibrium is governed by which law?

The solubility of a gas in a liquid is proportional to its pressure above the liquid, as stated by Henry's Law.

Le Chatelier’s Principle
Law of Mass Action
Raoult’s Law
Henry’s Law
4

The \( K_{sp} \) of \( \ce{AgBr} \) is \( 5.0 \times 10^{-13} \). What is \( [\ce{Ag+}] \) in a saturated solution with 0.05 M \( \ce{NaBr} \)?

For \( \ce{AgBr <=> Ag+ + Br-} \), \( K_{sp} = [\ce{Ag+}][\ce{Br-}] = 5.0 \times 10^{-13} \), \( [\ce{Br-}] \approx 0.05 \), \( [\ce{Ag+}] = \frac{5.0 \times 10^{-13}}{0.05} = 1.0 \times 10^{-11} \, \text{M} \).

\( 5.0 \times 10^{-13} \)
\( 1.0 \times 10^{-11} \)
\( 2.5 \times 10^{-12} \)
\( 7.1 \times 10^{-7} \)
2

Which species is a Lewis base?

A Lewis base donates an electron pair. \( \ce{CN-} \) has a lone pair and can donate it.

\( \ce{BCl3} \)
\( \ce{Al^{3+}} \)
\( \ce{CN-} \)
\( \ce{H+} \)
3

The \( K_{sp} \) of \( \ce{PbCl2} \) is \( 1.6 \times 10^{-5} \). What is its solubility in pure water in mol/L?

For \( \ce{PbCl2 <=> Pb^{2+} + 2Cl^-} \), \( K_{sp} = [\ce{Pb^{2+}}][\ce{Cl^-}]^2 = S (2S)^2 = 4S^3 = 1.6 \times 10^{-5} \). Solving, \( S^3 = 4 \times 10^{-6} \), \( S = (4 \times 10^{-6})^{1/3} \approx 1.6 \times 10^{-2} \).

\( 1.6 \times 10^{-2} \)
\( 2.0 \times 10^{-2} \)
\( 4.0 \times 10^{-3} \)
\( 8.0 \times 10^{-3} \)
1

A 0.01 M solution of \( \ce{KOH} \) is fully dissociated. What is its pH?

For \( \ce{KOH} \), \( [\ce{OH-}] = 0.01 \, \text{M} = 10^{-2} \), \( \text{pOH} = -\log(10^{-2}) = 2 \), \( \text{pH} = 14 - 2 = 12 \).

2
12
10
14
2

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