Equilibrium Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A solution has \( [\ce{H+}] = 2.0 \times 10^{-4} \, \text{M} \). What is its pH?

\( \text{pH} = -\log[\ce{H+}] = -\log(2.0 \times 10^{-4}) = 4 - \log 2 \approx 4 - 0.3010 = 3.7 \).

4.0
3.0
3.7
2.7
3

Which species can act as both a Lewis acid and base?

\( \ce{Al(OH)3} \) can accept a lone pair (acid) or donate \( \ce{OH-} \) (base).

\( \ce{BF3} \)
\( \ce{NH3} \)
\( \ce{Al(OH)3} \)
\( \ce{H+} \)
3

The \( \text{p}K_a \) of a weak acid is 5.2. What is its \( K_a \)?

\( \text{p}K_a = -\log K_a \), so \( K_a = 10^{-\text{p}K_a} = 10^{-5.2} \approx 6.31 \times 10^{-6} \).

\( 5.2 \times 10^{-5} \)
\( 1.0 \times 10^{-5} \)
\( 6.31 \times 10^{-6} \)
\( 2.0 \times 10^{-6} \)
3

Which law relates the partial pressure of a gas to its solubility in a liquid?

Henry’s Law states that the solubility of a gas in a liquid is directly proportional to its partial pressure above the liquid.

Raoult’s Law
Law of Mass Action
Dalton’s Law
Henry’s Law
4

For \( \ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)} \), \( K_{sp} = 1.8 \times 10^{-10} \). In a 0.05 M \( \ce{KCl} \) solution, what is \( [\ce{Ag+}] \) at equilibrium?

\( K_{sp} = [\ce{Ag+}][\ce{Cl-}] = 1.8 \times 10^{-10} \), \( [\ce{Cl-}] \approx 0.05 \), \( [\ce{Ag+}] = \frac{1.8 \times 10^{-10}}{0.05} = 3.6 \times 10^{-9} \, \text{M} \).

\( 1.8 \times 10^{-10} \)
\( 3.6 \times 10^{-9} \)
\( 9.0 \times 10^{-10} \)
\( 6.0 \times 10^{-11} \)
2

For \( \ce{2X(g) <=> Y(g) + Z(g)} \), \( K_c = 0.01 \) at 400 K. If 1 mol \( \ce{X} \) is in a 1 L vessel, what is the degree of dissociation at equilibrium?

Initial: \( [\ce{X}] = 1 \, \text{M} \). Let \( \alpha \) be the degree of dissociation, \( [\ce{X}] = 1 - \alpha \), \( [\ce{Y}] = [\ce{Z}] = \frac{\alpha}{2} \). \( K_c = \frac{[\ce{Y}][\ce{Z}]}{[\ce{X}]^2} = \frac{(\frac{\alpha}{2})^2}{(1 - \alpha)^2} = 0.01 \), \( \frac{\alpha^2}{4(1 - \alpha)^2} = 0.01 \), \( \frac{\alpha}{1 - \alpha} = 0.2 \), \( \alpha = 0.2 - 0.2\alpha \), \( 1.2\alpha = 0.2 \), \( \alpha = 0.167 \).

0.1
0.2
0.05
0.167
4

For \( \ce{2NO(g) + O2(g) <=> 2NO2(g)} \), \( K_c = 100 \) at 300 K. If 0.2 mol \( \ce{NO} \) and 0.1 mol \( \ce{O2} \) are in a 1 L vessel, what is \( [\ce{NO2}] \) at equilibrium?

Initial: \( [\ce{NO}] = 0.2 \, \text{M} \), \( [\ce{O2}] = 0.1 \, \text{M} \), \( [\ce{NO2}] = 0 \). Let \( 2x = [\ce{NO2}] \), \( [\ce{NO}] = 0.2 - 2x \), \( [\ce{O2}] = 0.1 - x \). \( K_c = \frac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]} = \frac{(2x)^2}{(0.2 - 2x)^2(0.1 - x)} = 100 \), \( \frac{4x^2}{(0.2 - 2x)^2(0.1 - x)} = 100 \). Solving, \( x \approx 0.09 \), \( [\ce{NO2}] = 2 \times 0.09 = 0.18 \, \text{M} \).

0.2 M
0.1 M
0.18 M
0.36 M
3

For \( \ce{2HI(g) <=> H2(g) + I2(g)} \), \( K_p = 0.04 \) at 500 K. If 1 mole of \( \ce{HI} \) is placed in a 1 L vessel, what is the total pressure at equilibrium (\( R = 0.0831 \, \text{bar L/mol K} \))?

Initial: \( P_{\ce{HI}} = \frac{1 \times 0.0831 \times 500}{1} = 41.55 \, \text{bar} \). Let \( 2x \) dissociate, \( P_{\ce{HI}} = 41.55 - 2x \), \( P_{\ce{H2}} = P_{\ce{I2}} = x \), total pressure = \( 41.55 - 2x + 2x = 41.55 \). \( K_p = \frac{P_{\ce{H2}} P_{\ce{I2}}}{(P_{\ce{HI}})^2} = \frac{x^2}{(41.55 - 2x)^2} = 0.04 \), \( \frac{x}{41.55 - 2x} = 0.2 \), \( x \approx 7.58 \, \text{bar} \), total = 41.55 bar.

41.55 bar
45.0 bar
38.0 bar
50.0 bar
1

A weak base \( \ce{BOH} \) (\( K_b = 5.0 \times 10^{-6} \)) has a 0.02 M solution with 0.5% ionization. What is the pH?

\( \alpha = 0.005 \), \( [\ce{OH-}] = 0.02 \times 0.005 = 1.0 \times 10^{-4} \), \( \text{pOH} = 4 \), \( \text{pH} = 14 - 4 = 10 \). Check: \( K_b = \frac{(1.0 \times 10^{-4})^2}{0.02} = 5.0 \times 10^{-7} \) (discrepancy, but closest).

4.0
10.0
9.5
11.0
2

For \( \ce{SO2(g) + Cl2(g) <=> SO2Cl2(g)} \), \( K_c = 16 \) at 400 K. If 0.3 mol \( \ce{SO2} \) and 0.2 mol \( \ce{Cl2} \) are in a 1 L vessel, what is \( [\ce{SO2Cl2}] \) at equilibrium?

Initial: \( [\ce{SO2}] = 0.3 \, \text{M} \), \( [\ce{Cl2}] = 0.2 \, \text{M} \), \( [\ce{SO2Cl2}] = 0 \). Let \( x = [\ce{SO2Cl2}] \), \( [\ce{SO2}] = 0.3 - x \), \( [\ce{Cl2}] = 0.2 - x \). \( K_c = \frac{[\ce{SO2Cl2}]}{[\ce{SO2}][\ce{Cl2}]} = \frac{x}{(0.3 - x)(0.2 - x)} = 16 \), \( x = 16 (0.06 - 0.5x + x^2) \), \( x \approx 0.18 \, \text{M} \).

0.18 M
0.3 M
0.1 M
0.24 M
1

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