Equilibrium Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A weak base \( \ce{NH3} \) (\( K_b = 1.8 \times 10^{-5} \)) is mixed with 0.1 M \( \ce{NH4Cl} \) in equal volumes. If the final \( [\ce{NH3}] = 0.05 \, \text{M} \), what is the pH?

\( [\ce{NH4+}] = 0.05 \, \text{M} \), \( \text{p}K_b = 4.74 \), \( \text{pOH} = 4.74 + \log \frac{0.05}{0.05} = 4.74 \), \( \text{pH} = 14 - 4.74 = 9.26 \).

4.74
9.26
5.0
10.0
2

Which species can act as both a Bronsted-Lowry acid and base?

\( \ce{H2PO4-} \) can donate \( \ce{H+} \) (acid) to form \( \ce{HPO4^{2-}} \) or accept \( \ce{H+} \) (base) to form \( \ce{H3PO4} \), making it amphoteric.

\( \ce{HCl} \)
\( \ce{NaOH} \)
\( \ce{H2PO4-} \)
\( \ce{SO4^{2-}} \)
3

For the reaction \( \ce{A(g) + 3B(g) <=> 2C(g)} \), \( K_c = 125 \) at 400 K. If 1 mole of \( \ce{A} \) and 4 moles of \( \ce{B} \) are placed in a 2 L vessel, what is \( [\ce{C}] \) at equilibrium?

Initial: \( [\ce{A}] = \frac{1}{2} = 0.5 \, \text{M} \), \( [\ce{B}] = \frac{4}{2} = 2 \, \text{M} \), \( [\ce{C}] = 0 \). Let \( 2x \) be moles of \( \ce{C} \) formed, so \( \ce{A} \) decreases by \( x \), \( \ce{B} \) by \( 3x \). At equilibrium: \( [\ce{A}] = 0.5 - x \), \( [\ce{B}] = 2 - 3x \), \( [\ce{C}] = x \). \( K_c = \frac{[\ce{C}]^2}{[\ce{A}][\ce{B}]^3} = \frac{(x)^2}{(0.5 - x)(2 - 3x)^3} = 125 \). Solving, test \( x = 0.4 \): \( \frac{(0.4)^2}{(0.1)(0.2)^3} = \frac{0.16}{0.0008} = 200 \) (too high), \( x = 0.35 \), \( \frac{(0.35)^2}{(0.15)(0.35)^3} = \frac{0.1225}{0.0064} \approx 19 \) (too low), \( x \approx 0.38 \), \( [\ce{C}] = 0.38 \, \text{M} \).

0.38 M
0.5 M
0.76 M
1.0 M
1

For \( \ce{Ag2S(s) <=> 2Ag+(aq) + S^{2-}(aq)} \), \( K_{sp} = 6.0 \times 10^{-50} \). If \( [\ce{Ag+}] = 1.0 \times 10^{-15} \, \text{M} \) in a saturated solution, what is \( [\ce{S^{2-}}] \)?

\( K_{sp} = [\ce{Ag+}]^2[\ce{S^{2-}}] = 6.0 \times 10^{-50} \). Given \( [\ce{Ag+}] = 1.0 \times 10^{-15} \), \( (1.0 \times 10^{-15})^2 [\ce{S^{2-}}] = 6.0 \times 10^{-50} \), \( 1.0 \times 10^{-30} [\ce{S^{2-}}] = 6.0 \times 10^{-50} \), \( [\ce{S^{2-}}] = 6.0 \times 10^{-20} \, \text{M} \).

\( 3.0 \times 10^{-20} \)
\( 6.0 \times 10^{-20} \)
\( 1.0 \times 10^{-15} \)
\( 6.0 \times 10^{-50} \)
2

For the equilibrium \( \ce{P(g) + Q(g) <=> R(g)} \), if \( K_c = 2.0 \) and initial moles of \( \ce{P} \) and \( \ce{Q} \) are 1 each in a 1 L vessel, what is \( [\ce{R}] \) at equilibrium?

Let \( [\ce{R}] = x \), \( [\ce{P}] = 1 - x \), \( [\ce{Q}] = 1 - x \). \( K_c = \frac{[\ce{R}]}{[\ce{P}][\ce{Q}]} = \frac{x}{(1 - x)^2} = 2.0 \). Solving, \( x = 2(1 - x)^2 \), let \( y = 1 - x \), \( 1 - y = 2y^2 \), \( 2y^2 + y - 1 = 0 \), \( y = 0.5 \), \( x = 1 - 0.5 = 0.5 \, \text{M} \).

1.0 M
0.25 M
2.0 M
0.5 M
4

The solubility of \( \ce{Ag2S} \) in 0.01 M \( \ce{Na2S} \) is \( 2.0 \times 10^{-16} \, \text{M} \). What is its \( K_{sp} \)?

For \( \ce{Ag2S <=> 2Ag+ + S^{2-}} \), \( [\ce{Ag+}] = 4.0 \times 10^{-16} \), \( [\ce{S^{2-}}] \approx 0.01 \), \( K_{sp} = (4.0 \times 10^{-16})^2 \times 0.01 = 1.6 \times 10^{-33} \).

\( 2.0 \times 10^{-16} \)
\( 8.0 \times 10^{-32} \)
\( 1.6 \times 10^{-33} \)
\( 4.0 \times 10^{-18} \)
3

For \( \ce{A2(g) <=> 2A(g)} \), \( K_c = 0.36 \) at 300 K. If 0.5 mol \( \ce{A2} \) is in a 1 L vessel, what is the degree of dissociation?

Initial: \( [\ce{A2}] = 0.5 \, \text{M} \), \( [\ce{A}] = 0 \). Let \( \alpha \) be the degree of dissociation, \( [\ce{A2}] = 0.5 (1 - \alpha) \), \( [\ce{A}] = \alpha \). \( K_c = \frac{[\ce{A}]^2}{[\ce{A2}]} = \frac{(\alpha)^2}{0.5 (1 - \alpha)} = 0.36 \), \( \alpha^2 = 0.18 (1 - \alpha) \), \( \alpha \approx 0.36 \).

0.5
0.2
0.6
0.36
4

For the reaction \( \ce{2A(g) <=> B(g) + C(g)} \), if \( K_c = 0.25 \) and at equilibrium \( [\ce{B}] = 0.1 \, \text{M} \), \( [\ce{C}] = 0.1 \, \text{M} \), what is \( [\ce{A}] \)?

\( K_c = \frac{[\ce{B}][\ce{C}]}{[\ce{A}]^2} = 0.25 \). Substituting, \( 0.25 = \frac{(0.1)(0.1)}{[\ce{A}]^2} = \frac{0.01}{[\ce{A}]^2} \), \( [\ce{A}]^2 = \frac{0.01}{0.25} = 0.04 \), \( [\ce{A}] = 0.2 \, \text{M} \).

0.2 M
0.4 M
0.1 M
0.05 M
1

The solubility of \( \ce{AgBr} \) is \( 7.1 \times 10^{-7} \, \text{M} \). What is its \( K_{sp} \)?

For \( \ce{AgBr <=> Ag+ + Br-} \), \( K_{sp} = [\ce{Ag+}][\ce{Br-}] = S^2 = (7.1 \times 10^{-7})^2 = 5.04 \times 10^{-13} \).

\( 7.1 \times 10^{-7} \)
\( 1.42 \times 10^{-6} \)
\( 5.04 \times 10^{-13} \)
\( 2.52 \times 10^{-13} \)
3

For \( \ce{2NO2(g) <=> 2NO(g) + O2(g)} \), \( K_p = 0.05 \) at 600 K. If 1 mole of \( \ce{NO2} \) is placed in a 2 L vessel, what is \( P_{\ce{O2}} \) at equilibrium (\( R = 0.0831 \, \text{bar L/mol K} \))?

Initial: \( P_{\ce{NO2}} = \frac{1 \times 0.0831 \times 600}{2} = 24.93 \, \text{bar} \). Let \( 2x \) dissociate, \( P_{\ce{NO2}} = 24.93 - 2x \), \( P_{\ce{NO}} = 2x \), \( P_{\ce{O2}} = x \), total pressure = \( 24.93 + x \). \( K_p = \frac{(P_{\ce{NO}})^2 P_{\ce{O2}}}{(P_{\ce{NO2}})^2} = \frac{(2x)^2 x}{(24.93 - 2x)^2} = 0.05 \), \( 4x^3 = 0.05 (24.93 - 2x)^2 \), \( x \approx 0.71 \), \( P_{\ce{O2}} = 0.71 \, \text{bar} \).

0.71 bar
1.0 bar
0.35 bar
1.5 bar
1

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