Equilibrium Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The \( K_a \) of a weak acid \( \ce{HX} \) is \( 2.5 \times 10^{-5} \). What is its \( \text{p}K_a \)?

\( \text{p}K_a = -\log K_a = -\log(2.5 \times 10^{-5}) = 5 - \log 2.5 \approx 5 - 0.398 = 4.6 \).

5.0
4.6
2.5
9.4
2

For \( \ce{COCl2(g) <=> CO(g) + Cl2(g)} \), \( K_c = 0.16 \) at 700 K. If 0.2 mol \( \ce{COCl2} \) and 0.1 mol \( \ce{CO} \) are in a 1 L vessel, what is \( [\ce{Cl2}] \) at equilibrium?

Initial: \( [\ce{COCl2}] = 0.2 \, \text{M} \), \( [\ce{CO}] = 0.1 \, \text{M} \), \( [\ce{Cl2}] = 0 \). Let \( x = [\ce{Cl2}] \), \( [\ce{COCl2}] = 0.2 - x \), \( [\ce{CO}] = 0.1 + x \). \( K_c = \frac{[\ce{CO}][\ce{Cl2}]}{[\ce{COCl2}]} = \frac{(0.1 + x)x}{0.2 - x} = 0.16 \), \( 0.1x + x^2 = 0.032 - 0.16x \), \( x^2 + 0.26x - 0.032 = 0 \), \( x \approx 0.11 \, \text{M} \).

0.11 M
0.2 M
0.05 M
0.16 M
1

For the equilibrium \( \ce{X2(g) <=> 2X(g)} \), if the initial pressure of \( \ce{X2} \) is 2 atm and at equilibrium the total pressure is 3 atm, what is \( K_p \)?

Let the pressure of \( \ce{X} \) at equilibrium be \( 2p \), \( \ce{X2} = 2 - p \), total pressure = \( (2 - p) + 2p = 2 + p = 3 \), \( p = 1 \). \( P_{\ce{X2}} = 1 \, \text{atm} \), \( P_{\ce{X}} = 2 \, \text{atm} \). \( K_p = \frac{(P_{\ce{X}})^2}{P_{\ce{X2}}} = \frac{(2)^2}{1} = 4 \).

2
4
1
0.5
2

The solubility of \( \ce{Ag2SO4} \) in 0.1 M \( \ce{Na2SO4} \) is \( 5.0 \times 10^{-5} \, \text{M} \). What is its \( K_{sp} \)?

For \( \ce{Ag2SO4 <=> 2Ag+ + SO4^{2-}} \), \( [\ce{Ag+}] = 2 \times 5.0 \times 10^{-5} = 1.0 \times 10^{-4} \), \( [\ce{SO4^{2-}}] = 0.1 + 5.0 \times 10^{-5} \approx 0.1 \). \( K_{sp} = (1.0 \times 10^{-4})^2 \times 0.1 = 1.0 \times 10^{-9} \).

\( 5.0 \times 10^{-5} \)
\( 2.0 \times 10^{-8} \)
\( 1.0 \times 10^{-9} \)
\( 4.0 \times 10^{-10} \)
3

What is the conjugate acid of \( \ce{NH3} \)?

The conjugate acid of \( \ce{NH3} \) is formed by accepting a proton (\( \ce{H+} \)), resulting in \( \ce{NH4+} \).

\( \ce{N2H4} \)
\( \ce{NH2-} \)
\( \ce{NH4+} \)
\( \ce{H3O+} \)
3

The \( K_b \) of a weak base \( \ce{B} \) is \( 3.2 \times 10^{-6} \). What is its \( \text{p}K_b \)?

\( \text{p}K_b = -\log K_b = -\log(3.2 \times 10^{-6}) = 6 - \log 3.2 \approx 6 - 0.505 = 5.495 \approx 5.5 \).

6.0
5.5
3.2
8.5
2

The \( K_{sp} \) of \( \ce{PbCl2} \) is \( 1.7 \times 10^{-5} \). What is the solubility of \( \ce{PbCl2} \) in a 0.1 M \( \ce{KCl} \) solution?

For \( \ce{PbCl2 <=> Pb^{2+} + 2Cl-} \), \( K_{sp} = [\ce{Pb^{2+}}][\ce{Cl-}]^2 = 1.7 \times 10^{-5} \). Let solubility = \( S \), \( [\ce{Cl-}] = 0.1 + 2S \approx 0.1 \), \( S (0.1)^2 = 1.7 \times 10^{-5} \), \( S = \frac{1.7 \times 10^{-5}}{0.01} = 1.7 \times 10^{-3} \, \text{M} \).

\( 1.7 \times 10^{-5} \)
\( 1.7 \times 10^{-3} \)
\( 4.25 \times 10^{-4} \)
\( 8.5 \times 10^{-3} \)
2

A weak base \( \ce{BOH} \) has a \( K_b = 4.0 \times 10^{-6} \). What is the pH of a 0.01 M solution of \( \ce{BOH} \)?

For \( \ce{BOH <=> B+ + OH-} \), \( K_b = \frac{x^2}{0.01 - x} \approx \frac{x^2}{0.01} = 4.0 \times 10^{-6} \). Solving, \( x = [\ce{OH-}] = \sqrt{4.0 \times 10^{-8}} = 2.0 \times 10^{-4} \). \( \text{pOH} = -\log(2.0 \times 10^{-4}) \approx 3.7 \), \( \text{pH} = 14 - 3.7 = 10.3 \).

10.3
9.7
11.0
3.7
1

For \( \ce{A2(g) + B2(g) <=> 2AB(g)} \), \( K_p = 9 \) at 600 K. If initial pressures are \( P_{\ce{A2}} = 1 \, \text{atm} \), \( P_{\ce{B2}} = 1 \, \text{atm} \), what is \( P_{\ce{AB}} \) at equilibrium?

Let \( 2x \) be \( P_{\ce{AB}} \), \( P_{\ce{A2}} = 1 - x \), \( P_{\ce{B2}} = 1 - x \), total pressure = \( (1 - x) + (1 - x) + 2x = 2 \). \( K_p = \frac{(P_{\ce{AB}})^2}{P_{\ce{A2}} P_{\ce{B2}}} = \frac{(2x)^2}{(1 - x)^2} = 9 \), \( \frac{4x^2}{(1 - x)^2} = 9 \), \( \frac{2x}{1 - x} = 3 \), \( 2x = 3 - 3x \), \( 5x = 3 \), \( x = 0.6 \), \( P_{\ce{AB}} = 2 \times 0.6 = 1.2 \, \text{atm} \).

1.0 atm
0.6 atm
1.2 atm
1.8 atm
3

For the reaction \( \ce{A(g) + 2B(g) <=> 3C(g)} \), \( K_c = 27 \) at 400 K. If 1 mole of \( \ce{A} \) and 3 moles of \( \ce{B} \) are placed in a 1 L vessel, what is \( [\ce{C}] \) at equilibrium?

Initial: \( [\ce{A}] = 1 \, \text{M} \), \( [\ce{B}] = 3 \, \text{M} \), \( [\ce{C}] = 0 \). Let \( 3x \) be moles of \( \ce{C} \) formed, so \( \ce{A} \) decreases by \( x \), \( \ce{B} \) by \( 2x \). At equilibrium: \( [\ce{A}] = 1 - x \), \( [\ce{B}] = 3 - 2x \), \( [\ce{C}] = 3x \). \( K_c = \frac{[\ce{C}]^3}{[\ce{A}][\ce{B}]^2} = \frac{(3x)^3}{(1 - x)(3 - 2x)^2} = 27 \), \( \frac{27x^3}{(1 - x)(3 - 2x)^2} = 27 \), \( \frac{x^3}{(1 - x)(3 - 2x)^2} = 1 \). Solving, \( x^3 = (1 - x)(3 - 2x)^2 \), test \( x = 0.5 \): \( (0.5)^3 = 0.125 \), \( (1 - 0.5)(3 - 1)^2 = 0.5 \times 4 = 2 \) (not equal). Solving numerically, \( x \approx 0.75 \), \( [\ce{C}] = 3 \times 0.75 = 2.25 \, \text{M} \).

2.25 M
1.5 M
3.0 M
0.75 M
1

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