Equilibrium Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A weak acid \( \ce{HB} \) (\( K_a = 1.0 \times 10^{-6} \)) is mixed with 0.03 M \( \ce{NaOH} \) in a 3:1 volume ratio (acid:base). If the final \( [\ce{HB}] = 0.08 \, \text{M} \), what is the pH?

Total volume = 4V, initial \( [\ce{HB}] = 0.107 \, \text{M} \), \( [\ce{B-}] = \frac{0.03V}{4V} = 0.0075 \, \text{M} \), \( \text{p}K_a = 6 \), \( \text{pH} = 6 + \log \frac{0.0075}{0.08} = 6 - 1.03 = 4.97 \).

4.97
6.0
5.5
4.0
1

The \( K_{sp} \) of \( \ce{Mg(OH)2} \) is \( 1.8 \times 10^{-11} \). What is the pH at which \( [\ce{Mg^{2+}}] = 1.0 \times 10^{-4} \, \text{M} \) in a saturated solution?

For \( \ce{Mg(OH)2 <=> Mg^{2+} + 2OH-} \), \( K_{sp} = [\ce{Mg^{2+}}][\ce{OH-}]^2 = 1.8 \times 10^{-11} \), \( (1.0 \times 10^{-4})[\ce{OH-}]^2 = 1.8 \times 10^{-11} \), \( [\ce{OH-}]^2 = 1.8 \times 10^{-7} \), \( [\ce{OH-}] = 1.34 \times 10^{-4} \), \( \text{pOH} = 3.87 \), \( \text{pH} = 14 - 3.87 = 10.13 \).

7.0
10.13
3.87
11.0
1

The solubility of \( \ce{BaSO4} \) in 0.05 M \( \ce{Na2SO4} \) is \( 2.0 \times 10^{-9} \, \text{M} \). What is its \( K_{sp} \)?

For \( \ce{BaSO4 <=> Ba^{2+} + SO4^{2-}} \), \( [\ce{Ba^{2+}}] = 2.0 \times 10^{-9} \), \( [\ce{SO4^{2-}}] \approx 0.05 \), \( K_{sp} = (2.0 \times 10^{-9}) \times 0.05 = 1.0 \times 10^{-10} \).

\( 2.0 \times 10^{-9} \)
\( 4.0 \times 10^{-10} \)
\( 1.0 \times 10^{-10} \)
\( 5.0 \times 10^{-11} \)
3

For \( \ce{2A(g) <=> B(g)} \), \( K_c = 0.04 \) at 300 K. If 0.5 mol \( \ce{A} \) is in a 1 L vessel, what is \( [\ce{B}] \) at equilibrium?

Initial: \( [\ce{A}] = 0.5 \, \text{M} \), \( [\ce{B}] = 0 \). Let \( x = [\ce{B}] \), \( [\ce{A}] = 0.5 - 2x \). \( K_c = \frac{[\ce{B}]}{[\ce{A}]^2} = \frac{x}{(0.5 - 2x)^2} = 0.04 \), \( x = 0.04 (0.5 - 2x)^2 \), \( \sqrt{x} = 0.2 (0.5 - 2x) \), \( x \approx 0.04 \, \text{M} \).

0.5 M
0.1 M
0.04 M
0.08 M
3

A weak base \( \ce{BOH} \) (\( K_b = 2.0 \times 10^{-5} \)) has a 0.1 M solution with pH 10.5. What is the degree of ionization?

\( \text{pOH} = 14 - 10.5 = 3.5 \), \( [\ce{OH-}] = 10^{-3.5} \approx 3.16 \times 10^{-4} \), \( \alpha = \frac{[\ce{OH-}]}{[\ce{BOH}]} = \frac{3.16 \times 10^{-4}}{0.1} = 3.16 \times 10^{-3} = 0.00316 \).

0.01
0.005
0.001
0.00316
4

For \( \ce{XY(s) <=> X+(aq) + Y-(aq)} \), if \( K_{sp} = 9.0 \times 10^{-6} \), what is \( [\ce{X+}] \) in mol/L?

\( K_{sp} = [\ce{X+}][\ce{Y-}] = S^2 = 9.0 \times 10^{-6} \), \( S = \sqrt{9.0 \times 10^{-6}} = 3.0 \times 10^{-3} \), so \( [\ce{X+}] = 3.0 \times 10^{-3} \, \text{M} \).

\( 9.0 \times 10^{-6} \)
\( 4.5 \times 10^{-3} \)
\( 1.5 \times 10^{-3} \)
\( 3.0 \times 10^{-3} \)
4

Which species acts as a Lewis acid?

A Lewis acid accepts an electron pair. \( \ce{BF3} \) has an incomplete octet and can accept an electron pair, making it a Lewis acid.

\( \ce{NH3} \)
\( \ce{OH-} \)
\( \ce{F-} \)
\( \ce{BF3} \)
4

For \( \ce{A(g) + B(g) <=> 2C(g)} \), \( K_p = 1 \) at 300 K. If the total pressure at equilibrium is 3 atm and \( P_{\ce{A}} = 1 \, \text{atm} \), what is \( P_{\ce{C}} \)?

Total pressure = \( P_{\ce{A}} + P_{\ce{B}} + P_{\ce{C}} = 3 \), \( P_{\ce{B}} + P_{\ce{C}} = 2 \), \( K_p = \frac{(P_{\ce{C}})^2}{P_{\ce{A}} P_{\ce{B}}} = \frac{(P_{\ce{C}})^2}{1 \cdot (2 - P_{\ce{C}})} = 1 \), \( P_{\ce{C}}^2 = 2 - P_{\ce{C}} \), \( P_{\ce{C}} \approx 1.414 \, \text{atm} \).

1.0 atm
2.0 atm
0.5 atm
1.414 atm
4

For \( \ce{A(g) + 2B(g) <=> 2C(g)} \), if \( K_c = 4 \) and \( [\ce{C}] = 0.8 \, \text{M} \), \( [\ce{B}] = 0.4 \, \text{M} \) at equilibrium, what is \( [\ce{A}] \)?

\( K_c = \frac{[\ce{C}]^2}{[\ce{A}][\ce{B}]^2} = 4 \), \( 4 = \frac{(0.8)^2}{[\ce{A}](0.4)^2} = \frac{0.64}{[\ce{A}] \times 0.16} \), \( 4 \times 0.16 [\ce{A}] = 0.64 \), \( 0.64 [\ce{A}] = 0.64 \), \( [\ce{A}] = 1.0 \, \text{M} \).

0.8 M
0.4 M
2.0 M
1.0 M
4

A weak base \( \ce{BOH} \) has \( K_b = 4.0 \times 10^{-4} \). If its 0.02 M solution has a pH of 10.8, what is the percentage ionization?

\( \text{pH} = 10.8 \), \( \text{pOH} = 14 - 10.8 = 3.2 \), \( [\ce{OH-}] = 10^{-3.2} \approx 6.31 \times 10^{-4} \). \( [\ce{BOH}] = 0.02 - 6.31 \times 10^{-4} \approx 0.0194 \). \( K_b = \frac{[\ce{OH-}]^2}{[\ce{BOH}]} = \frac{(6.31 \times 10^{-4})^2}{0.0194} \approx 2.05 \times 10^{-5} \), but given \( K_b = 4.0 \times 10^{-4} \), ionization \( \alpha = \frac{[\ce{OH-}]}{[\ce{BOH}]_{\text{initial}}} = \frac{6.31 \times 10^{-4}}{0.02} = 0.03155 \), % = 3.155%.

2.0%
3.16%
4.0%
5.0%
2

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