Equilibrium Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A weak acid \( \ce{HA} \) (\( K_a = 1.6 \times 10^{-6} \)) is mixed with 0.05 M \( \ce{NaA} \) and 0.1 M \( \ce{HCl} \) in equal volumes. What is the pH?

After mixing: \( [\ce{HCl}] = 0.05 \, \text{M} \), \( [\ce{A-}] = 0.025 \, \text{M} \), \( [\ce{HA}] \) negligible due to common ion effect. \( K_a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \), \( [\ce{H+}] \approx 0.05 \), \( \text{pH} = -\log(0.05) = 1.3 \).

6.0
2.0
1.3
4.8
3

In the equilibrium \( \ce{2P(g) <=> Q(g) + R(g)} \), adding \( \ce{Q} \) at constant volume will shift the equilibrium in which direction?

According to Le Chatelier’s principle, adding \( \ce{Q} \) increases its concentration, shifting the equilibrium left to reduce \( \ce{Q} \) by forming more \( \ce{P} \).

Right
No shift
Depends on \( K_c \)
Left
4

For \( \ce{2NOCl(g) <=> 2NO(g) + Cl2(g)} \), \( K_p = 1.8 \times 10^{-2} \) at 500 K. If 2 moles of \( \ce{NOCl} \) are placed in a 2 L vessel, what is the total pressure at equilibrium (\( R = 0.0831 \, \text{bar L/mol K} \))?

Initial: \( P_{\ce{NOCl}} = \frac{2 \times 0.0831 \times 500}{2} = 41.55 \, \text{bar} \). Let \( 2x \) mol dissociate, \( P_{\ce{NOCl}} = 41.55(1 - x) \), \( P_{\ce{NO}} = 41.55x \), \( P_{\ce{Cl2}} = 41.55 \times \frac{x}{2} \), total pressure = \( 41.55(1 - x + x + \frac{x}{2}) = 41.55(1 + \frac{x}{2}) \). \( K_p = \frac{(P_{\ce{NO}})^2 P_{\ce{Cl2}}}{(P_{\ce{NOCl}})^2} = \frac{(41.55x)^2 (41.55 \frac{x}{2})}{[41.55(1 - x)]^2} = 1.8 \times 10^{-2} \), \( \frac{41.55x^3}{2(1 - x)^2} = 0.018 \), \( x^3 = 8.67 \times 10^{-4} (1 - x)^2 \), \( x \approx 0.09 \), total pressure = \( 41.55 \times 1.045 = 43.42 \, \text{bar} \).

43.42 bar
41.55 bar
45.0 bar
39.0 bar
1

A weak acid \( \ce{HY} \) (\( K_a = 2.0 \times 10^{-5} \)) is mixed with 0.01 M \( \ce{NaOH} \) in a 2:1 volume ratio (acid:base). If the final \( [\ce{HY}] = 0.04 \, \text{M} \), what is the pH?

Let volumes be 2V and V, total volume = 3V. Moles: \( \ce{HY} = 0.04 \times 3V \), initial \( [\ce{HY}] = 0.06 \, \text{M} \), moles \( \ce{NaOH} = 0.01V \), \( [\ce{Y-}] = \frac{0.01V}{3V} = 0.00333 \, \text{M} \), remaining \( [\ce{HY}] = 0.04 \). \( \text{pH} = 4.7 + \log \frac{0.00333}{0.04} = 4.7 - 1.08 = 3.62 \).

3.62
4.7
5.0
2.5
1

In a closed container, the vapor pressure of a liquid reaches a constant value at a fixed temperature. What does this indicate?

The constant vapor pressure indicates that the rate of evaporation equals the rate of condensation, establishing a dynamic equilibrium between the liquid and its vapor.

Dynamic equilibrium
Static equilibrium
No reaction occurring
Complete evaporation
1

For \( \ce{AB(s) <=> A+(aq) + B-(aq)} \), if \( K_{sp} = 4.0 \times 10^{-8} \), what is \( [\ce{B-}] \) in mol/L?

\( K_{sp} = [\ce{A+}][\ce{B-}] = S^2 = 4.0 \times 10^{-8} \), \( S = \sqrt{4.0 \times 10^{-8}} = 2.0 \times 10^{-4} \), so \( [\ce{B-}] = 2.0 \times 10^{-4} \, \text{M} \).

\( 4.0 \times 10^{-8} \)
\( 2.0 \times 10^{-4} \)
\( 1.0 \times 10^{-4} \)
\( 8.0 \times 10^{-8} \)
2

For \( \ce{A(g) <=> 2B(g)} \), \( K_c = 0.25 \) at 500 K. If 0.4 mol \( \ce{A} \) is in a 1 L vessel, what is \( [\ce{B}] \) at equilibrium?

Initial: \( [\ce{A}] = 0.4 \, \text{M} \), \( [\ce{B}] = 0 \). Let \( 2x = [\ce{B}] \), \( [\ce{A}] = 0.4 - x \). \( K_c = \frac{[\ce{B}]^2}{[\ce{A}]} = \frac{(2x)^2}{0.4 - x} = 0.25 \), \( 4x^2 = 0.1 - 0.25x \), \( x \approx 0.125 \), \( [\ce{B}] = 0.25 \, \text{M} \).

0.4 M
0.1 M
0.25 M
0.5 M
3

A weak base \( \ce{BOH} \) (\( K_b = 2.5 \times 10^{-5} \)) has a 0.1 M solution with \( \alpha = 0.005 \). What is the pH?

\( [\ce{OH-}] = 0.1 \times 0.005 = 5.0 \times 10^{-4} \), \( \text{pOH} = 3.3 \), \( \text{pH} = 14 - 3.3 = 10.7 \).

3.3
11.0
10.0
10.7
4

The \( K_w \) of water at 310 K is \( 2.9 \times 10^{-14} \). What is the pH of pure water at this temperature?

\( K_w = [\ce{H+}][\ce{OH-}] = 2.9 \times 10^{-14} \). In pure water, \( [\ce{H+}] = \sqrt{2.9 \times 10^{-14}} \approx 1.7 \times 10^{-7} \), \( \text{pH} = -\log(1.7 \times 10^{-7}) \approx 6.77 \).

7.0
6.0
6.77
14.0
3

For \( \ce{CO(g) + 2H2(g) <=> CH3OH(g)} \), \( K_c = 10 \) at 600 K. If 0.2 mol \( \ce{CO} \) and 0.4 mol \( \ce{H2} \) are in a 1 L vessel, what is \( [\ce{CH3OH}] \) at equilibrium?

Initial: \( [\ce{CO}] = 0.2 \, \text{M} \), \( [\ce{H2}] = 0.4 \, \text{M} \), \( [\ce{CH3OH}] = 0 \). Let \( x = [\ce{CH3OH}] \), \( [\ce{CO}] = 0.2 - x \), \( [\ce{H2}] = 0.4 - 2x \). \( K_c = \frac{[\ce{CH3OH}]}{[\ce{CO}][\ce{H2}]^2} = \frac{x}{(0.2 - x)(0.4 - 2x)^2} = 10 \), \( x \approx 0.15 \, \text{M} \).

0.2 M
0.15 M
0.1 M
0.3 M
2

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0