Equilibrium Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For \( \ce{A(g) + B(g) <=> 2C(g)} \), \( K_p = 4 \) at 300 K. If the total pressure at equilibrium is 3 atm and \( P_{\ce{A}} = P_{\ce{B}} \), what is \( P_{\ce{C}} \)?

Let \( P_{\ce{A}} = P_{\ce{B}} = x \), \( P_{\ce{C}} = 3 - 2x \). \( K_p = \frac{(P_{\ce{C}})^2}{P_{\ce{A}} P_{\ce{B}}} = \frac{(3 - 2x)^2}{x^2} = 4 \), \( \frac{3 - 2x}{x} = 2 \), \( 3 - 2x = 2x \), \( 3 = 4x \), \( x = 0.75 \), \( P_{\ce{C}} = 3 - 2 \times 0.75 = 1.5 \, \text{atm} \).

1.0 atm
2.0 atm
0.75 atm
1.5 atm
4

A weak base \( \ce{NH3} \) (\( K_b = 1.8 \times 10^{-5} \)) is mixed with 0.02 M \( \ce{NH4Cl} \) in a 1:1 volume ratio. If the final \( [\ce{NH3}] = 0.05 \, \text{M} \), what is the pH?

\( [\ce{NH4+}] = 0.01 \, \text{M} \), \( \text{p}K_b = 4.74 \), \( \text{pOH} = 4.74 + \log \frac{0.01}{0.05} = 4.74 - 0.7 = 4.04 \), \( \text{pH} = 14 - 4.04 = 9.96 \).

4.74
9.96
10.5
9.0
2

For \( \ce{Ag2CrO4(s) <=> 2Ag+(aq) + CrO4^{2-}(aq)} \), \( K_{sp} = 1.1 \times 10^{-12} \). In a 0.005 M \( \ce{AgNO3} \) solution, what is \( [\ce{CrO4^{2-}}] \) in a saturated solution?

\( K_{sp} = [\ce{Ag+}]^2[\ce{CrO4^{2-}}] = 1.1 \times 10^{-12} \). \( [\ce{Ag+}] = 0.005 + 2[\ce{CrO4^{2-}}] \approx 0.005 \), \( (0.005)^2 [\ce{CrO4^{2-}}] = 1.1 \times 10^{-12} \), \( [\ce{CrO4^{2-}}] = \frac{1.1 \times 10^{-12}}{2.5 \times 10^{-5}} = 4.4 \times 10^{-8} \, \text{M} \).

\( 1.1 \times 10^{-12} \)
\( 4.4 \times 10^{-8} \)
\( 2.2 \times 10^{-7} \)
\( 5.5 \times 10^{-9} \)
2

A weak acid \( \ce{HX} \) (\( K_a = 1.0 \times 10^{-5} \)) and its salt \( \ce{NaX} \) are mixed in equal volumes of 0.2 M and 0.1 M solutions, respectively. What is the pH?

After mixing: \( [\ce{HX}] = \frac{0.2}{2} = 0.1 \, \text{M} \), \( [\ce{X-}] = \frac{0.1}{2} = 0.05 \, \text{M} \). \( \text{pH} = \text{p}K_a + \log \frac{[\ce{X-}]}{[\ce{HX}]} = 5 + \log \frac{0.05}{0.1} = 5 - 0.301 = 4.7 \).

5.0
4.3
4.7
5.3
3

For the equilibrium \( \ce{2A(g) <=> B(g) + C(g)} \), if \( K_c = 0.04 \) and \( [\ce{A}] = 0.4 \, \text{M} \) at equilibrium, what is \( [\ce{B}] \)?

\( K_c = \frac{[\ce{B}][\ce{C}]}{[\ce{A}]^2} = 0.04 \), \( [\ce{B}] = [\ce{C}] = x \), \( 0.04 = \frac{x^2}{(0.4)^2} = \frac{x^2}{0.16} \), \( x^2 = 0.0064 \), \( x = 0.08 \, \text{M} \).

0.4 M
0.16 M
0.08 M
0.02 M
3

A weak base \( \ce{NH3} \) (\( K_b = 1.8 \times 10^{-5} \)) is 1% ionized in a solution. What is \( [\ce{NH3}] \)?

Let \( [\ce{NH3}] = C \), \( \alpha = 0.01 \), \( [\ce{OH-}] = C \alpha = 0.01C \). \( K_b = \frac{(0.01C)^2}{C(1 - 0.01)} \approx \frac{0.0001C^2}{0.99C} = 1.01 \times 10^{-4}C = 1.8 \times 10^{-5} \), \( C = \frac{1.8 \times 10^{-5}}{1.01 \times 10^{-4}} \approx 0.178 \, \text{M} \).

0.1 M
0.178 M
0.018 M
1.0 M
2

A weak acid \( \ce{HY} \) (\( K_a = 1.0 \times 10^{-5} \)) has a 0.2 M solution mixed with 0.1 M \( \ce{NaY} \) in a 1:1 volume ratio. What is the pH?

After mixing: \( [\ce{HY}] = 0.1 \, \text{M} \), \( [\ce{Y-}] = 0.05 \, \text{M} \), \( \text{p}K_a = 5 \), \( \text{pH} = 5 + \log \frac{0.05}{0.1} = 5 - 0.301 = 4.7 \).

5.0
4.0
5.3
4.7
4

A solution has a pH of 2.5. What is \( [\ce{H+}] \) in mol/L?

\( \text{pH} = -\log [\ce{H+}] \), \( 2.5 = -\log [\ce{H+}] \), \( [\ce{H+}] = 10^{-2.5} \approx 3.16 \times 10^{-3} \).

\( 1.0 \times 10^{-2} \)
\( 5.0 \times 10^{-3} \)
\( 3.16 \times 10^{-3} \)
\( 2.5 \times 10^{-3} \)
3

For \( \ce{CH4(g) + H2O(g) <=> CO(g) + 3H2(g)} \), \( K_c = 0.1 \) at 800 K. If 0.5 mol \( \ce{CH4} \) and 0.5 mol \( \ce{H2O} \) are in a 2 L vessel, what is \( [\ce{H2}] \) at equilibrium?

Initial: \( [\ce{CH4}] = [\ce{H2O}] = 0.25 \, \text{M} \), \( [\ce{CO}] = [\ce{H2}] = 0 \). Let \( 3x = [\ce{H2}] \), \( [\ce{CO}] = x \), \( [\ce{CH4}] = [\ce{H2O}] = 0.25 - x \). \( K_c = \frac{[\ce{CO}][\ce{H2}]^3}{[\ce{CH4}][\ce{H2O}]} = \frac{x (3x)^3}{(0.25 - x)^2} = 0.1 \), \( 27x^4 = 0.1 (0.25 - x)^2 \), \( x \approx 0.06 \), \( [\ce{H2}] = 0.18 \, \text{M} \).

0.25 M
0.18 M
0.06 M
0.3 M
2

A 0.05 M solution of a weak base \( \ce{B} \) has a pH of 10. What is its \( K_b \)?

\( \text{pH} = 10 \), \( \text{pOH} = 14 - 10 = 4 \), \( [\ce{OH-}] = 10^{-4} \). For \( \ce{B + H2O <=> BH+ + OH-} \), \( K_b = \frac{[\ce{OH-}]^2}{[\ce{B}]} = \frac{(10^{-4})^2}{0.05} = \frac{10^{-8}}{0.05} = 2.0 \times 10^{-7} \).

\( 1.0 \times 10^{-6} \)
\( 2.0 \times 10^{-7} \)
\( 5.0 \times 10^{-8} \)
\( 4.0 \times 10^{-7} \)
2

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