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Equilibrium

Multiple Choice Questions

671.

5 mL of 0.4 N NaOH is mixed with 20 mL of 0.1 N HCl.The pH of the resulting solution will be

672.

The pH of the solution obtained by mixing 100 ml of a solution of pH = 3 with 400 mL of a solution of pH = 4 is

7 - log 2.8
4 - log 2.8
5 - log 2.8
3 - log 2.8

673.

The equilibrium constant of the reaction

A (s) + 2B²⁺ (aq) $⇌$ A²⁺ (aq) + 2B (s);

E $_{cell}^{°} = 0.0295 V is [\frac{2.303 RT}{F} = 0.059]$

2 × 10²
3 × 10²
2 × 10⁵
10

674.

An example for a neutral buffer is

ammonium hydroxide and ammonium chloride
acetic acid and sodium acetate
acetic acid and ammonium hydroxide
citric acid and sodium citrate

675.

Conjugate base of H₂PO $_{4}^{2 -}$

H₃PO $_{4}^{-}$
HPO $_{4}^{2 -}$
H₃PO₄
PO $_{4}^{3 -}$

676.
For Cr₂O $_{7}^{2 -}$ + 14 H⁺ + 6e^- → 2Cr³⁺ + 7H₂O; E° = 1.33 V. At [Cr₂O $_{7}^{2 -}$ ] = 4.5 millimole, [Cr³⁺] = 15 millimole, E is 1.067 V. The pH of the solution is nearly equal to

2

3

5

4

Cr₂O $_{7}^{2 -}$ + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

$E = E ° - \frac{0.059}{n} \log \frac{[{Cr}^{3 +}]^{2} [H_{2} O]^{7}}{[{Cr}_{2} O_{7}^{2 -}] [H^{+}]^{14}} 1.067 = 1.33 - \frac{0.059}{6} \log \frac{{(15 \times 10^{- 3})}^{2} (1)^{7}}{(4.5 \times 10^{- 3}) [H^{+}]^{14}} = \frac{0.059}{6} \log [\frac{225 \times 10^{- 6}}{(4.5 \times 10^{- 3}) [H^{+}]^{14}}] = 0.263 = \log [\frac{225 \times 10^{- 6}}{(4.5 \times 10^{- 3}) [H^{+}]^{14}}] = \frac{0.263 \times 6}{0.059} = 26.74 \log [\frac{50 \times 10^{- 3}}{[H^{+}]^{14}}] = 26.74$

log (50 × 10^-3) - log (H⁺)¹⁴ = 26.74

-1.3010 - 14 log [H⁺] = 26.74

-14 log [H⁺] = 26.74 + 1.3010

+14 pH = 28.041

pH = $\frac{28.041}{14}$ { $∵$ pH = -log [H⁺]}

= 2

677.

Identify a species which is not a Bronsted acid but a Lewis acid

BF₃
H $_{3}^{+}$ O
NH₃
HCl

678.

The pH of 10^-8 M HCl solution is

8
6.9586
more than 8
slightly more than 7

679.

For the equilibrium,

CaCO₃ (s) $⇌$ CaO (s) + CO₂ (g) ;

K_p = 1.64 atm at 1000 K

50 g of CaCO₃ in a 10 L closed vessel is heated to 1000 K. Percentage of CaCO₃ that remains unreacted at equilibrium is (Given, R = 0.082 L atm K^-1 mol^-1).