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3 years ago

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Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc

for the reverse reaction.
(1) Al(s) + NaOH(aq) + H2O(l) ⇋ Na[Al(OH)4](aq) + H2(g) Kc for balanced reaction = 11
(2) H2O(l) + SO3(g) ⇋ H2SO4 (aq) Kc for balanced reaction = 0.0123
(3) P4(s) + O2(g) ⇋ P4O6(s) Kc for balanced reaction = 1.56

1 answer:

marysya [2.9K]3 years ago

5 0

Explanation:

1) $2 Al(s) + 2 NaOH(aq) + 6 H_2O(l) \longleftrightarrow 2 Na[Al(OH)_4](aq) + 7 H_2(g)$

$Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{11}=0.091$

2) $H_2O(l) + SO_3(g) \longleftrightarrow H_2SO_4 (aq)$

$Kc=\frac{[H_2SO_4]}{[SO_3]^2}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{0.0123}=81.3$

3) $P_4(s) + 3 O_2(g) \longleftrightarrow P_4O_6(s)$

$Kc=\frac{1}{[O_2]^3}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{1.56}=0.641$

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Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

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$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

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