Chemical and Ionic Equilibrium - Result Question 110
54. One mole of nitrogen is mixed with three moles of hydrogen in a four litre container. If 0.25 per cent of nitrogen is converted to ammonia by the following reaction
$ N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g) \text {, then } $
calculate the equilibrium constant, $K _c$ in concentration units. What will be the value of $K _c$ for the following equilibrium?
$ \frac{1}{2} N _2(g)+\frac{3}{2} H _2(g) \rightleftharpoons NH _3(g) $
$(1981,4 M)$
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Solution:
(i) $CH _3 COOH \rightleftharpoons CH _3 COO^{-}+H^{+}$
$ C(1-\alpha) \quad C \alpha \quad C \alpha $
If no $HCl$ is present,
$ \begin{aligned} {[HCl] } & =\frac{0.2}{2}=0.10 M \\ {\left[CH _3 COOH\right] } & =0.10 M \end{aligned} $
The major contributor of $H^{+}$in solution is $HCl$.
$ \begin{aligned} K _a & =\frac{C \alpha(0.1)}{C(1-\alpha)}=1.75 \times 10^{-5} \\ \alpha & =1.75 \times 10^{-4} \end{aligned} $
(ii) $mmol$ of $NaOH$ added $=\frac{6}{40} \times 1000=150$
$ mmol \text { of } HCl=500 \times 0.2=100 $
$mmol$ of $CH _3 COOH=500 \times 0.2=100$
After neutralisation, $mmol$ of $CH _3 COOH=50$
$mmol$ of $CH _3 COONa=50$
$ pH=p K _a=4.75 $
BETA
