Question

$H_3{PO}_4$ is a tribasic acid, it undergoes ionization as:

$H_3{PO}_4\leftrightharpoons H^{+}+H_2{PO}_4^{-};K_1$
$H_2{PO}_4^{-}\rightleftharpoons H^{+}+H{PO}_4^{2-};K_2$
$H{PO}_4^{2-}\rightleftharpoons H^{+}+{PO}_4^{3-};K_3$

Then, equilibrium constant for the following reaction will be:
$H_3{PO}_4\rightleftharpoons 3H^{+}+{PO}_4^{3-}$

• A. $K_1{K}_2{K}_3$
• B. $\frac{K_1{K}_2}{K_3}$
• C. $\frac{K_1{K}_3}{K_2}$
• D. $\frac{K_1{K}_2}{{K_3}^2}$

Answer 1

The correct option is A $K_1{K}_2{K}_3$

$Explanation:$

$H_{3}P{O}_4\rightleftharpoons H^{+}+H_{2}P{O}_4^{-};K_1\to \left(1\right)$

$H_{2}P{O}_4^{-}\rightleftharpoons H^{+}+HP{O}_4^{-2};K_2\to \left(2\right)$

$HP{O}_4^{-2}\rightleftharpoons H^{+}+P{O}_4^{-3};K_3\to \left(3\right)$

Add equation (1) and (2)

$H_{3}P{O}_4\rightleftharpoons H^{+}+H_{2}P{O}_4^{-};K_1$

$+$ $H_{2}P{O}_4^{-2}\rightleftharpoons H^{+}+HP{O}_4^{-2};K_2$

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$=H_{3}P{O}_4\rightleftharpoons 2H^{+}+HP{O}_4^{-2};K_1{K}_2\to \left(4\right)$

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$H_{3}P{O}_4\rightleftharpoons 2H^{+}+HP{O}_4^{-2};K_1{K}_2$

$+$ $HP{O}_4^{-2}\rightleftharpoons H^{+}+P{O}_4^{-3};K_3$

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$H_{3}P{O}_4\rightleftharpoons 3H^{+}+P{O}_4^{-3};K_1{K}_2{K}_3\to \left(5\right)$

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$\therefore K=K_1{K}_2{K}_3$

When the reaction can be expressed as the sum of two other reactions, the $K_c$ of the overall reaction is equal to the product of equilibrium constants of individual reaction.

Thus $K=K_1{K}_2{K}_3$

Hence the correct answer is option $A$