Question

use the following information to answer the next question.
hydrogen phosphate, h₃po₄(s), is a chemical substance, which, in pure anhydrous form, exists as white solid. it melts at 42.35°c to form a viscous colourless liquid. in aqueous solution, it ionizes as follows:
h₃po₄ (aq) + h₂o(l) ⇌ h₃o⁺(aq) + h₂po₄⁻ (aq)
h₂po₄⁻(aq) + h₂o(l) ⇌ h₃o⁺(aq) + hpo₄²⁻(aq)
hpo₄²⁻ (aq) + h₂o(l) ⇌ h₃o⁺(aq) + po₄³⁻(aq)
the concentration of the following substances involved in the reaction are:
h₃o⁺(aq) = x mol/l
po₄³⁻(aq) = y mol/l
h₃po₄(aq) = z mol/l
the equilibrium constant for the overall reaction is:
○ x³yz
○ (x²y)/z
○ x²yz
○ (x³y)/z
○ z/(x³y)

Explanation:

Step1: Determine the overall reaction

First, we add the three given ionization reactions:

1. $\ce{H_{3}PO_{4}(aq) + H_{2}O(l) <=> H_{3}O^{+}(aq) + H_{2}PO_{4}^{-}(aq)}$
2. $\ce{H_{2}PO_{4}^{-}(aq) + H_{2}O(l) <=> H_{3}O^{+}(aq) + HPO_{4}^{2-}(aq)}$
3. $\ce{HPO_{4}^{2-}(aq) + H_{2}O(l) <=> H_{3}O^{+}(aq) + PO_{4}^{3-}(aq)}$

When we add these reactions, the intermediate species ($\ce{H_{2}PO_{4}^{-}}$ and $\ce{HPO_{4}^{2-}}$) cancel out. The overall reaction is:
$\ce{H_{3}PO_{4}(aq) + 3H_{2}O(l) <=> 3H_{3}O^{+}(aq) + PO_{4}^{3-}(aq)}$

Step2: Write the equilibrium constant expression

For a reaction $\ce{aA + bB <=> cC + dD}$, the equilibrium constant $K$ is given by $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$. In the overall reaction, the concentration of liquid water ($\ce{H_{2}O(l)}$) is considered constant and is incorporated into the equilibrium constant, so we exclude it from the expression.

For the overall reaction $\ce{H_{3}PO_{4}(aq) + 3H_{2}O(l) <=> 3H_{3}O^{+}(aq) + PO_{4}^{3-}(aq)}$, the equilibrium constant $K$ is:
$K = \frac{[\ce{H_{3}O^{+}}]^3[\ce{PO_{4}^{3-}}]}{[\ce{H_{3}PO_{4}}]}$

Given $[\ce{H_{3}O^{+}}] = x$, $[\ce{PO_{4}^{3-}}] = y$, and $[\ce{H_{3}PO_{4}}] = z$, substituting these values into the equilibrium constant expression gives:
$K = \frac{x^3 y}{z}$

Answer:

$\boldsymbol{\frac{x^3 y}{z}}$ (corresponding to the option $\frac{x^3 y}{z}$)