Question

write the equilibrium constant expression for this reaction:
3h^{+}(aq)+po_{4}^{3 - }(aq)\to h_{3}po_{4}(aq)

Explanation:

Step1: Recall the formula for equilibrium constant

For a reaction \(aA + bB
ightleftharpoons cC + dD\), the equilibrium constant \(K=\frac{[C]^c[D]^d}{[A]^a[B]^b}\), where \([X]\) represents the molar concentration of species \(X\).

Step2: Identify the reactants and products

In the reaction \(3H^{+}(aq)+PO_{4}^{3 - }(aq)
ightleftharpoons H_{3}PO_{4}(aq)\), the reactants are \(H^{+}\) and \(PO_{4}^{3 - }\) with coefficients \(3\) and \(1\) respectively, and the product is \(H_{3}PO_{4}\) with coefficient \(1\).

Step3: Write the equilibrium constant expression

Using the formula from Step 1, we substitute the species and their coefficients. So, \(K=\frac{[H_{3}PO_{4}]}{[H^{+}]^{3}[PO_{4}^{3 - }]}\)

Answer:

\(K=\frac{[H_{3}PO_{4}]}{[H^{+}]^{3}[PO_{4}^{3 - }]}\)