In a saturated solution of \(\mathrm{Na}_{3} \mathrm{PO}_{4},\left[\mathrm{Na}^{+}\right]=0.30 \mathrm{M} .\) What is the molar solubility of \(\mathrm{Na}_{3} \mathrm{PO}_{4} ?\)

Short Answer

Expert verified
The molar solubility of \(Na_{3}PO_{4}\) in a saturated solution is \(0.10M\).

Step by step solution

01

Write the Dissociation Equation

First, let's write down the dissociation equation for the given molecule \(Na_{3}PO_{4}\). In water, it dissociates into its ions: \(Na_{3}PO_{4} \leftrightarrow 3Na^{+} + PO_{4}^{3-}\).
02

Write the Solubility Product Expression

The solubility product expression for this reaction is given by the concentrations of the ions raised to their stoichiometric coefficients. We get \(K_{sp} = [Na^{+}]^{3} \cdot [PO_{4}^{3-}]\).
03

Solve for \(Na_{3}PO_{4}\) Molar Solubility

Given that \([Na^{+}]= 0.30M\) and from the dissociation, for each one \(Na_{3}PO_{4}\) that dissolves, three \(Na^{+}\) ions are formed and one \(PO_{4}^{3-}\) ion is formed. Thus, \([PO_{4}^{3-}] = 0.30M / 3 = 0.10M\). Hence, the molar solubility of \(Na_{3}PO_{4}\) in saturated solution equals the \(PO_{4}^{3-}\) concentration, which is \(0.10M\).

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