Question

The ionic compound, Na3PO4, will fully dissolve in water. If 19.68 g is dissolved in 504.0 mL water, find the following ion concentrations.

part 1:The concentration of the Na+ is: ____M

part 2: The concentration of the PO43- is: ___M

Answer 1

Answer: The concentration of $Na^+\text{ and }PO_4^{3-}$ is 0.72 M and 0.24 M respectively.

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of sodium phosphate = 19.68 g

Molar mass of sodium phosphate = 164 g/mol

Volume of solution = 504.0 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{19.68\times 1000}{164\times 504.0}\\\\\text{Molarity of solution}=0.24M$

We are given:

An ionic compound having chemical formula $Na_3PO_4$. Its ionization reaction follows:

$Na_3PO_4(aq.)\rightarrow 3Na^+(aq.)+PO_4^{3-}(aq.)$

1 mole of sodium phosphate produces 3 moles of sodium ions and 1 mole of phosphate ions

Concentration of sodium ions = $(3\times 0.24)=0.72M$

Concentration of phosphate ion = $(1\times 0.24)=0.24M$

Hence, the concentration of $Na^+\text{ and }PO_4^{3-}$ is 0.72 M and 0.24 M respectively.

Answer 2

Taking into account the definition of molarity, the concentration of the Na⁺ is 0.72 M and the concentration of the PO₄³⁻ is 0.24 M.

Definition of molarity

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$molarity=\frac{number of moles}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$.

Molarity of Na₃PO₄

In this case, you know:

• number of moles= 19.68 g× $\frac{1 mole}{164 g}$= 0.12 moles (being 164 the molar mass of Na₃PO₄)
• volume= 504 mL= 0.504 L

Replacing in the definition of molarity:

$molarity=\frac{0.12 moles}{0.504 L}$

Solving:

molarity= 0.24 M

Concentration of sodium ions and phosphate ion

The ionization reaction is:

Na₃PO₄ → 3 Na⁺ + PO₄³⁻

You can observe that 1 mole of sodium phosphate produces 3 moles of sodium ions and 1 mole of phosphate ions.

Knowing the concentration of Na₃PO₄, the concentration of the corresponding ions can be calculated by:

• Concentration of the Na⁺: 3×  0.24 M= 0.72 M
• Concentration of the PO₄³⁻: 1×  0.24 M= 0.24 M

In summary, the concentration of the Na⁺ is 0.72 M and the concentration of the PO₄³⁻ is 0.24 M.

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