8. (5 pts) What is the concentration (M) of sodium ions in a 0.35 M sodium phosphate solution?

[tex]\[ \text{(0.35 M Na}_3\text{PO}_4 \text{ per 1 L solution)} \][/tex]



Answer :

To determine the concentration of sodium ions in a 0.35 M sodium phosphate (NaPO₄) solution, follow these steps:

1. Understand the Dissociation of Sodium Phosphate (NaPO₄):
Sodium phosphate dissociates completely in water to produce sodium (Na⁺) ions and phosphate (PO₄³⁻) ions. The balanced dissociation equation is:
[tex]\[ \text{NaPO₄} \rightarrow 3\text{Na}^+ + \text{PO₄}^{3-} \][/tex]

2. Determine the Mole Ratio:
From the dissociation equation, we can observe that one molecule of sodium phosphate produces three sodium ions. This gives us a mole ratio of 1:3 between sodium phosphate and sodium ions.

3. Calculate the Concentration of Sodium Ions:
Given that the concentration of sodium phosphate is 0.35 M, and knowing the mole ratio from step 2, we can calculate the concentration of sodium ions.
[tex]\[ \text{Concentration of Na⁺ ions} = \text{Concentration of NaPO₄} \times \text{Number of Na⁺ ions per NaPO₄ molecule} \][/tex]
[tex]\[ \text{Concentration of Na⁺ ions} = 0.35\, \text{M} \times 3 \][/tex]

4. Compute the Result:
[tex]\[ \text{Concentration of Na⁺ ions} = 1.0499999999999998\, \text{M} \][/tex]

Therefore, the concentration of sodium ions in a 0.35 M sodium phosphate solution is approximately 1.05 M.