Respuesta :
Answer: The mass of [tex]Na_3PO_4[/tex] that will be needed is 30.75 g.
Explanation:
The chemical equation for the ionization of sodium phosphate follows:
[tex]Na_3PO_4\rightarrow 3Na^++PO_4^{3-}[/tex]
We are given:
Total concentration of Sodium ions in the solution = 1.5 M
So, concentration of sodium phosphate will be = [tex]\frac{1.5}{3}=0.5M[/tex]
To calculate the moles of a solute, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
Volume of solution = 375 mL = 0.375 L (Conversion factor: 1 L = 1000 mL)
Molarity of the solution = 0.5 moles/ L
Putting values in above equation, we get:
[tex]0.5mol/L=\frac{\text{Moles of }Na_3PO_4}{0.065L}\\\\\text{Moles of }Na_3PO_4=0.1875mol[/tex]
Now, calculating the mass of sodium phosphate using equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of sodium phosphate = 164 g/mol
Moles of sodium phosphate = 0.1875 mol
Putting values in above equation, we get:
[tex]0.1875mol=\frac{\text{Mass of sodium phosphate}}{164g/mol}\\\\\text{Mass of sodium phosphate}=30.75g[/tex]
Hence, the mass of sodium phosphate used to prepare a solution will be 30.75 grams.
