C(Na+) = C(phosphate) x (n(Na+)/n(phosphate))
where:
C(Na+) is the concentration of sodium ions in mol/L
C(phosphate) is the concentration of phosphate in mol/L
n(Na+) is the number of sodium ions in the phosphate molecule
n(phosphate) is the number of phosphate ions in the phosphate molecule
In this case, the concentration of phosphate is given as 2 Molarity. Since the phosphate ion has 3 sodium ions, n(Na+)/n(phosphate) is 3. Therefore:
C(Na+) = 2 Molarity x 3 = 6 Molarity
To calculate the number of sodium ions present in 500 milliliters of this solution, we can use the formula:
N(Na+) = C(Na+) x V
where:
N(Na+) is the number of sodium ions
C(Na+) is the concentration of sodium ions in mol/L
V is the volume of the solution in liters
Substituting the given values:
N(Na+) = 6 Molarity x 0.5 L = 3 moles
Therefore, there are 3 moles of sodium ions present in 500 milliliters of 2 Molarity aqueous phosphate.
