Question

Phosphorus is present in seawater to the extent of \(0.07 \mathrm{ppm}\) by mass. If the phosphorus is present as phosphate, \(\mathrm{PO}_{4}{ }^{3-}\), calculate the corresponding molar concentration of phosphate in seawater.

Short answer

The molar concentration of phosphate in seawater is approximately \(2.26 \times 10^{-6} \, \mathrm{mol/L}\).

Step-by-step solution

Convert ppm to mass concentration

First, let's convert the given concentration of phosphorus from ppm to a more usable unit. 1 ppm is equivalent to 1 mg of phosphorus in 1 L of seawater. So the mass concentration of phosphorus is: 0.07 ppm = 0.07 mg/L

Calculate the mass of phosphate ion

Now, we need to find out the mass of the phosphate ion, \(\mathrm{PO}_{4}{ }^{3-}\), in which the phosphorus is present. The molecular weight of phosphate ion (\(\mathrm{PO}_{4}{ }^{3-}\)) can be calculated as follows: Molecular weight of $\mathrm{P} = 30.97 \, \mathrm{g/mol} \\ Molecular weight of \(\mathrm{O} = 16.00 \, \mathrm{g/mol}\) Molecular weight of \(\mathrm{PO}_{4}{ }^{3-} \) = 1 \times (30.97 \, \mathrm{g/mol}) + 4 \times (16.00 \, \mathrm{g/mol}) = 94.97 \, \mathrm{g/mol}$

Convert mass concentration of phosphorus to phosphate

Next, we need to convert the mass concentration of phosphorus (0.07 mg/L) to the mass concentration of phosphate ion. We can do this by using the ratio of the molecular weight of phosphate ion and the molecular weight of phosphorus: Mass concentration of \(\mathrm{PO}_{4}{ }^{3-} \) = Mass concentration of \(\mathrm{P} \times \left(\frac{\mathrm{Molecular \, weight \, of \, PO}_{4}{ }^{3-}}{\mathrm{Molecular \, weight \, of \, P}}\right)\) Mass concentration of \(\mathrm{PO}_{4}{ }^{3-} \) = 0.07 \, \mathrm{mg/L} \times \left(\frac{94.97 \, \mathrm{g/mol}}{30.97 \, \mathrm{g/mol}}\right) = 0.215 \, \mathrm{mg/L}$

Calculate the molar concentration of phosphate

Finally, we can calculate the molar concentration of phosphate by dividing the mass concentration of phosphate by its molecular weight: Molar concentration of \(\mathrm{PO}_{4}{ }^{3-} \) = \(\frac{\mathrm{Mass \, concentration \, of \, PO}_{4}{ }^{3-}}{\mathrm{Molecular \, weight \, of \, PO}_{4}{ }^{3-}}\) Molar concentration of \(\mathrm{PO}_{4}{ }^{3-} \) = \(\frac{0.215 \, \mathrm{mg/L}}{94.97 \, \mathrm{g/mol}} \times \frac{1 \, \mathrm{g}}{1000\, \mathrm{mg}} \times \frac{1 \, \mathrm{L}}{1 \, \mathrm{L}} = 2.26 \times 10^{-6} \, \mathrm{mol/L}\) Thus, the molar concentration of phosphate in seawater is approximately \(2.26 \times 10^{-6} \, \mathrm{mol/L}\).