The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. If this arsenic is present as arsenate, \(\mathrm{AsO}_{4}^{3-},\) what mass of sodium arsenate would be present in a 1.00-L sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$ \mathrm{Ppb}=\frac{\mathrm{g} \text { solute }}{\mathrm{g} \text { solution }} \times 10^{9} $$

According to the U.S. standard, the drinking water should contain no greater than 10 ppb of arsenic. Here, we consider 1.00 L of water, which is equivalent to 1.00 kg. Using the given definition of ppb, we can find the mass of arsenic in the water sample: $$ \mathrm{Ppb}=\frac{\mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}\times 10^{9}, $$ Rearrange the equation to find the mass of arsenic solute: $$ \mathrm{g} \text { solute }=\frac{\mathrm{Ppb} \times \mathrm{g} \text { solution }}{10^{9}}. $$ Insert the values for ppb and mass of solution: $$ \mathrm{g} \text { arsenic }=\frac{10 \times 1,000}{10^{9}}=10^{-5} \,\mathrm{g}. $$

Next, we need to convert the mass of arsenic into moles, using the molar mass of arsenic (As): $$ \mathrm{moles} =\frac{\mathrm{mass}}{\mathrm{molar \, mass}}. $$ The molar mass of arsenic is \(\mathrm{74.92\,g/mol}\), so: $$ \mathrm{moles\, As} =\frac{10^{-5}\,\mathrm{g}}{74.92\,\mathrm{g/mol}}=1.34 \times 10^{-7}\,\mathrm{mol}. $$

Since each molecule of sodium arsenate contains one arsenic atom, the number of moles of sodium arsenate in the sample is equal to the number of moles of arsenic: $$ \mathrm{moles\, Na}_{3}\mathrm{AsO}_{4} = 1.34 \times 10^{-7} \,\mathrm{mol}. $$

To obtain the mass of sodium arsenate in the sample, we will multiply the number of moles by the molar mass of sodium arsenate. The molar mass of sodium arsenate is: $$ \mathrm{3 \times (molar \, mass \, of \, Na) + molar \, mass\, of\, As+4 \times (molar\, mass\,of\, O) }. $$ The molar mass of sodium (Na), arsenic (As), and oxygen (O) are \(\mathrm{22.99\,g/mol}\), \(\mathrm{74.92\,g/mol}\), and \(\mathrm{16.00\,g/mol}\), respectively. Thus, $$ \mathrm{molar \, mass\, of\, Na}_{3}\mathrm{AsO}_{4} =\mathrm{3 \times 22.99 + 74.92 + 4 \times 16.00}=207.89\, \mathrm{g/mol}. $$ Now, we can find the mass of sodium arsenate in the sample: $$ \mathrm{mass\, of\, Na}_{3}\mathrm{AsO}_{4} = (\mathrm{moles}\,\mathrm{of\,Na}_{3}\mathrm{AsO}_{4}) \times (\mathrm{molar\,mass\,of\,Na}_{3}\mathrm{AsO}_{4})= (1.34 \times 10^{-7}\,\mathrm{mol}) \times (207.89\,\mathrm{g/mol})= 2.79\times 10^{-5}\,\mathrm{g}. $$ So, the mass of sodium arsenate in a 1.00 L sample of drinking water that just meets the U.S. standard is \(2.79 \times 10^{-5}\, \mathrm{g}\).