The new standard for arsenate in drinking water, mandated by the Safe Drinking Water Act, requires that by January, 2006, public water supplies must contain no greater than 10 parts per billion arsenic.

Assuming that this arsenic is present as arsenate what mass of sodium arsenate would be present in a 1.10 sample of drinking water that just meets the standard?

When you are asking a chemistry question the biggest gaffe of all is to omit the units. With no units on 1.10 there is no way to answer the question.

Sorry. I didn't realize it wasn't there. it is 1.1 Liters.

I think I searched every corner of google. But even I cannot find the answer to this question ?!!!

So frustrating

To determine the mass of sodium arsenate present in a 1.10 sample of drinking water that meets the standard, we need to use the concentration of arsenic in the water and the molar mass of sodium arsenate.

First, let's convert the concentration from parts per billion (ppb) to moles per liter (molarity). Since 1 ppb is equivalent to 1 microgram per liter (μg/L), we can convert it to moles per liter using the molar mass of arsenic.

The molar mass of arsenic is 74.92160 g/mol.

So, for a concentration of 10 ppb (= 10 μg/L), we can calculate the molarity as follows:

10 μg/L * (1 g/1000 mg) * (1 mg/1000 μg) * (1 mol/74.92160 g) = x mol/L

Simplifying the calculation:

x = (10 * 10^-6) / 74.92160 mol/L
x = 1.33 * 10^-7 mol/L

Next, we need to find the moles of sodium arsenate (Na3AsO4) present in the sample. The balanced chemical equation for sodium arsenate is:

2 Na3AsO4 + 3 H2O → 6 NaOH + As2O3

From the balanced equation, we can see that for every 2 moles of Na3AsO4, there is 1 mole of arsenic (As).

Therefore, to find the moles of sodium arsenate (Na3AsO4):

1.33 * 10^-7 mol/L * (2 mol Na3AsO4 / 1 mol As) = 2.66 * 10^-7 mol/L Na3AsO4

Finally, to calculate the mass of sodium arsenate present in the sample, we need to multiply the moles of sodium arsenate by its molar mass.

The molar mass of sodium arsenate (Na3AsO4) is calculated as follows:

(22.98977 g/mol * 3) + (74.92160 g/mol * 1) + (15.999 g/mol * 4) = 307.90937 g/mol

So, the mass of sodium arsenate in the 1.10 sample of drinking water that meets the standard is:

2.66 * 10^-7 mol/L * 307.90937 g/mol = 8.18 * 10^-5 g