Question

The newest US 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 (ppb) arsenic.If this arsenic is present as arsenate, AsO3?4, what mass of sodium arsenate would be present in a 1.70L sample of drinking water that just meets the standard?Parts per billion is defined on a mass basis asppb = gsolutegsolution

Answer 1

Answer:

$4.71\times 10^{-5} g$of sodium arsenate would be present in a 1.70 L sample of drinking water that just meets the standard.

Explanation:

Mass of sodium arsenate =x

Mass of water = M

Density of water = d = 1000 g/L

Volume of the water , V= 1.70 L

M = $d\times V= 1000 g/L\times 1.70 L=1,700 g=1.700 kg$

1 Parts per billion  = 1 μg/kg = $\frac{10^{-9} kg}{1 kg}$

$10 ppb = \frac{x}{1.700 kg}$

$x=10\times 10^{-9} \times 1.700 kg=1.700\times 10^{-8} kg$

$x=1.700\times 10^{-8} kg=1.700\times 10^{-8}\times 10^3 g=1.700\times 10^{-5}g$

(1 kg =1000 g)

Moles of arsenic =$\frac{1.700\times 10^{-5}g}{75 g/mol}=2.267\times 10^{-7} mol$

1 mole of sodium arsenate has 1 mol of arsenic atom.Then $2.267\times 10^{-7} mol$ of arsenic will found in:

$1\times 2.267\times 10^{-7} mol=2.267\times 10^{-7} mol$ of sodium arsenate.

Mass of $2.267\times 10^{-7} mol$ of sodium arsenate:

$208 g/mol\times 2.267\times 10^{-7} mo=4.71\times 10^{-5} g$

$4.71\times 10^{-5} g$of sodium arsenate would be present in a 1.70 L sample of drinking water that just meets the standard.