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 1.40 L sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as ppb = (gsolute/gsolution)×10^9
Hey! So in this problem, we will be using the ppb equation to solve for the missing grams of solute. We know that we have 10 ppb and we're given 1.40 L of sample. It's very important to understand that the 1.40 L of sample are only of drinking water, meaning that is the solvent. Remember that a solution is made up of a solute and a solvent.
First thing we want to do is convert the 1.40 L to grams since we want grams of the solution on the denominator. Again, 1.40 L is the solvent. So we do the conversion above and we use the density of water, which is 1 g/ mL to find the grams of solvent. We then use the equation for ppb to find the grams of solute missing. We plug in an X for what we don't know (mass of solute). In the denominator, we must have solute + solvent =solution, so we plug in X (solute) + 14000 g (solvent). We then just solve for X and get the grams of solute. Hope this helps!