# How much would a dosage of 1g strontium chloride hexahydrate raise the PPM of strontium in a total water volume of 100L?

Sep 8, 2015

That much strontium chloride hexahydrate in that much water would increase the concentration of strontium ions by $\text{3.3 ppm}$.

#### Explanation:

PPM concentration is defined as one part solute per one million parts solvent.

To determine the concentration in PPM, you need to divide the mass of the solute in grams by the mass of the solvent in grams, and multiply the ratio by ${10}^{6}$.

This is exactly what you would do to get the percent concentration, the only difference being that you multiply said ratio by ${10}^{2}$, not ${10}^{6}$.

The important thing to remember about hydrates is that when you're adding them to water, you're essentially adding a combination of solute and solvent to your existing volume of water.

To find the increase in strontium levels, you need to first determine two things

• the percent composition of strontium chloride in strontium chloride hexahydrate
• the percent composition of strontium in strontium chloride

Let's start with the first one. Use the molar masses of the hydrate and of the anhydrous salt to get the percent composition of strontium chloride in the hydrate

(158.526color(red)(cancel(color(black)("g/mol"))))/(266.618color(red)(cancel(color(black)("g/mol")))) * 100 = "59.46% SrCl"""_2

This means that 1 g of strontium chloride hexahydrate would contain

1color(red)(cancel(color(black)("g hydrate"))) * ("59.46 g SrCl"""_2)/(100color(red)(cancel(color(black)("g hydrate")))) = "0.5946 g SrCl"""_2

Now use strontium's's molar mass to find out the percent composition of strontium in strontium chloride

(87.62color(red)(cancel(color(black)("g/mol"))))/(158.526color(red)(cancel(color(black)("g/mol")))) * 100 = "55.27% Sr"

This means that 0.5946 g of strontium chloride would contain

0.5946color(red)(cancel(color(black)("g SrCl"""_2))) * "55.27 g Sr"/(100color(red)(cancel(color(black)("g SrCl"""_2)))) = "0.3286 g Sr"

Now, to get the concetration of calcium in ppm, you need to divide this mass by the total mass of the solute. Since you're adding less than one gram of water to a volume of 100 L of water, you can safely assume that the volume will not change.

The total mass of water will be - assuming a density of $\text{1000 g/L}$, will be

100color(red)(cancel(color(black)("L"))) * "1000 g"/color(red)(cancel(color(black)("L"))) = 10^5"g water"

The concentration in ppm for the strontium ions will be

(0.3286color(red)(cancel(color(black)("g"))))/(10^5color(red)(cancel(color(black)("g")))) * 10^6 = color(green)("3.3 ppm")