# I want to increase the NO3 concentration in a 40 Liters tank using Ca(NO3)2.4H2O. The concentration of the CaNO3 solution is 500g/L. How much solution do I need to add increase the concentration by 2ppm? Thanks for helping out!

## How much calcium would be added at the same time?

Apr 15, 2017

You must add 0.3 mL of the calcium nitrate solution. This will include 25 mg of calcium.

#### Explanation:

Let's express the concentration of 2 ppm as grams per litre.

$\text{2 ppm" = (2 × 10^"-6" color(white)(l)"g")/(1 color(red)(cancel(color(black)("g water")))) × (1000 color(red)(cancel(color(black)("g water"))))/"1 L" = "0.002 g/L" = "2 mg/L}$

For 40 L of water, you will need

40 color(red)(cancel(color(black)("L"))) × "2 mg"/(1 color(red)(cancel(color(black)("L")))) = "80 mg" = "0.08 g" of nitrate ion.

The molar mass of $\text{Ca"("NO"_3)_2·"4H"_2"O}$ is 236·15 g.

Of this, 124.01 g comes from the nitrate ions.

Hence, to get 0.08 g of nitrate, you will have to use

0.08 color(red)(cancel(color(black)("g NO"_3^"-"))) × (236.15 "g Ca"("NO"_3)_2·"4H"_2"O")/(124.01 color(red)(cancel(color(black)("g NO"_3^"-")))) = "0.15 g Ca"("NO"_3)_2·"4H"_2"O"

The required volume of concentrated salt solution is

0.15 color(red)(cancel(color(black)("g salt"))) × "1 L salt"/(500 color(red)(cancel(color(black)("g salt")))) = "0.0003 L salt" = "0.3 mL salt"

There are 40.078 g of $\text{Ca}$ in $\text{236.15 g Ca"("NO"_3)_2·"4H"_2"O}$.

∴ The mass of calcium added is

0.15 color(red)(cancel(color(black)("g Ca"("NO"_3)_2·"4H"_2"O"))) × "40.078 g Ca"/(236.15 color(red)(cancel(color(black)("g Ca"("NO"_3)_2·"4H"_2"O")))) = "0.025 g Ca" = "25 mg Ca" .