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" .