Question

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?

Answer 1

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.

`"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 `"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 `"Ca"` in `"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"`.