[Samotonaki]

Hello,

I have some difficulty calculating values after dissolving a substance. I've gotten different answers from my environment, but I'm not sure who is calculating it the right way. I hope someone here can shed some light on the issue.

We dissolve 81,53g of KN03 in 500ml of demineralised water. We then dilute 1ml of the solution into 100L of water. In the end, what will the ppm of K and N be in the 100L of water? What is the correct way to calculate these values?

[Hunter2]

First calculate the concentraton in g/l
If you know this then you also know how much is in 1 ml. This amount is transferred into 100 l.
The relationship of the elements is connected via molar mass.

Calculate the amount of KNO3 into mol then you get also the mol of K and N. Calculate this back to gramm..

[Borek]

This is a bit tricky, depending on how accurate result you need.

As a first approximation - if you dissolve some amount of substance into 500 mL and you take 1 mL of the solution, it is effectively as if you took just 1/500 of the initial mass of the substance (so 81,53g/500) - then just divide by the final volume to find concentration.

In reality dissolving 81,53g of KNO₃ in 500 mL of water will produce more than 500 mL of the solution (finding out exact volume requires using density tables, in this particular case it will be around 530 mL), so you are not taking 1/500 but a bit less.

That being said the way we work here is: show the numbers you have gotten so far and we will start from there.

[Samotonaki]

Thank you both for your replies. I eventually found out that I was calculating the wrong values. I was supposed to calculate the NO3 value, not the N value. Doing that brings me to the expected change of 1 ppm and leads me to believe that now I finally have the correct K value. Below I will share the calculation I have now. Basically, the result should be accurate in 2 decimals. @Borek does your explanation then change anything?

1ppm = 1mg/L, thus 1 ppm increase = 100mg/100L

K = 39,1 g/mol
NO3 = 14 + 3*16 = 62 g/mol

Total = 101,1 g/mol

Mass NO3 in powder 62/101,1 = 0,6133 | 81,53 * 0,6133 = 50g = 50000mg
Mass K in powder 39,1/101,1 = 0,387 | 81,53 * 0,387 = 31,54g = 31540mg

Concentration of N in the solution when adding 1 mL to 100L of demineralised water should then make 50000mg/500ml = 100mg/ml = 1ppm. Concentration of K 31540mg/500ml = 63,08mg/ml = 0,63ppm

[Hunter2]

First consider what Borek said.
It change in this way you have larger volume as 0,5 l. Your values are smaller.

Better is to make a solution in this way. Put   81,53 g in an measured flask  and add water until you get 0,5 l otherwise need the specific gravity, if you dissolve the salt in 0,5 l. The volume will then more as 0,5 l.
If you follow so then you have a solution of 500 ml of it.
You calculate the concentration what means 81,53g/0,5 l = 163,06 g/l
This is equal 163,06 mg/ml.
You dissolve 1 ml  in 100 l means 1,63 mg/l
In mol it's  1,63 mg/l / 101 mg/mmol = 0,016 mmol/l
K = 39,1 mg/mmol * 0,016 mmol/l = 0,63 mg/l = 0,63 ppm K
N = 14 mg/mmol * 0,016 mmol/l = 0,22 mg/l = 0,22 ppm N
NO3- 62 mg/mmol * 0,016 mmol/l = 1,00 m/l = 1 ppm NO3-

[Borek]

Taking the density changes into account the procedure based on mixing 81.53 g of KNO₃ with 500 mL of water produces 0.94 ppm of NO₃^- in the final solution.

For 1 ppm you need to dissolve KNO3 in a somewhat lesser amount of water and the fill it up to 500 mL.

Mistaking volume of solvent with the final volume is a very common error.