Help understanding NT Labs liquid ammonia test

[Edward Shave]

NT Labs say their ammonia test results are expressed as nitrogen content rather than concentration of ion. If I want the concentration of ion I have to multiply by 1.12.
I am plugging the test result into the free ammonia calculator I found here.
My question is... What type of result is the free ammonia calculator expecting? Do I need to multiply by 1.12?

 

[dw1305]

Hi all,

NT Labs say their ammonia test results are expressed as nitrogen content rather than concentration of ion. If I want the concentration of ion I have to multiply by 1.12.

I assume it is just nitrogen content of NH3, although it should be 1.21, rather 1.12?

It is worked out from the RAM of nitrogen (N) = 14 and hydrogen (H) = 1, so the RMM of NH3 = (14 + 3) = 17 and 14/17 = 1.214. (NH4+ would be "multiply by 1.29"). These tests usually add an acid to convert all the TAN to ammonium (NH4+), because dissolved gases are problematic to measure. If it adds a base (usually NaOH) it will convert all the NH4+ to NH3.

I am plugging the test result into the free ammonia calculator I found <"here">.

This does something slightly different from the way you want to use it, it tells you what proportion of the Total Ammoniacal Nitrogen (TAN) is as dissolved ammonia gas (NH3). It should give you the same result as the table below.

TAN comprises NH3 ("un-ionized" and toxic to aquatic life) and the ammonium (NH4+) ion ("ionized" and relatively benign). The proportion of the TAN that is ionized or un-ionized is <"dependent upon temperature and pH">.

cheers Darrel

 

[Edward Shave]

Hi Darrel, many thanks for your detailed reply. I got the 1.12 figure from this article. However if I was a betting man and going by past experience I'd put my money on you being right. LOL

I get that there is total ammonia and free ammonia (which I think is NH3?) but as usual with chemistry they can never leave it at that...! There are always If's & but's LOL.
Given that we can get the percentage from your chart (in my case 0.5%) I'm still unsure of which number to plug in to get the actual concentration of NH3. In other words do I need to multiply by 1.21 or not?

 

[MichaelJ]

FYI Here is another free ammonia (NH3) calculator. (gives the same results as above table btw. i.e. free_ammonia = table1(temp,pH) / 100 x total_ammonia )

 

[Edward Shave]

FYI Here is another free ammonia (NH3) calculator. (gives the same results as above table btw. i.e. free_ammonia = table1(temp,pH) / 100 x total_ammonia )

Thanks for that. This helped in as much as it precisely specified exactly what to enter as the total ammonia value. i.e.
"Total (NH3+NH4)-Nitrogen Concentration (mg/L)"
As near as I can figure (as a non chemist) that is the same form as returned by my NT Labs test.
If anyone knows better please let me know...

 

[dw1305]

Hi all,

I got the 1.12 figure from this article. However if I was a betting man and going by past experience I'd put my money on you being right.

You would have got your money, it should be 1.21.

I get that there is total ammonia and free ammonia (which I think is NH3?) but as usual with chemistry they can never leave it at that...!

That is right, free ammonia = NH3. To take pH and temperature out of the equation scientists use "Total Ammoniacal Nitrogen" (TAN). I think TAN helps as a concept.

Some aquarists will argue that you can ignore NH4+, because it isn't toxic, but I look at a bit like the relationship between kinetic (~ NH3) and potential energy (~NH4+) when you have a large mass on an inclined plane. It might not be a problem, but if the mass starts moving it is going to make a hell of a mess. Same with free ammonia (NH3), if you end up with a lot of it, it is going to be total "wipe-out".

The real answer is to get rid of the TAN <"as rapidly as possible"> and have systems with built in capacity to deal with <"rogue large masses">.

This helped in as much as it precisely specified exactly what to enter as the total ammonia value. i.e.
"Total (NH3+NH4)-Nitrogen Concentration (mg/L)"
As near as I can figure (as a non chemist) that is the same form as returned by my NT Labs test.

It is.

cheers Darrel