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4dKH Solution

Luis Batista

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19 Jun 2014
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Portugal
Hello

I have been read about solutions of 4dKH for the drop checker but they all are a little big in the amounts of water to use. I dont want to be left with 4 or 5 L of a 40dKH solution.

So i scale it down to this:

500ml Distilled water + 0,6gr sodium bicarbonate = 500ml solution 40dKH

100ml de 40dKH + 900ml Distilled water = 1000ml solution 4dKH

or even more scale down

50ml de 40dkh + 450ml Distilled water = 500ml solution 4dkh

It's the math accurate?
Just have to cut in half all the amounts to scale it down even more?

cheers,

Luis Batista
 
Last edited:

dw1305

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Hi all,
Yes the maths is right and you can do this, but because the sodium bicarbonate (NaHCO3) is really cheap to buy I'd recommend making up a fairly strong initial solution and then using "serial dilution" to give you the strength you want. The main advantage of "serial dilution" is you don't need as accurate a balance. I always like to deal with larger weights and volumes because you don't have to be so precise.

For NaHCO3 12g in 100 litre of RO water is 4dKH, so 12g in 10 litre is 40dKH (or 6g in 5 litres or 0.6g in 500cm3).

How about starting with 6g in in 500cm3? This would be a 400 dKH stock solution, but this is still quite a dilute solution, so 1cm3 would weight ~1g meaning that you don't need to be able to measure volumes accurately. Dilution by 5 in 500 (stock sol:H2O) would give you 4dKH, and you would only need a balance that can weigh 5g fairly accurately.

cheers Darrel
 

Luis Batista

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Hi all,
Yes the maths is right and you can do this, but because the sodium bicarbonate (NaHCO3) is really cheap to buy I'd recommend making up a fairly strong initial solution and then using "serial dilution" to give you the strength you want. The main advantage of "serial dilution" is you don't need as accurate a balance. I always like to deal with larger weights and volumes because you don't have to be so precise.

For NaHCO3 12g in 100 litre of RO water is 4dKH, so 12g in 10 litre is 40dKH (or 6g in 5 litres or 0.6g in 500cm3).

How about starting with 6g in in 500cm3? This would be a 400 dKH stock solution, but this is still quite a dilute solution, so 1cm3 would weight ~1g meaning that you don't need to be able to measure volumes accurately. Dilution by 5 in 500 (stock sol:H2O) would give you 4dKH, and you would only need a balance that can weigh 5g fairly accurately.

cheers Darrel


Thanks Darrel

Next time i make a solution i give it a try

Cheers

Luis
 

Luis Batista

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Thread starter
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19 Jun 2014
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Portugal
Hi all,
Yes the maths is right and you can do this, but because the sodium bicarbonate (NaHCO3) is really cheap to buy I'd recommend making up a fairly strong initial solution and then using "serial dilution" to give you the strength you want. The main advantage of "serial dilution" is you don't need as accurate a balance. I always like to deal with larger weights and volumes because you don't have to be so precise.

For NaHCO3 12g in 100 litre of RO water is 4dKH, so 12g in 10 litre is 40dKH (or 6g in 5 litres or 0.6g in 500cm3).

How about starting with 6g in in 500cm3? This would be a 400 dKH stock solution, but this is still quite a dilute solution, so 1cm3 would weight ~1g meaning that you don't need to be able to measure volumes accurately. Dilution by 5 in 500 (stock sol:H2O) would give you 4dKH, and you would only need a balance that can weigh 5g fairly accurately.

cheers Darrel


Hello Darrel

I want to make this solution that you mention so lets see if i understand it right.

6g sodium bicarbonate (NaHCO3) + 500ml H2O = 500ml of 400dKH stock solution.
But now what´s next?
You say:" Dilution by 5 in 500".
I dont fully understand that math :)
It´s 5ml of 400dkh stock solution + 500ml H20 to give 505ml of 4dKH solution?

how many ml of the 400dKH to how many H2O to give the 4dKH solution?

Thanks for your help
 
Last edited:

dw1305

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Hi all,
I see I never replied to this one.
It´s 5ml of 400dkh stock solution + 500ml H20 to give 505ml of 4dKH solution?
No, it is 5ml of 400dKH stock sol. + 495cm3 H2O. Because we are dealing with a very dilute solution 500cm3 = 500g.

cheers Darrel
 

GreenNeedle

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Always best to do the large 5ltr first and dilute down for accuracy. You won't be left with wastage though. Save 100ml of it for future use and use the other 4.9 litres to make up your fert solutions or top ups etc.
 

Luis Batista

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Hi all,
I see I never replied to this one. No, it is 5ml of 400dKH stock sol. + 495cm3 H2O. Because we are dealing with a very dilute solution 500cm3 = 500g.

cheers Darrel


Thanks Darrel!

I have access to a pharmaceutical grade scale and wish I could do an even smaller solution.
So, if :

12g in 100L gives 100L of 4dKH solution

0,06g in 0,5L gives 500ml of 4dKH solution
0,012g in 0,1L gives 100ml of 4dKH solution

That´s about right, yes?

Once again, Thank You for your help

Best regards,

Luis Batista
 
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rebel

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Hi guys,

I made a 4dkh solution with baking soda (NaHCO3)
0.12g in 500ml of distilled water

When I use my API test kit it gives me a dkH of 12!!

I know the maths is wrong somewhere. My scales are accurate to 0.001g. I suspect my math is wrong or baking soda is contaminated or perhaps my measuring of ml is wrong.

Any ideas as to how to approach this problem to get some answers?
 

Manuel Arias

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I made a 4dkh solution with baking soda (NaHCO3)
0.12g in 500ml of distilled water

When I use my API test kit it gives me a dkH of 12!!
Should be 0.06gr in 500ml to make 4dK


Well, maybe my maths are a bit old, but if 0.12 in 500ml gives 12 dkH, 0.06 (half of it) in the same volume should provide 6 dkH...not 4. ;)

In any case, this is more or less easy to compute:

1dkH is equivalent to 17.848 mg/l of CaCO3.

4dkH are then equivalent to 71.392 mg/l of CaCO3, or 35.696 mg dissolved in 500 ml.

On the other side, CaCO3 has a molecular weight of 100,0869 g/mol, and NaHCO3 is 84,0066 g/mol. CO3 part has a molecular weight of 60,0089 g/mol, so the proportion of carbonates in CaCO3 is of 0.599 and in NaHCO3 is of 0.714. The amount of carbonates in 35.696 mg of CaCO is then 21.382 mg, which means that to get the same amount of carbonates, we need 29.947 mg of NaHCO3, i.e. 0.03 gr. of NaHCO3, not 0.06.

So, the recipe should be:

500 ml distilled water + 0.03gr of NaHCO3

If you have a precision balance, this should not be a problem at all.

Hope this help.
 

Antoni

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I would personally do a larger amount, as you can be more precise and to get the correct amount of salt. If you do 100mls only, every tenth of a gram less or more, due to a calibration error or human error will affect the solution.
 

Manuel Arias

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I would personally do a larger amount, as you can be more precise and to get the correct amount of salt. If you do 100mls only, every tenth of a gram less or more, due to a calibration error or human error will affect the solution.

Quite true. A way to do this is to dissolve 6gr in 1 litre of distilled water. Then taking 100 ml of this solution, and add 900 ml of distilled water. And then, taking again 100 ml of the resulting solution and adding again 900ml of distilled water. ;) Easy.
 

rebel

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I would personally do a larger amount, as you can be more precise and to get the correct amount of salt. If you do 100mls only, every tenth of a gram less or more, due to a calibration error or human error will affect the solution.
Well, maybe my maths are a bit old, but if 0.12 in 500ml gives 12 dkH, 0.06 (half of it) in the same volume should provide 6 dkH...not 4. ;)

In any case, this is more or less easy to compute:

1dkH is equivalent to 17.848 mg/l of CaCO3.

4dkH are then equivalent to 71.392 mg/l of CaCO3, or 35.696 mg dissolved in 500 ml.

On the other side, CaCO3 has a molecular weight of 100,0869 g/mol, and NaHCO3 is 84,0066 g/mol. CO3 part has a molecular weight of 60,0089 g/mol, so the proportion of carbonates in CaCO3 is of 0.599 and in NaHCO3 is of 0.714. The amount of carbonates in 35.696 mg of CaCO is then 21.382 mg, which means that to get the same amount of carbonates, we need 29.947 mg of NaHCO3, i.e. 0.03 gr. of NaHCO3, not 0.06.

So, the recipe should be:

500 ml distilled water + 0.03gr of NaHCO3

If you have a precision balance, this should not be a problem at all.

Hope this help.

Manual I really appreciate your reply and long explanation. I will be measuring white powder with precision scales tonight. Maybe I'll have a wild look in my eye and have a little powder on my upper lip as well..... :)

Will report back using both methods.

I am keen to calibrate my API kH test kit as well; within limits of course.
 

Luis Batista

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Well, maybe my maths are a bit old, but if 0.12 in 500ml gives 12 dkH, 0.06 (half of it) in the same volume should provide 6 dkH...not 4. ;)

In any case, this is more or less easy to compute:

1dkH is equivalent to 17.848 mg/l of CaCO3.

4dkH are then equivalent to 71.392 mg/l of CaCO3, or 35.696 mg dissolved in 500 ml.

On the other side, CaCO3 has a molecular weight of 100,0869 g/mol, and NaHCO3 is 84,0066 g/mol. CO3 part has a molecular weight of 60,0089 g/mol, so the proportion of carbonates in CaCO3 is of 0.599 and in NaHCO3 is of 0.714. The amount of carbonates in 35.696 mg of CaCO is then 21.382 mg, which means that to get the same amount of carbonates, we need 29.947 mg of NaHCO3, i.e. 0.03 gr. of NaHCO3, not 0.06.

So, the recipe should be:

500 ml distilled water + 0.03gr of NaHCO3

If you have a precision balance, this should not be a problem at all.

Hope this help.
Thanks for that!

Luís Batista

Sent from my Nokia 3310!
 

Hanuman

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Thailand
1dkH is equivalent to 17.848 mg/l of CaCO3.
4dkH are then equivalent to 71.392 mg/l of CaCO3, or 35.696 mg dissolved in 500 ml.

On the other side, CaCO3 has a molecular weight of 100,0869 g/mol, and NaHCO3 is 84,0066 g/mol. CO3 part has a molecular weight of 60,0089 g/mol, so the proportion of carbonates in CaCO3 is of 0.599 and in NaHCO3 is of 0.714. The amount of carbonates in 35.696 mg of CaCO is then 21.382 mg, which means that to get the same amount of carbonates, we need 29.947 mg of NaHCO3, i.e. 0.03 gr. of NaHCO3, not 0.06.
I am reviving this threat because I spend quite some time scratching my head with the calculation provided above which were not matching my own calculations. It seems a shortcut was made.
1 dkH is equivalent to 17.848 mg/l of CaCO₃ (Calcium carbonate)
but

1 dkH is equivalent to 30 mg/l of NaHCO₃ (Sodium bicarbonate)
500 ml distilled water + 0.03gr of NaHCO3
The above would result in 2 dKH not 4dKH.
Therefore 0.06g of NaHCO3 in 500 ml of water would be required to bring the dKH to 4.

Also if one was to use Na₂CO₃ (Sodium carbonate) instead of NaHCO₃ (Sodium bicarbonate) the following would be required:
0.0378g of Na₂CO₃ in 500 ml of water would bring the dKH to 4.

Sources:
 
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