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Re: [APD] CO2 revelations

Paul M Wallace wrote:
> The math should read 246.5mg of NaHCO3 in the prior post.
> 1dkh = 17.8575 mg/liter CaCO3
> or (17.8575-7.143)= 10.7145mg/liter CO3
>   so 10.7 *(mw NaHCO3 84)/(mw CO3 60)= 14.98 mg/liter NaHCO3
> So 149.8 mg NaHCO3 in a liter should be 10dkh.

Keep in mind that CO3 form two bicarbonate ions in water and tests 
measure CaCO3 equivalents, so you need twice the NaHCO3.

Jerry Baker
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