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Re: [APD] CO2 revelations
Jerry Baker wrote:
>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.
Excellent! Here I was grunting away with Herculean effort to make this
work (hunting out my Merk CD was the first step). However every time I
get around to using any chemistry, my memory of it has slipped to the
wayside. Thanks much!! Now I can go back to being a mental lounge
lizard again. Off to get some distilled water....(no distiller... dammit).
Speaking of which, anyone know a good chemistry textbook that I should
keep my eyes open for at used book sales and the like? It would have to
be Canadian, as I doubt I would come across many southern school texts
here. It would have to spell out the basics in human speak, then shift
gears to keep me interested.
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