Reef Chemistry Question of the Day #88 Baking Soda 1

[Randy Holmes-Farley]

Reef Chemistry Question of the Day #88

A stock solution is made by dissolving 8.401 grams of baking soda (sodium bicarbonate; molecular weight = 84.01 g/mole) into enough pure freshwater at 25° C to make 1 L total volume. This solution has a pH of 8.300 and an alkalinity of 100.00 meq/L (~280 dKH). 1.000 mL of this solution is added to 99.000 mL of natural seawater (pH 8.200; alkalinity = 2.500 meq/L or 7.00 dKH). The alkalinity of the final solution is closest to:


A. 3.500 meq/L (9.80 dKH)
B. 3.475 meq/L (9.73 dKH)
C. 2.515 meq/L (7.04 dKH)
D. 2.500 meq/L (7.00 dKH)

Good luck!




















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[tonizzy22]

Answer is B ((1mL×100meq/L)+(99mL×2.5meq/L))/100mL=3.475meq/L or ((1mL×280dKH)+(99mL×7dKH))/100mL= 9.73dKH

 

[Skydvr]

I agree with Tonizz's answer and rationale.

 

[Steven@]

I also agree with the science of tonizzy22

 

[reefins]

Ya i'm down with Tonizz's answer too.. nailed it!

 

[DFW]

B. (99x2.5) + (1x100)= 347.5
347.5/100=3.475

 

[Randy Holmes-Farley]

And the answer is...B. 3.475 meq/L (9.73 dKH)

Good job folks!

Alkalinity of combined solutions is normally the weighted average of the two alkalinities.