Reef Chemistry Question of the Day #121 Alkalinity and mixing

[Randy Holmes-Farley]

Reef Chemistry Question of the Day #121

Suppose that the water in a reef aquarium has an alkalinity of 2.5 meq/L (7 dKH), and a new batch of artificial seawater at the same temperature has an alkalinity of 4 meq/L (11 dKH). If 10% of the water in the aquarium is replaced by the new water, the final alkalinity in the aquarium will be closest to which of the following values?


A. 2.65 meq/L (7.4 dKH)
B. 2.85 meq/L (8.0 dKH)
C. 3.05 meq/L (8.5 dKH)
D. Not enough information is provided to properly answer

Good luck!






























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[beaslbob]

alk is a linear measure.
the final tank is 90% 7 and 10% 11.
.9*7=6.3
.1*11=1.1
6.3+1.1=7.4
So the answer is D because we don't know how much co2 changed during the water change.:squigglemouth:
otherwise A:wink:
(or at least I fell for Dr. Randy Holmes-Farley's "trick" on this one.) oh:

 

[Randy Holmes-Farley]

CO2 has no impact on seawater alkalinity.

 

[glb]

7.4. 90% of the water is 7 and 10% of the water is 11. 7 x.9 + 11 x .1 = 6.3+1.1 = 7.4

 

[beaslbob]

7.4. 90% of the water is 7 and 10% of the water is 11. 7 x.9 + 11 x .1 = 6.3+1.1 = 7.4

copy kat!!!!!!!!

 

[beaslbob]

CO2 has no impact on seawater alkalinity.

Oh yea that was pH.
Guess I'm wrong again.
how about c:wink:
naaaaaaaaa

 

[glb]

copy kat!!!!!!!!

Did I copy you? I promise my thinking was original but if you did it first I will give credit where credit is due. [emoji5]️

 

[beaslbob]

Did I copy you? I promise my thinking was original but if you did it first I will give credit where credit is due. [emoji5]️

Great minds think alike.
Or at least attended the same math class in school.

 

[glb]

Great minds think alike.
Or at least attended the same math class in school.

Yes. An old time, old math class!!!

 

[dodgerblew]

I'll go with b

 

[fishjelly]

Assume 100 L of water volume. Removing 10% of this will leave 90 L remaining.

90 L x 2.5 meq/L = 225 meq in the tank

Replacing the water with 10 L of the new batch:

10 L x 4 meq/L = 40 meq

225 meq + 40 meq = 265 meq

And with the total volume being 100 L again:

265 meq / 100 L = 2.65 meq/L, so A?

 

[dodgerblew]

Impressive. Now seeing it done this way I would say A. So, question for you, are you a mathematician, engineer or a math related profession? I always sucked at word problems

 

[DFW]

A!!

 

[Greenstreet.1]

C [emoji16]

 

[Randy Holmes-Farley]

And the answer is...A. 2.65 meq/L (7.4 dKH)

As noted above, one can calculate alkalinity by the weighted average of the solutions being mixed.

So
0.9 x 7 dKH + 0.1 x 11 dKH = 7.4 dKH
or
0.9 x 2.5 meq/L + 0.1 x 4 meq/L = 2.65 meq/L

Not only is this calculation method worth knowing, but folks are often surprised at how little a water change of this size impacts alkalinity, even if the starting salt water is much higher in alkalinity.

 

[tcoyle]

Hence why my 5 gallon of water changes with RSCP mixed at 12, does very little in my 70 gallon tank. When I switched, my tank was around 8dk. I dose about 70ml a day of baking soda to keep it a stable 8.5 even with water changes.