Reef Chemistry Question of the Day #148 Surface Area

[Randy Holmes-Farley]

Reef Chemistry Question of the Day [HASHTAG]#148[/HASHTAG]

Which of the following has the highest surface area for potential dissolution of silica/silicate?

A. All of the interior glass surfaces of a standard 120 gallon aquarium
B. 5 pounds of fine silica sand, assuming it is made of spheres of 0.1 mm diameter
C. 10 pounds of silica sand, assuming it is made of spheres of 0.5 mm diameter
D. 25 pounds of coarse silica sand, assuming it is made of spheres of 2 mm diameter

Good luck!
































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[john.m.cole3]

B

 

[JimWelsh]

Well, let's figure each one out:

A) I don't know of any international "standard" for aquarium sizes, but I'll assume that you mean a rectangular aquarium 48" long x 24" wide x 24" deep. For this exercise, we can safely disregard the thickness of the glass, so the aquarium has 2 panes that are 48" x 24" (front and back), two panes that are 24" x 24" (left and right), and one more pane that is 48" x 24" (bottom), for a total of 3 x 48 x 24 = 3456 square inches plus 2 x 24 x 24 = 1152 square inches, for a grand total of 4608 square inches. Converting to SI units, that's 4608 * 2.54 * 2.54 = 29,729, or 2.97E+04, square centimeters (cm^3).

B) Assuming a silica density of 2.648 g/cm^3, then 5 lbs. of silica sand has a total volume of V = 5 lbs. * 454 g / 2.646 = 857.25 cm^3. Each "grain" has an individual volume v = 4/3*pi*r^3, and an individual surface area a = 4*pi*r^2, where r = 0.1 / 2 / 1000 cm, so v = 5.24E-13 cm^3, and a = 3.14E-08 cm^2. Therefore, the number of "grains" of sand in the 5 lbs. N = 857.25 cm^3 / 5.24E-13 cm^3 = 1.64E+15 grains, and the total surface area A = n * a, or 1.64E+15 * 3.14E-08 cm^2 = 5.15E+07 cm^2.

To summarize: V = 857.25 cm^3, v = 5.24E-13 cm^3, a = 3.14E-08 cm^2, N = 1.64E+15, and A = 5.15E+07 cm^2.

C) In this case, V = 1714.5 cm^3, v = 6.54E-11 cm^3, a = 7.85E-07 cm^2, N = 2.62E-13, and A = 2.06E+07 cm^2.

D) This time, V = 4286 cm^3, v = 4.19E-09 cm^3, a = 1.26E-05 cm^2, N = 1.02E+12, and A = 1.29E+07 cm^2.

The answer is B.

 

[AllSignsPointToFish]

B by a long shot. B has over 960 sq. ft. of surface area.

 

[beaslbob]

according to my highly approximate reasoning based upon highly questionable assumptions with my dangerous and experimental techniques:

b

 

[beaslbob]

all of which produced:
a: 32
b: 962.5263
c: 385.0105
d 240.6316

 

[beaslbob]

which kinda agree with other answers here also.

 

[beaslbob]

Hey

Doesn't it depend on what shape the crains are? Spherical then all the above but cubical then just the outside area and the interior area isn't touching the water.

 

[Randy Holmes-Farley]

And the answer is... B. 5 pounds of fine silica sand, assuming it is made of spheres of 0.1 mm diameter

Very good analysis, folks. I won't go through the math since others have done a fine job of that.

This issue came up previously in a discussion I was having with someone who was trying to prove that silica sand could not be a source of silica/silicate to an aquarium.

He said that silica was insoluble (based on an MSDS statement, which is not correct), and that anyway, the glass of aquariums isn't dissolving so why should silica sand be any different?

One way it is different, I noted, is that much higher surface area of the sand.

Happy reefing!

 

[Randy Holmes-Farley]

Hey

Doesn't it depend on what shape the crains are? Spherical then all the above but cubical then just the outside area and the interior area isn't touching the water.

A perhaps better way to blast the question is to ask what the actual surface area of the aquarium glass is, and not just the apparent physical dimensions. It may start pretty flat, but after snails and sea urchins have been chomping away, it is likely rough enough to have a big impact on the actual surface area.

Even in new glass, it is not perfectly smooth, and the actual roughness vs smoothness can have an impact on some of its properties:

https://pubs.acs.org/cen/news/87/i51/8751notw5.html

 

[beaslbob]

Hey

Doesn't it depend on what shape the crains are? Spherical then all the above but cubical then just the outside area and the interior area isn't touching the water.

I wonder if I meant whooping or other crains.

 

[JimWelsh]

I notice that my answer differs in from the others' answers when converted to square feet. Aside from my bonehead mistake of thinking there are 1000 mm in a cm (d'oh!), making my sand surface area calculations be off by two orders of magnitude, you guys used the bulk density of sand, which is about 1.523 g/cm^3, but that doesn't account for sand's porosity. In other words, that bulk density includes all the air around in the spaces between the sand grains. I used the density of silica quartz absent the air.

 

[Randy Holmes-Farley]

I notice that my answer differs in from the others' answers when converted to square feet. Aside from my bonehead mistake of thinking there are 1000 mm in a cm (d'oh!), making my sand surface area calculations be off by two orders of magnitude, you guys used the bulk density of sand, which is about 1.523 g/cm^3, but that doesn't account for sand's porosity. In other words, that bulk density includes all the air around in the spaces between the sand grains. I used the density of silica quartz absent the air.

I trusted you, Jim

So for clarity, here's the calculation for 0.1 mm:

density of silica = 2.65 grams per cm^3

5 pounds = 2268 grams.

So the 5 pounds contains 2268 grams which has a solid volume of 2268 grams / 2.65 g/cm^3 = 855 cm^3

now we need to figure out how many particles that is.

Each 0.1 mm particle has a radius of 0.05 mm = 0.005 cm, and a volume of 4/3 pi r^3 = 4/3 * 3.14*(0.005 cm)^3 = 0.00000052 cm^3

So how many particles do we have? It is just 855 cm^3 total volume divided by 0.00000052 cm^3 per particle = 1,633,000,000 particles.

That's quite a few! More than a billion!

Each 0.1 mm particle has this surface area: 4 pi r^2 = 4*3.14*(0.005 cm)^2 = 0.00031 cm^2

So the total surface area = surface area per particle times total particles = 0.00031 cm^2 *1,633,000,000 = 513,000 cm^2 or 552 sq feet.

The larger particle sizes have proportionally lower surface are per gram. So 1 mm particles have 1/10 as much surface area as 0.1 mm particles of the same mass. So for the larger particles the increased mass doesn't offset the rise in surface area.

So B wins.