Master Reef Chemist Certification Question!!! (yes, you want this title)

[Chris86]

Okay, if I understand this correctly, he is basically asking us what happens when you dilute a carbonate buffer with pure water. Part of the equilibrium for carbonate involves CO2 coming into equilibrium with the environment. So if the question is asking about the pH prior to gas exchange, the aqueous CO2 does not come into equilibrium with the atmosphere after dilution.

There are two things that contribute to pH in a carbonate buffer. The total amount of carbon dioxide and the alkalinity. If the volume of the solution is doubled, then the both of these are cut in half. This will result in an increase in pH until gas exchange happens.

I’m tired. I think I explained this well. Lol

 

[Newb73]

So it seems PH is a measurement of hydrodgen ACTIVITY not concentration.

If pure H20 gives up H it trends towards acidic and if it absorbs H it tends towards base. Since i have observed low ph in my rodi I assume it readily gives up H when it is that pure.

I am still not sure what effect introducing NACL into the mix has.....(honestly i sort of hate chemistry) is it to break off even more H? That could explain a DROP in PH from the original mix.

 

[Newb73]

It seems nacl cleves into na and cl and then simply reorients the h20 two ways via it's poles with no net loss of H.

 

[Chris86]

Okay, if I understand this correctly, he is basically asking us what happens when you dilute a carbonate buffer with pure water. Part of the equilibrium for carbonate involves CO2 coming into equilibrium with the environment. So if the question is asking about the pH prior to gas exchange, the aqueous CO2 does not come into equilibrium with the atmosphere after dilution.

There are two things that contribute to pH in a carbonate buffer. The total amount of carbon dioxide and the alkalinity. If the volume of the solution is doubled, then the both of these are cut in half. This will result in an increase in pH until gas exchange happens.

I’m tired. I think I explained this well. Lol

One small detail I forgot to include. The pK value of bicarbonate is 6.1. At pH 8, the percent of the buffer that is bicarbonate is nearly 100%. This is why an increase in volume causes an increase in pH.

 

[Chris86]

One small detail I forgot to include. The pK value of bicarbonate is 6.1. At pH 8, the percent of the buffer that is bicarbonate is nearly 100%. This is why an increase in volume causes an increase in pH.

Okay one last thing lol

When you increase the volume with pure water, you actually increase the amount of carbonic acid. This is because there is a lot of CO2 in aqueous solution.

Increasing the amount of carbonic acid at this pH, considering the pK value of bicarbonate, ends up increasing the concentration of bicarbonate and increases the pH

 

[Haubfather07]

At 35 ppt salinity pH is 8.1 and the density is 1.0264. So I just set up a simple ratio looking for x. 1.0264/8.1pH = 1.009/x.

X= 7.96, which is none of the options.......

 

[reeferfoxx]

At 35 ppt salinity pH is 8.1 and the density is 1.0264. So I just set up a simple ratio looking for x. 1.0264/8.1pH = 1.009/x.

X= 7.96, which is none of the options.......

It would be B "Above 7.7"

 

[Haubfather07]

It would be B "Above 7.7"

“About 7.7” didn’t say above. While 7.9 I guess is close, the relative distance between options made me think that wasn’t close enough to call it truly B.

 

[reeferfoxx]

“About 7.7” didn’t say above. While 7.9 I guess is close, the relative distance between options made me think that wasn’t close enough to call it truly B.

Ahh shoot you're right! ;Bucktooth A's a better option anyway

 

[Haubfather07]

Ahh shoot you're right! ;Bucktooth A's a better option anyway

Learned more about chemistry in this post then I did in all the classes I took in high school hahaha. New at the Reef game too, so I’m just happy to read through the discussion.

 

[Frogger]

How can the ph raise when you add water with a pH of 7? If this was the case I would just lower my salinity if I wanted to increase my pH.

So when my pH drops below 8 just add RO water and I am back in the game.

I guess if I splash lye in my eye I shouldn't flush with water. (I am just joking to make a point)

 

[Belgian Anthias]

When the salinity is lowered the PH will rise, this when the salt (NaCl)° is removed out of the liquid. In this case no salt is removed from the solution which means the same amount of hydrogen is needed to keep it in solution, I suppose there will be no H+ on the move caused by the salinity drop. Will there be other big changes in H+ caused by adding clear neutral water to the solution?
The seawater content is diluted by a factor of 3 but nothing is removed. What chemical reactions ( not bio-chemical as the solution is supposed to be clean) may change the H+ availability caused by adding H2O?

 

[Frogger]

In my haste to get an answer in I made a slight calculation error. I have taken a second look at it. You need to add 1.9l of ro water to 1l of saltwater to get a total volume of 2.9l not 2.9l of ro water to lower the salinity to .009. I still stick to my answer of B as we are dealing with a logarithm scale. I used litres as a hypothetical volume to help make the calculation make sense.

 

[ReefBeta]

I think you don't have the correct answer in your options.

Reasoning:
pH is measure of hydorgen ion concentration in the liquid, unit as mol/L. I think mixing water will result in the same total moles of hydorgen ion (if this is not true, then my calculation will be wrong). So target pH is calculated from ratio of total moles of hydorgen ion to total volume of water.

Calculation:
saltwater: 35 ppt, PH 8.1
ro water: 0 ppt, PH 7

Let's note the total weight of saltwater as x (kg), so salt weight is x*(0.035)
Total weight of RO water as y (kg), salt weight is 0

From target water to be 12ppt we have:
(0.035x)/(x + y) = 0.012
0.035x = 0.012x + 0.012y
0.023x=0.012y
y=23/12x ~= 2x

volume of saltwater: x/1.0264 (L)
volume of ro water: y (L)
hydorgen ion in sw: 10^(-8.1) * x/1.0264 (mol)
hydorgen ion in ro: 10^(-7) * y (mol)

target hydorgen ion: (10^(-8.1) * x/1.0264 + 10^(-7) * y) (mol)
target volume: (x+y)/1.009 (L)

target PH = log( (10^(-8.1) * x/1.0264 + 10^(-7) * y) / ((x+y)/1.009) )
=log( (10^(-8.1) /1.0264 + 10^(-7) * 2) / (3/1.009) )
=-7.156

so the resulting pH = 7.156, closet Answer is D, about 7.3, but it's still pretty far away ...

 

[ReefBeta]

duplicate post

 

[TheEngineer]

I think A. Pure water has no buffering capacity.

 

[Randy Holmes-Farley]

A lot of great discussion folks!

One of the first things a chemist looking at pH in seawater would say is that is very complicated if you really want to understand every detail. It's a mess, actually. Nearly every element in the periodic table is present in seawater, and they do not need to be acids or bases to be involved.

We (reefers) usually don't care about the background details. But we want to be close enough to get answers that matter.

When I begin to explain the answer later today or tomorrow morning, I'll start by discussing very simple "models" of the problem (say, starting with RO/DI and sodium hydroxide solution at pH 8.1) and then stepwise increase the complexity.

Note there are two big jumps in complexity that get close enough to true seawater.

 

[kecked]

RodI has no buffering and thus will take on the pH of whatever the buffering is from the resulting solution. So long as there is nothing to disturb by using up the buffering alkalinity it shouldn’t change. You said no air exchange so you’d have to account for the dissolved CO2 in the RodI and then account for how it uses up the alkalinity in the buffering.

So to answer this I need to calculate the buffering available and the dissolved co2 as carbonic acid and the other varieties but mainly the carbonic acid.

My feeling is it would be slightly less that the seawater you start with. How much requires an equilibrium calculation.

I don’t have time to answer this right so I’ll come back but here’s a good reference. http://ion.chem.usu.edu/~sbialkow/Classes/3600/Overheads/Carbonate/CO2.html

So about 335ppm co2 is a ph around 5.35 in the RodI. That’s from the above reference. I’ll finish later.

 

[Chuk]

Roughly 3:1 dilution
h+ in SW is 10^8.1*1L
and freshwater is 10^7x3L
So log((10^8.1+3x10^7))/4~7.5 which is C

dang did my dilution wrong quick off the top of the head i did total volume not added volume it should be 2:1 so that changes to B. Rookie mistake.

 

[DBH2017]

B