pH question

[cal6404]

Hi,

I haven't done chemistry in a long time so this is probably quite basic but I just want to check my working for a pH problem:

There are two streams that enter a mixing tank. Stream 1 flows at 92 m3/d and has a pH of 11. Stream 2 flows at 105.6 m3/d and has a pH of 2.45. I need to know what will be the resulting pH in the mix tank after the streams combine.

How I got my answer:

H+ + OH- <-> H2O
I need to find the final H+ concentration

stream 1)
Assuming concentration of H+ is insignificant
mol of OH- per day = (10^-(14-11))*92.5 = 0.0925

stream 2)
Assuming concentration of OH- is insignificant
mol of H+ per day = (10^-(2.45))*105.6 = 0.3747

H+ and OH- react
0.3747 - 0.0925 = 0.282 mol of H+ after reaction

resulting pH = -log10(0.282/(92.5+105.6)) = 2.846

 

[georgeverghese]

The way H+ and OH- ions influence solution behaviour is considerably affected by the other cations and anions in solution - the mix solution does not behave in the log sum way. If it really does, it can only be coincidence.
Interesting perspective on things though.

 

[25362]

Assuming both are aqueous solutions are of equal compositions except their pH, I get a combined pH = 2.7, as follows:

pH = -log[sub]10[/sub](92[&times;]10[sup]-11[/sup]+105.6[&times;]10[sup]-2.45[/sup])[&divide;](92+105.6) = 2.7​

Please, correct me if I'm wrong.

 

[cal6404]

I realise something stupid. I could just go and take a sample of both streams, mix them in the lab and measure the resulting pH. This being said, understanding the theory is still important.

25462, I don't think you can use your method because you didn't take into account H+ and OH- reacting together. It is valid if you mix two acids and neutralisation doesn't occur. Pretty close to my answer anyway.

 

[georgeverghese]

These calcs would be valid only if the cations and anions and solvent in streams 1 and 2 were the same.

 

[bimr]

In addition to the formula for the reaction between these chemicals, you need to know the molarity. Molarity tells the number of moles of solute in exactly one liter of a solution.

Two pieces of information are necessary to calculate the molarity of a solute in a solution:

• The moles of solute present in the solution.
• The volume of solution (in liters) containing the solute.

To calculate the pH of an aqueous solution you need to know the concentration of the hydronium ion in moles per liter (molarity). The pH is then calculated using the expression:

pH = - log [H3O+].

Example: Find the pH of a 0.0025 M HCl solution. The HCl is a strong acid and is 100% ionized in water. The hydronium ion concentration is 0.0025 M. Thus:

pH = - log (0.0025) = - ( - 2.60) = 2.60

 

[cal6404]

Thanks guys, you can consider this problem as solved now (I played around in the lab for a while and got a decent result). Still a bit vague about the theory behind it but all good. Not sure if there is a button to press on this page to say that I have been helped and I do not require further feedback

 

[cal6404]

hmmm I could get the lab guys to run an inorganics test on both samples to get full specs on ions. Would then know total composition and resulting reactions!

 

[25362]

There are some points to consider:

[&bullet;] whether there is a buffer effect present
[&bullet;] that the streams are thoroughly mixed
[&bullet;] that there is no additional reaction between the acidic and the alkaline streams such as, for example, the release of CO[sub]2[/sub].
[&bullet;] that the reaction has been carried out at 25[sup]o[/sup]C for pK[sub]w[/sub] = 14. Its variation with temperature is:

at 0ºC it is 14.94
at 10ºC it is 14.53
at 20ºC it is 14.16
at 30ºC it is 13.83
at 40ºC it is 13.53

Therefore, your conclusion of actually measuring the final pH is IMHO the correct one.