# Calculate the amount to bring down the pH

I have 60,000 LBS of a NaOH solution with a pH of 12.6. I need to be able to calculate the amount of 25% H2SO4 to add to bring the pH down to below 10. Is this possible to calculate?

It sure is. First thing I would do is calculate the molarity of the NaOH solution. From there, figuring out the rest is fairly simple, especially since NaOH and H2SO4 are strong bases/acids.

I apologize for not being a chemist, could you elaborate?

60,000 lbs NaOH solution at pH 12.6 I believe is equal to ≈ .04 mol solution and 60,000 lbs ≈ 2.7215x10^4 L solution @ .04 mol/L (doesn't include adding in the moles NaOH so not completely accurate)

2.7215x10^4 L solution × .04 mol/L ≈ 1089 mol NaOH

1089 mol NaOH will need 544 mol H2SO4 to nuetralize it since sulfuric acid releases two hydrogen ions per molecule and NaOH only one OH.

From here we just need to know is the 25% H2SO4 by weight or volume and we can complete the calculation.

Someone on the board should check my numbers too! :)

I decided to make it hard on myself and take the mass of NaOH into account and ended up with 1084 mol NaOH, so Mesa's numbers look good.

Borek
Mentor
Someone on the board should check my numbers too! :)

They look reasonable.

The H2SO4 is 25% by weight. Thank you so much.

epenguin
Homework Helper
Gold Member
We have answered the question you asked but whether that solves your problem we don't know as we don't know what it is. Why do you want it specifically "below 10"?

3 or 4 are below 10, and when you have added one drop too much you will probably get down to something like that - these are small H+ concentrations. A drop too liitle and you will be above 10. The pH changes precipitously at neutralisation. I don't know whether you don't want to have it too acid, or to use no acid more than necessary.

The way you state the problem suggests you do not know the concentration of acid very accurately. So if your problem is how to calculate moderately well so you know better than ballpark what volumes to use you have been answered, but if it matters to add enough but not too much, then you will have to not trust the calculations except for this setting-up, and you will be having to measure pH or using an indicator as you do it. And do the same next time as your next bottle may not be the exact same concentration whatever it says on the label.

Epenquin is right, (should have noticed that lol) because the pH scale is lograthmic. You can look at it this way:

A pOH of 12.6 is about .04mols NaOH/Litre whereas 10 is .0001mols NaOH/Litre.
So to completely nuetralise your NaOH (assuming everything is near perfectly measured) you would need 544mol H2SO4; for a pH of 10 you need 542.9mol so you can see a tiny change at this levels makes a huge difference in pH.

Just for numbers sake, that works out to about 18.865L of 25% H2SO4 by weight (about28.78M). Someone on the board should probably check these numbers too,

When I had my plant I used my burette, a scale, oxalic acid, phenolphthalein, deionized water and some basic glassware to make calibrated titration solution. I believe you could use the same equipment to check your situation too and get more accurate measurements. (someone feel free to jump in if Im missing something)

Small scale trial and error is also a good idea.

Borek
Mentor
Small scale trial and error is also a good idea.

Actually it is the best idea, as we know nothing about accuracy of the initial numbers. They give a simple way of finding the order of magnitude type answer, but not the exact value.

...as we know nothing about accuracy of the initial numbers. They give a simple way of finding the order of magnitude type answer, but not the exact value.

Lue, taking samples and testing is sounding like a good way to tackle this problem, make as accurate measurements as you can and see what change in pH you get. Come back with the data and also to what degree of accuracy your equipment is and we can give you a better estimate based on that. We also need to know your pH range.

epenguin
Homework Helper
Gold Member
I should have said that as well as not knowing exactly the concentrations of H2SO4 you also do not know accurately that of NaOH either as it comes in rather messy pellets with unknown amount of water that has deliquesced into them. For this reason weighed NaOH is never used as 'primary standard'.

I couldn't find a good density for 25% wt H2SO4, but did find that baterry acid is typically a little more concentrated- it does appear that the molarity would not be as high as 28M since 100% H2SO4 is given as 18M.

Also at these volumes we want to emphasize safety as adding a relatively concentrated acid like this is going to be exothermic and the stirring rate is going to have to be relatively high to get a good incorporation. The relative amount of water in the NaOH solution may be sufficient to transfer most of the reaction heat; otherwise cooling would be necessary.

Doing the small scale titration and then scale up is the appropriate; do it in a thermos and measure delta T and some idea of the heat of reaction can be derived too.

Borek
Mentor
I couldn't find a good density for 25% wt H2SO4

1.1783 g/mL, 3.00M.

Clarification to my above post MrSid...

for a pH of 10 you need 542.9mol so you can see a tiny change at this levels makes a huge difference in pH.

Just for numbers sake, that works out to about 18.865L of 25% H2SO4 by weight (about28.78M). Someone on the board should probably check these numbers too,

1.1783 g/mL, 3.00M.

Therefore 181 liters of 25% H2SO4 for neutralizing the ideal 544moles of NaOH.

I couldn't find a good density for 25% wt H2SO4, but did find that baterry acid is typically a little more concentrated- it does appear that the molarity would not be as high as 28M since 100% H2SO4 is given as 18M.

Damn I screwed that one up lol!
I am getting ≈ 18M for 100% H2SO4 too, this is why it is a good idea to check my numbers.

I'm going to have to break this down and see where the heck I messed up so bad, maybe I better go sign up for an 082 MAT while I'm at it! :)