# H2SO4 and pH fluctuations

1. Jul 17, 2016

### usED

1. The problem statement, all variables and given/known data
pH fluctuations need to be determined at a maximum dosage of 50mg/L of 93% sulphuric acid .

2. Relevant equations
If the start pH of the water sample is 8, how much will pH be lowered by adding of 50mg/L of 93% sulphuric acid, and what is the correlation between the different mass/volumes added and pH fluctuations.

3. The attempt at a solution

First, I have tried to find out what would be the pH of 50mg/l of 93% H2SO4 and from given data i have calculated molarity:
Concentration * Density or Specific Quantity / Molar Mass = Molarity
(0.930) *(0.05g/L):( 98.079 g/mol)=0.0004741

H2SO4 --> 2 H+ + SO4-2
H2SO4 is a strong acid so concentration of [H+]=Molarity
pH=-log[H+]=-log(2x0.0004741)= 3.0231

However, this is where the calculation stops, since i don't know how to calculate the final pH

I am even wondering was this right way to start the calculation..

2. Jul 17, 2016

### epenguin

I do t know what you mean by 'fluctuations', the word seems irrelevant, you just want to calculate a pH from an acid concentration.

Your calculation is quite reasonable, as far as I can see correct, from your assumptions.

But one assumption is slightly wrong. Only in its first dissociation H2SO4 ⇔ H+ + HSO4- is sulphuric acid super-strong. The second dissociation HSO4- ⇔ H+ + SO42- is only 'moderately strong' with a pKa near 2.

But not a lot of people know that.

And then again your solution is quite dilute so in this case your second proton is likely to be mostly dissociated too. And moerover you can reason (even if it sounds circular it's OK) the pH is around 3 so that second proton is going to be something like 90% dissociated, so your error is small. Depending what you need it for it may not be worth the trouble of calculating it more exactly, which is not very difficult.

Note that if it were more concentrated, about 1M, only the first dissociation above would be significant and you would have to leave out that factor 2 that you put in the calculations. That would be not too bad as approximation at 0 .1 M; at intermediate concentrations you have a more complicated calculation but I don't know if you are up with this sort of thing yet.

Last edited: Jul 18, 2016
3. Jul 18, 2016

### usED

Hi epenguin, thanks for that. You are right, i don't need more detailed calculations since the errors will be small.
The thing is, I wanted to know is there any way to consider the start pH while doing this calculation because it's not going to be same if i add certain amount of H2SO4 to water that has pH=7 or pH=14..?

Further, I tried to get to the formula which will help me to calculate backwards the mass of H2SO4 starting with the pH, and my first calculation should have been a starting point to check any further calculations, but I keep bumping into the wall...

4. Jul 18, 2016

### epenguin

Water has pH 7. I would not call something with pH 14 'water', I would call it 1M alkali, e.g 1M NaOH. For calculating the result of adding acid to that, you just have to calculate the amount of alkali neutralised by the acid. Subract. From whatever is left over, or in excess you calculate [OH-] or [H+].

For calculating backwards there is no new formula – it is just the same formula and the calculation is done backwards! Maybe you need to do it to just one step at a time to see it.

(Edit: unless you ha le forgotten what logarithns really are and that if x = log10 y, then y = 10x.)

Last edited: Jul 19, 2016