Chemistry How much NaOH I have to add to increase pH?

  • Thread starter Thread starter bluecalculator
  • Start date Start date
  • Tags Tags
    increase Ph
AI Thread Summary
To increase the pH of two 5M H2SO4 solutions, calculations must account for the diprotic nature of sulfuric acid. The initial step involves calculating the volume of 5M NaOH needed to raise the pH from approximately 0 to 2, followed by a second step to reach pH 7 for one solution and pH 11 for the other. It's important to consider that the pKa2 value is close to 2, affecting the calculations. Additionally, the discussion highlights the importance of understanding the practical application of these pH adjustments. Accurate calculations require clarity on the desired outcomes and the role of sulfuric acid in the process.
bluecalculator
Messages
2
Reaction score
0
Homework Statement
How much NaOH I have to add?
Relevant Equations
No relevant equations
Hi everyone, I'm stuck in solving this problem. I have two solution of H2SO4 5M that needs two consequential pH increasing: the first from the natural pH of H2SO4 5M (around zero) to pH 2 and then from pH 2 to pH 7. The second solution from the natural pH of H2SO4 5M to pH 2 and then from pH 2 to pH 11. I would like to understand how can I calculate the volume of NaOH (let's say 5M) I have to add for each step.
at the beginning, to calculate the Volume from pH 0 to 2 I was considering this formula:

[H3O+](after adding NaOH)=(V(H2SO4)×[H2SO4](initial)−V(NaOH)added×[NaOH])/V(H2SO4)+V(NaOH)added

But somebody told me this calculation is wrong because I have to consider that H2SO4 is diprotic.

The value I used are:
[H3O+](after adding NaOH)= 10^-2
V(H2SO4)= 1L
[H2SO4](initial)= 5 M

Thank you !
 
Last edited:
Physics news on Phys.org
You should take into account not only fact the acid is diprotic, but also - especially around pH 2 - that pKa2 is close to 2.

I don't see any difference between the first and the second solution.
 
Yes sorry..a typo. Now it's correct. Thank you, I'm going to try!
 
You want to get your solution to pH7? The pH of pure water is 7, you wouldn't start with sulphuric acid. That is the pH of freshly deionised water is 7, if you leave it standing around without preventing it, it will absorb CO2 from the air and decrease to about 4.5. Also 0.001 M NaOH should be pH 11 - what do you need the sulphuric acid for?

When we get this sort of question from someone sounding rather unfamiliar we can really only help (maybe) if they tell us the practical thing they are trying to achieve or overall problem to solve.
 
Last edited:
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top