Calculating the pH of 350L of Sulfuric Acid (H2SO4)

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SUMMARY

The discussion revolves around calculating the pH of a 350L solution of sulfuric acid (H2SO4) with a hydrogen ion concentration of 2.0 x 10-4 mol/L. The correct pH calculation yields a value of approximately 3.69, although considerations regarding water's auto-ionization at low concentrations are noted. Additionally, participants clarify that the volume of the solution is relevant for stoichiometric calculations when determining the mass of sodium hydroxide (NaOH) needed for neutralization, which is calculated to be 2.8g based on the reaction 2 NaOH + H2SO4 → Na2SO4 + 2 H2O.

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  • Understanding of pH calculations and the formula pH = -log[H3O+]
  • Knowledge of sulfuric acid dissociation and its two protons
  • Familiarity with stoichiometry and molarity calculations
  • Basic understanding of neutralization reactions involving acids and bases
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  • Research the auto-ionization of water and its impact on pH calculations
  • Learn about the dissociation reactions of strong acids, specifically sulfuric acid
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  • #31
Can you write NaOH in dissociated form?
 
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  • #32
sure
NaOH ----> Na+ + OH-
 
  • #33
OK, now, what would happen when you add H+ to the mix?

H+ + Na+ + OH- -> ?
 
  • #34
i don't know...but why we will add H+
 
  • #35
Why? Because the question is about reaction between H+ and OH-. You have a solution of H+ and you add solution of OH- (Na+ and SO42- don't take part in the reaction, they are so called spectators).

What simple, electrically neutral molecule can be a product of reaction between H+ and OH-? Try to guess, it is so simple you will hate yourself you have not realized it earlier.
 
  • #36
sulphuric has 2 protons.

so ph =-log(0.0004)

=-(log4-4)
=-(0.6021-4)
=3.3979
 
  • #37
1 hydrogen has 1 proton but in sulphuric there are 2 hydrogen
 
  • #38
raviyank said:
sulphuric has 2 protons.

so ph =-log(0.0004)

=-(log4-4)
=-(0.6021-4)
=3.3979

This is wrong, please reread the question.

And read forum rules, giving final answers is against them. Not to mention giving wrong final answers.
 
  • #39
Hello Borek o:)
please check this:

CA*VA=CB*VB
2*10-4*350=mB/40
mB=2*10-4*350*40 = 2.8g

That's it ?
 
  • #40
2.8g of NaOH it is.
 
  • #41
Borek said:
2.8g of NaOH it is.

Yay :-p thank you Borek o:)
 

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