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

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Discussion Overview

The discussion revolves around calculating the pH of a 350L solution of sulfuric acid (H2SO4) and determining the mass of sodium hydroxide (NaOH) required to neutralize it. Participants explore various aspects of acid-base chemistry, including the effects of concentration, the dissociation of sulfuric acid, and stoichiometry in neutralization reactions.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the pH using the formula pH = -log[H3O+] and arrives at a value of 3.69 based on a hydrogen ion concentration of 2.0 ∙ 10-4 mol/L.
  • Another participant suggests that the auto-ionization of water might need to be considered due to the low concentration of the acid.
  • Concerns are raised about the relevance of the volume of the solution in the calculations, with some arguing it may not be necessary for pH determination.
  • Participants discuss the stoichiometry of the neutralization reaction between sulfuric acid and sodium hydroxide, with varying calculations of the mass of NaOH needed.
  • There is confusion regarding the mass of sulfuric acid versus the mass of the solution, with multiple participants attempting to clarify the distinction.
  • One participant expresses frustration and confusion over the problem, indicating a lack of understanding of the concepts involved.
  • Several participants engage in a back-and-forth about the correct approach to calculating the number of moles of H+ and the corresponding moles of NaOH required for neutralization.
  • There are multiple calculations presented for the mass of NaOH needed, with some participants correcting each other’s methods and results.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem, with multiple competing views on the relevance of volume, the calculations for pH, and the stoichiometry of the neutralization reaction. Disagreements persist regarding the interpretation of the problem and the necessary steps to arrive at a solution.

Contextual Notes

Participants highlight the importance of understanding the concentration of sulfuric acid in relation to the total mass of the solution, as well as the implications of acid dissociation on pH calculations. There are unresolved issues regarding the correct application of stoichiometric principles and the role of water's pH in the calculations.

Who May Find This Useful

This discussion may be useful for students studying acid-base chemistry, particularly those working on homework problems involving pH calculations and neutralization reactions.

  • #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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