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.

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Homework Statement


Sulfuric acid (H2SO4) were collected from the waste acid container and the volume is 350 l.

Homework Equations


a) Calculate the acid pH of the solution when the hydrogen ion concentration is 2.0 ∙ 10-4 mol / l.


The Attempt at a Solution


i think it's simple no ?
PH=-log H3O+
PH= -log 2*10-4
PH=3.69
 
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Maybe. Its either that simple or you have to take into account the auto ionization of water. Since your acid is at a low enough concentration, the water may come into play. Also, there is a chemistry forum that you might have more luck posting on.

EDIT: I can't remember for sure, but I think you're answer is right since its at 10^-7 mol/L that you have to take into account the water's pH
 
where did that 10^-7 came from? and why they gave us the volume of 350L
 
I don't know. You need to find the concentration at which you need to take into account water's pH. I vaguely remember 10^-7 mol/L as being something important, but as I said, I can't remember for sure. Google it or something.
 
See pH of a strong acid/base.

The question is either that simple, or there is something more to it, a lot depends on what you were doing lately. For sure volume is irrelevant. It is possible that they wanted you to take ionic strength of the solution into account. If you have no idea what ionic strength and activity coefficients are (and you have not slept during a lecture), 3.96 is right (although it probably contains too many significant figures).
 
ok thank you, i think they ment only that 3.96
however in part B they asked :what is the mass of sodium hydroxide NaOH needed to neutralize the effect of the acid solution..?
 
Here the volume will come handy, as for stoichiometry you need number of moles.
 
Borek said:
Here the volume will come handy, as for stoichiometry you need number of moles.

is it ok like this:
2 NaOH + H2SO4 ----->Na2SO4 + 2 H2O
m-----------350000g
2*40g------98.07g

m=285510.34g
that's 285.51 Kg
 
350 kg of solution is not equivalent to 350 kg of sulfuric acid.
 
  • #10
Borek said:
350 kg of solution is not equivalent to 350 kg of sulfuric acid.

Oups

2 NaOH + H2SO4 ----> Na2SO4 + 2H2O

m-------------------------350000g
2*40g--------------------178.07g

m=157241.53 g
that's 157.24 Kg
 
  • #11
You are still completely wrong and guessing at random won't get you far.

350 kg is not mass of anything that you need to use. You need to use CONCENTRATION to calculate amount of substance.
 
  • #12
I'm lost :(
can you please tell me the answer, and i will try to find out how it came..Please
 
  • #13
No, I will try to walk you through the problem.

First of all - if the concentration of H+ is 2x10-4 M, what is the concentration of sulfuric acid?
 
  • #14
I really don't know but i can try this:
concentration of sulfuric acid = n/V

n= m/M = 350000/98.07 = 3568.87 moles

concentration of sulfuric acid = n/V = 3568.87/350 = 10.19 mol/l
 
  • #15
You are again trying to calculate number of moles of acid assuming its mass is 350 kg - it is not. You have 350 kg of solution, not of the acid.

What is dissociation reaction of sulfuric acid in water? Can you write it? How many moles of H+ are produced for each mole of sulfuric acid put into water?
 
  • #16
i don't know.
 
  • #17
Seems like your teachers take money for nothing then.

Let's try different approach.

How many moles of NaOH are required to neutralize 1 mole of H+?
 
  • #18
I really don't know, I'm sorry...
i have learned in french and this is kind of confusing for me..i hate to say this..but I'm going to give up about this problem..before it confuse the concepts that i already know and work with.
 
  • #19
How does NaOH react with acid? What are products?
 
  • #20
Borek said:
How does NaOH react with acid? What are products?

Na2SO4 and 2H2O and H3O+ i think
 
  • #21
No, just salt and water. H3O+ is a product of acid dissociation (H3O+ or H+ - these are in many ways equivalent).

Do you know what net ionic reaction is?

Can you write balanced reaction equation for a reaction of H+ with NaOH?
 
  • #22
Does this law help?
concentration of acid * volume of acid = concentration of base * volume of base
 
  • #23
This is not a law and in general it is wrong - but it is not far from what you need to solve the question.

Take a look here: http://www.titrations.info/titration-calculation - while it is about titration, it nicely addresses everything needed to solve the problem.
 
  • #24
i can't use much of Ca*Va = Cb*Vb
but i can write it: na = nb
na = mb/Mb
mb = na*Mb
mb = 2*40 = 80g
 
  • #25
Stop using masses for everything.

Can you calculate number of moles of H+? You know concentration and volume.
 
  • #26
n=C*V = 2*10-4*350= 0.07mol
 
  • #27
OK, now how many moles of NaOH will react with 0.07 moles of H+? Do you know what the reaction is? Reaction equation?
 
  • #28
2 NaOH + H2SO4 ---> Na2SO4 + 2 H2O
 
  • #29
This reaction is correct, but it is not what I am asking about.

I am not asking about the reaction with sulfuric acid, but about reaction with H+.
 
  • #30
Ahhh i don't know that one :eek:
 

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