What is the pH of NaOH & H2SO4 Mixture?

  • Thread starter Thread starter PhysicBeginner
  • Start date Start date
  • Tags Tags
    Ph Scale
Click For Summary

Discussion Overview

The discussion revolves around calculating the pH of a mixture of sodium hydroxide (NaOH) and sulfuric acid (H2SO4). Participants explore the implications of neutralization reactions, the concept of excess reactants, and the resulting pH of the solution after mixing. The scope includes theoretical reasoning and mathematical calculations related to acid-base chemistry.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether the resulting solution should have a neutral pH of 7, given that the reaction is a neutralization.
  • Another participant clarifies that neutral pH occurs only when equal moles of NaOH and H2SO4 are present, emphasizing the need to determine which reactant is in excess.
  • There is a discussion about whether to find the excess of OH- ions or NaOH, and how to incorporate the pH of water into the final calculation.
  • One participant expresses confusion after calculating a pH of around 2, questioning the relationship between pH values and acidity.
  • Another participant corrects the misunderstanding about the stoichiometry of the reaction, noting that 1 mole of H2SO4 reacts with 2 moles of NaOH.
  • Participants request calculations to clarify how many moles of each reactant are consumed and what remains in excess.
  • One participant suggests that the correct pH is around 13, indicating a basic solution, and points out potential confusion between pH and pOH.
  • Another participant advises calculating the initial pH of the NaOH solution before adding sulfuric acid, suggesting that the final pH will still be basic but lower than the initial value.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the exact pH of the resulting solution, with some suggesting it could be around 2 and others indicating it is closer to 12.9. There is ongoing debate about the calculations and the implications of the stoichiometry of the reaction.

Contextual Notes

Participants express uncertainty regarding the calculations of moles and the resulting pH, highlighting potential misunderstandings about the relationship between pH and the concentrations of the reactants. There are also mentions of the importance of considering the final volume of the solution in calculations.

PhysicBeginner
Messages
35
Reaction score
0
Hi everyone i have comed across a fairly difficult question in chemisty that's been baffling me.

A 51.1 mL of 0.153M NaOH was mixed with 27.5mL of 0.0147M H2SO4. What is the pH of the resulting solution.

What I want to ask is shouldn't the resulting solution, water, be neutral with pH of 7 since this reaction is a neutralization, and if not how do i find the pH?
 
Physics news on Phys.org
Yes this reaction is a neutralization, but only if you had the exact same amount of NaOH and H2SO4 would the solution have a neutral pH. You need to find out what is in excess and how many moles of it will be left over after the reaction takes place. From there, take the pH.
 
then do i need to find the excess in OH- ions or the excess of NaOH? Also after I've found the pH do i add the pH of 7 from water to it as well since that number alone is under 7 which shouldn't happen because its the base that's in excess.
 
Last edited:
well find the excess moles NaOH, that will tell you how many moles OH- you have and find the pH from there
 
but i got a pH of around 2, but isn't acids the ones that are below pH of 7?
 
Cesium said:
Yes this reaction is a neutralization, but only if you had the exact same amount of NaOH and H2SO4 would the solution have a neutral pH.

That's not true because 1 mole of H2S04 reacts with 2 moles of NaOH.
 
So can anyone tell me if the pH is around 2?
 
PhysicBeginner said:
So can anyone tell me if the pH is around 2?

No, it is not around 2.
Can you post and show how exactly you calculated the pH? That is, can you show how many moles of NaOH and H2SO4 were consumed and how many moles were excess?
 
Last edited:
Correct pH is about 13 (not exactly 13!). I suppose you have mistaken pH with pOH and you forgot about final volume being sum of volumes.
 
  • #10
Look at the problem again, note that you are starting with a VERY basic solution. To see this, calculate the pH of your NaOH solution *before* you added any sulfuric acid to it. You should get a starting pOH of less than 1 and a starting pH of approximately 13.2 for your .153M OH^- solution. Try this. If you can't do this calculation then stop and look into how to calculate pH from pOH and vice versa because that might be where the problem is.

The next step is to see what the pOH is after you have added the acid to it following the advice others gave you above (especially note that 1M Sulfuric Acid neutralizes *2*M OH^-). If I did the calculation right, you should get a final concentration of .089M OH^- (note that adding the acid nuetralized some of the OH molecules so that you now have a solution with a lower concentration of OH ions). You can then solve for the pOH and from there find the pH. I have a pH for this at around 12.9 (which indicates that the solution is still basic, but less basic than before the sulfuric acid was added to it).
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
17K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 22 ·
Replies
22
Views
7K
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K