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PH scale Question

  • #1
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?
 

Answers and Replies

  • #2
274
0
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.
 
  • #3
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 aswell since that number alone is under 7 which shouldn't happen because its the base thats in excess.
 
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  • #4
274
0
well find the excess moles NaOH, that will tell you how many moles OH- you have and find the pH from there
 
  • #5
but i got a pH of around 2, but isn't acids the ones that are below pH of 7?
 
  • #6
siddharth
Homework Helper
Gold Member
1,127
0
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.
 
  • #7
So can anyone tell me if the pH is around 2?
 
  • #8
siddharth
Homework Helper
Gold Member
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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?
 
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  • #9
Borek
Mentor
28,327
2,714
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
281
1
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).
 

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