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Calculating the molar concentration of H3O+ and pH of solutions

  • Thread starter NYK
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  • #1
NYK
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Homework Statement


Calculate the the molar concentration of H3O+ ions and the pH of the following solutions:
a) 25.0 cm3 of 0.144 M HCl(aq) was added to 25.0 cm3 of 0.125 M NaOH(aq)
b) 25.0 cm3 of 0.15 M HCl(aq) was added to 35.0 cm3 of 0.15 M KOH(aq)
c) 21.2 cm3 of 0.22 M HNO3(aq) was added to 10.0 cm3 of 0.30 M NaOH(aq)

Homework Equations


pH = -log[H3O+] Handerson - Hasselbach eqn.

The Attempt at a Solution


[/B]
I have only been able to solve part a)

I did that by multiplying the molairty of the strong acid/base by the total volume

then: [H3O+] = [HCl] - [NaOH] = 9.5 x 10-3mol

then took the negative log of that number to find the pH = 2.0

which are the correct answers, but when I do that process for part b) and c) which are also strong acid/base combos, the answers are no where near correct.

Any help will be appreciated.
 

Answers and Replies

  • #2
Borek
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In the second case - what is the limiting reagent? What is left in the solution after the neutralization reaction took place?
 
  • #3
NYK
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In the second case - what is the limiting reagent? What is left in the solution after the neutralization reaction took place?
In the second case the LR is the acid. When i do that calculation I find that the [H3O+] = .04875 mol/L

the answer is 21 mmol/L and a pH = 12.3
 
  • #4
Borek
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In the second case the LR is the acid.
Good.

When i do that calculation I find that the [H3O+] = .04875 mol/L

the answer is 21 mmol/L and a pH = 12.3
I am afraid neither of these numbers is correct.

First of all - you said acid was the limiting reagent. If so, how come there is mo much H3O+ left?

21 mmol/L of what?

pH of 12.3 is quite close - but it is possible to easily give a better answer.
 
  • #5
epenguin
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Strong acids and bases, so Henderson - Hasselbach eqn. never came into it.
 

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