Solving pH of Acid-Base Mixture - Adrian's Revision

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Discussion Overview

The discussion revolves around calculating the resulting pH of a mixture of hydrochloric acid (HCl) solutions with different pH values. Participants explore the mathematical approach to determine the pH after mixing equal volumes of these solutions, addressing concepts related to concentration and logarithmic relationships in acid-base chemistry.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Adrian presents a problem involving the mixing of HCl solutions with pH values of 4.0 and 5.0, seeking guidance on how to approach the calculation.
  • Some participants suggest calculating the concentration of hydrogen ions ([H+]) instead of averaging the pH values directly.
  • There is a discussion about the correct units for [H+], with clarification that it should be in moles per liter (M).
  • One participant emphasizes that concentrations cannot simply be added together, referencing the principles of dilution and mixing solutions.
  • Another participant provides a method for calculating the total moles of H+ from both solutions and suggests using this to find the total concentration before calculating pH.
  • Adrian proposes a calculation involving the total moles and concentration, arriving at a pH value of 4.26.
  • Participants confirm that the approach to the calculation is correct, though there is no explicit verification of the final pH result.

Areas of Agreement / Disagreement

Participants generally agree on the method of calculating pH through concentration but do not reach a consensus on the final numerical result. There is acknowledgment of the logarithmic nature of pH and the importance of proper units, but some disagreement exists regarding the addition of concentrations.

Contextual Notes

Limitations include potential misunderstandings about the addition of concentrations and the logarithmic relationship of pH, which may affect the accuracy of the calculations presented.

Who May Find This Useful

This discussion may be useful for students preparing for exams in chemistry, particularly those focusing on acid-base reactions and pH calculations.

adoado
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Hello all,

I am trying to revise over acids and bases before my exams, but there is one question I am stuck on, and I am not sure how to approach it. Any hints would be great ^^

"100mL of HCL of pH of 4.0 is mixed with 100mL of HCL of PH of 5.0. Whats the resulting pH of the solution formed".

Its probably really easy...

This is not homework, just revision I decided to attempt.

Cheers,
Adrian
 
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You might be tempted to say (4.0 + 5.0) / 2 = 4.5, it's not that easy. It is straightforward though: just calculate the concentration of H+.

What is the relation between that and pH?
 
Thanks for the reply ^^

You might be tempted to say (4.0 + 5.0) / 2 = 4.5, it's not that easy

lol if only...

Hows this:

[H+]1 = 10-4
[H+]2 = 10-5

pH = -log[10-4 + 10-5]

Close? ^^
 
Closer, yes :)
Can you include the units of [H+] ?
It's not moles, as your calculation now suggests.
 
Remember that [H+] is concentration, which is logarithmic. You do seem to be getting closer though.
 
No, pH is logarithmic. Concentration would be antilogarithmic :wink:

--
 
Thanks for the replies everyone.

Right, so the units would be moles per litre or M

Apart from that, would the addition of the two concentrations and then taking the negative log yield the correct answer? ^^
 
No, you can't just add concentrations. It would violate one of the most basic laws that governs physics and chemistry.

See diluting and mixing solutions for details of the correct approach.

--
 
Don't guess at the answer, work it out. You have a concentration with 10-4 mol/l and you take 100 ml. Then you add to that 100 ml of a concentration with 10-5 mol/l. You get how many with what concentration?
 
  • #11
Ok, hows this:

n = cV so

n1 = 0.1 x 10-4 = 10-5

n2 = 0.1 x 10-5 = 10-6

Total moles = n1 + n2
so total concentration = total moles / .200Litres.

C = (10-5 + 10-6) / 0.200

pH = -log(C)

pH = 4.26 ?
 
  • #12
Not checking math - approach is perfect :smile:

--
methods
 
  • #13
And the number sounds familiar from when I checked it a few days ago when you first posted the question.
It is also what you would expect: mixing equal amounts of pH 4 and pH 5 solutions gives something with pH between 4 and 5, but lower than 4.5 due to the logarithmic scale.
 
  • #14
Cool, thanks everyone for helping :)

Thanks!
Adrian
 

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