What Is the Resulting pH of Mixing These Solutions?

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SUMMARY

The resulting pH of mixing 200 mL of a solution with H3O+ concentration of 1 * 10-4 and 100 mL of a solution with H3O+ concentration of 2 * 10-11 is approximately 10, not 4 as initially calculated. The discrepancy arises from the need to account for the autoionization of water, which contributes an additional H3O+ concentration of 10-7 in each solution. The problem is fundamentally flawed as it lacks clarity on the nature of the solutions involved.

PREREQUISITES
  • Understanding of pH calculation using the formula pH = -log[H3O+]
  • Knowledge of molarity and how to calculate moles (NH3O+ = M * L)
  • Familiarity with the concept of autoionization of water
  • Basic principles of acid-base chemistry, including the behavior of strong acids and bases
NEXT STEPS
  • Research the concept of autoionization of water and its impact on pH calculations
  • Study the differences between strong acids (e.g., HCl) and strong bases (e.g., NaOH) in solution
  • Learn how to construct solutions to achieve specific pH levels through mixing
  • Explore advanced pH calculation techniques for mixed solutions with varying concentrations
USEFUL FOR

Chemistry students, educators, and anyone involved in laboratory work or acid-base titration experiments will benefit from this discussion.

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Homework Statement



You mix the below two solutions. What is the resulting pH?

200 mL solution with H_{3}O^{+} = 1 * 10^{-4}
100 mL solution with H_{3}O^{+} = 2 * 10^{-11}

Homework Equations



pH = -log[H_{3}O^{+}]

N_{H_{3}O^{+}} = M * L

The Attempt at a Solution



I first found the number of moles of hydronium ion on each of the two solutions by multiplying volume in liters by the molarity of the solution. For the first solution, the number of moles of hydronium ion is 2 * 10^{-5}. The second solution has 2 * 10^{-12} moles of hydronium ion.

I added the number of moles of hydronium ion and divided by the volume of the mixture - 300 mL or 0.3 L. This yields the new hydronium ion molarity.

I then took the negative logarithm of the hydronium ion molarity and found the pH to be approximately 4.

However, the key says the pH is instead approximately 10, which I noticed to be 14 - 4 (pK_{w} = 14 = pH + pHO).

Why is the solution basic instead of acidic?
 
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Qube said:

Homework Statement



You mix the below two solutions. What is the resulting pH?

200 mL solution with H_{3}O^{+} = 1 * 10^{-4}
100 mL solution with H_{3}O^{+} = 2 * 10^{-11}

Homework Equations



pH = -log[H_{3}O^{+}]

N_{H_{3}O^{+}} = M * L

The Attempt at a Solution



I first found the number of moles of hydronium ion on each of the two solutions by multiplying volume in liters by the molarity of the solution. For the first solution, the number of moles of hydronium ion is 2 * 10^{-5}. The second solution has 2 * 10^{-12} moles of hydronium ion.

I added the number of moles of hydronium ion and divided by the volume of the mixture - 300 mL or 0.3 L. This yields the new hydronium ion molarity.

I then took the negative logarithm of the hydronium ion molarity and found the pH to be approximately 4.

However, the key says the pH is instead approximately 10, which I noticed to be 14 - 4 (pK_{w} = 14 = pH + pHO).

Why is the solution basic instead of acidic?

You were drawn maybe by the wording into thinking oh that's just water with this and other water with that H3O+ concentration. But water is water for which [H3O+] is 10-7. So what you've got in those solutions is dilute acid and dilute alkali. First think how acid one and how alkaline the other is and you will see which of the answers, yours or the official one, is reasonable.

Suppose the one is HCl and the other is NaOH and work out the pH. I get 10.
 
Hmm, so I forgot the auto ionization of water. But wouldn't that be included in the hydronium ion concentrations given? Or should I add 10^-7 for each solution?
 
Actually this problem is faulty and has no reasonable answer as worded. It doesn't say anything about the other compounds in the solution, which is a crucial thing.

I can construct for you example with solutions that will give after mixing ANY pH between 4 and 10.7. Name pH and I will tell you how to do it. Seriously.
 
That's true, I just made the simplest assumption about what the solutions were, which turns out to give the stated right answer.
It should also have been stated they were 10-4 M etc.

Quite often on this forum we have to guess what the questions really were. :biggrin:
 
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