I am sure I am doing this Acid/Base problem wrong

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SUMMARY

The discussion centers on calculating the pH of a solution formed by mixing 34.6 ml of 3.45 x 10^-4 M H2SO4 with 41.7 ml of 4.56 x 10^-4 M Al(OH)3. Participants emphasize the importance of updating concentrations based on the total volume of 76.3 ml and understanding the neutralization reaction between the acid and base. The challenge lies in the insolubility of Al(OH)3, which complicates the calculation of remaining acid or base and the resulting pH. The solubility product of Al(OH)3 is crucial for accurately solving the problem.

PREREQUISITES
  • Understanding of acid-base neutralization reactions
  • Knowledge of pH calculations and the relationship between [H+] and [OH-]
  • Familiarity with the concept of limiting reagents in chemical reactions
  • Basic grasp of solubility products, particularly for Al(OH)3
NEXT STEPS
  • Research the RICE table method for equilibrium calculations
  • Learn about the solubility product constant (Ksp) for Al(OH)3
  • Study how to calculate pH from concentrations of strong acids and bases
  • Explore limiting reagent problems in acid-base chemistry
USEFUL FOR

Chemistry students, educators, and anyone involved in acid-base chemistry who seeks to deepen their understanding of pH calculations and the implications of solubility in reactions.

Renee Crosby
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What is the pH of a solution made from 34.6 ml if 3.45*10^-4 M of H2SO4 added to 41.7 ml of 4.56*10^-4 M of Al(OH)3?

I know how to find H+ and OH- and pH I also know how to use x mol / L, but I cannot get them to click together. Everyone keeps telling me to use RICE but my professor never explained what this was to us.

I keep trying different ways to start this problem and they all look wrong. This is the only problem that I don't even know how to put together from the start.
 
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First update the concentrations of acid and base in the total solution of 76,3ml
Then write down the chemical neutralization reaction between the acid and the base.
Then calculate how much acid (or it might be base) that will be left. If it is acid that will be left then you can find [H+]* (and [OH-] will follow)
If it is base that will be left then you can find [OH-] and then [H+].

* have in mind that extra water will be produced from the neutralization reaction so you should update the concentration of the acid left (or base left).
 
What if you were told to start it just as a limiting reagent problem?

But I don't like this question. Unless you are expected to ignore the Al(OH)3 insolubility and to treat the solution as if Al(OH)3 was completely dissociated (which is not true) finding the answer is pretty tricky.
 
Borek said:
What if you were told to start it just as a limiting reagent problem?

But I don't like this question. Unless you are expected to ignore the Al(OH)3 insolubility and to treat the solution as if Al(OH)3 was completely dissociated (which is not true) finding the answer is pretty tricky.
Can we say impossible to solve? The solubility product of Al(OH)3 is absolutely essential.
 

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