PH and pOH and [H+] and [OH-] Problems

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Discussion Overview

The discussion revolves around calculating pH and pOH in various chemical scenarios, particularly involving strong acids, weak bases, and neutralization reactions. Participants seek assistance with specific problems related to these concepts, including the mixing of solutions and the equilibrium of weak bases.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents three problems involving pH calculations for mixed solutions of HCl and KOH, expressing uncertainty about the methods to solve them.
  • Another participant calculates the pH of a .400M Sr(OH)2 solution, questioning why the concentration used is .800M instead of .400M, suggesting it may be due to the presence of two hydroxide ions.
  • Responses to the initial problems suggest calculating the number of moles of HCl in mixed solutions to determine the resulting concentration and pH, and emphasize the importance of stoichiometry in neutralization reactions.
  • Participants mention using equilibrium constants for weak bases, specifically referencing the Kb of aniline and how to derive it from pH calculations.

Areas of Agreement / Disagreement

There is no clear consensus on the methods for solving the problems presented, as participants express varying levels of understanding and seek clarification on different aspects of the calculations.

Contextual Notes

Some assumptions regarding solution volumes and the behavior of weak bases are mentioned, but not all details are fully resolved, leaving room for interpretation and further discussion.

Who May Find This Useful

Students preparing for chemistry tests, particularly those focusing on acid-base chemistry and pH calculations.

justinkoko
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1. Two HCl(aq) solutions have pH= 2.00 and pH= 3.00, respectively. If equal volumes of these two solutions are combined, the pH of the resulting solution wil be ...

Now I know the answer is 2.26 but I don't know how to actually solve it..

2.Aniline(C6H5NH2) is a weak base. At 25C, a 0.100M aqueous solution has a pH of 8.80. The Kb of aniline is..

Now I know the answer is 4.0x10^-10 but I don't know how to actually solve it..

3. 60.0ml of .100 M KOH is mixed with 40.0ml of .250M HCl at 25C. Assuming the volumes are additive, what is the pH of the resulting solution?

1.4 but I don't know how to solve it..

Please tell me how to do those..I have a chem test tomorrow :bugeye:

TIA..
 
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Calculate the pH of .400M Sr(OH)2

-log(.800M) = .09691

pH= 14-pOH= 13.9


Why is it .800 instead os .400M?

Is it because there are TWO OH s?
 
justinkoko said:
Calculate the pH of .400M Sr(OH)2

-log(.800M) = .09691

pH= 14-pOH= 13.9


Why is it .800 instead os .400M?

Is it because there are TWO OH s?

Nevermind this. I figured it out.

I still need help on the problems in the first thread.
 
1. Assume you are mixing 1L of each solution. Calculate number of moles of HCl in each solution - if you add these numbers you will know how much acid is present. You also know final solution volume. That gives concentration, pH calculation should be easy from this moment on ;)

2. Aniline reacts with water to produce OH-. Equlibrium of this reaction is described by Kb constant. Check out how you did pH calculation for weak base (or how it is described in your textbook). You must use exactly the same equation and the same simplifying assumptions, just solve for Kb, not H+ (or OH-).

3. Simple stoichiometry and limiting reagent question - reagent that is left after the neutralization will define pH. If there are exactly the same amounts of reagents autoionization of water will define pH. So check what was left, pH calculation will be just pressing log button on your calculator...

Chemical calculators at
 
Borek said:
1. Assume you are mixing 1L of each solution. Calculate number of moles of HCl in each solution - if you add these numbers you will know how much acid is present. You also know final solution volume. That gives concentration, pH calculation should be easy from this moment on ;)

2. Aniline reacts with water to produce OH-. Equlibrium of this reaction is described by Kb constant. Check out how you did pH calculation for weak base (or how it is described in your textbook). You must use exactly the same equation and the same simplifying assumptions, just solve for Kb, not H+ (or OH-).

3. Simple stoichiometry and limiting reagent question - reagent that is left after the neutralization will define pH. If there are exactly the same amounts of reagents autoionization of water will define pH. So check what was left, pH calculation will be just pressing log button on your calculator...


Borek
--

Genius!

Thanks for the help! :smile:
 

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