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

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To find the pH of mixed HCl solutions with pH 2.00 and 3.00, calculate the moles of HCl in each solution and determine the concentration in the final volume, leading to a pH of 2.26. For aniline, use the pH of the weak base solution to derive Kb, which is 4.0x10^-10. When mixing KOH and HCl, identify the limiting reagent to find the resulting pH, which is 1.4. The confusion regarding Sr(OH)2 arises from its dissociation producing two OH- ions, resulting in an effective concentration of 0.800M for pOH calculations. Understanding these principles will aid in solving similar chemistry problems effectively.
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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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