Problem 6D.13 asks:
Rank the following solutions in order of increasing pH.
a) 1.0 x 10^-5 M HCl
b) 0.20 M CH3NH3Cl (aq)
c) 0.20 M CH3COOH (aq)
d) 0.20 M C6H5NH2 (aq)
Can someone explain how we are supposed to find the pH for parts b-d? Or, if we do not need to solve for the specific pH, how can we tell which ones will have a higher or lower pH?
Textbook Problem 6D.13
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Re: Textbook Problem 6D.13
I actually got tripped up with this too. Turns out we're supposed to use tables 6c.1 and 6c.2 where they list out the Ka and Kb values for different acids and bases. From there we can solve for the pH for each molecule. Hope this helps!
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Re: Textbook Problem 6D.13
Hi! For these you need the tables from 6C.1 and 6C.2 to get the Ka and Kb values of the acids and bases, and then set up the ICE table and all.
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Re: Textbook Problem 6D.13
Amy Shimizu 1J wrote:Problem 6D.13 asks:
Rank the following solutions in order of increasing pH.
a) 1.0 x 10^-5 M HCl
b) 0.20 M CH3NH3Cl (aq)
c) 0.20 M CH3COOH (aq)
d) 0.20 M C6H5NH2 (aq)
Can someone explain how we are supposed to find the pH for parts b-d? Or, if we do not need to solve for the specific pH, how can we tell which ones will have a higher or lower pH?
Substance D, C6H5NH2 (aniline) is listed as a base under Table 6C.2. You can also tell it's a base because it's an amine, which are basic compounds. Amines have a nitrogen with a lone pair that can be donated. So, we don't have to calculate its pH, we already know it will have the greatest one.
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Re: Textbook Problem 6D.13
I think you would have to get the Ka or Kb values from a table and set up an ICE table in order to get the pH.
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