A few chemistry

I got the rest of the homework, but these are confusing me so much...ugghhhh.

1. The pH of a .400 M solution of iodic acid, HlO3, is .726 at 25 degrees C. What is the Ka(acid constant) at this temperature?

2. The pH of a .150 M solution of HClO is found to be 4.55 at 25 degrees C. What is KAfor HClO at this temp?

3. CH3CH2CH2NH2 is a weak base. At equilibrium, .039 M solution of it has an OH- concentration of 3.74 x 10^-3 M. What is the pH of this solution and Kb(base constant) for this this weak base?

1.
HIO3 ==> H^+ + IO3^-

pH = 0.726. Use pH = -log(H^+) to calculate (H^+).
Then set up Ka expression.
(H^+) = (IO3^-) and (HIO3) = 0.4-(H^+).
Plug into Ka expression and solve for Ka.

2. Same process for #2.

3. Same process but this is Kb instead of Ka.
Post your work if you get stuck or want to check your answers.

Ok, so on #1, I did the antilog, and got 5.32, so I plugged it in.
5.32=IO3^-
HIO3=.4-5.32, =-4.92

But now I don't get what to do...

No, No. How in the world did you get 5.32?
pH = -log(H^+)
0.726 = -log(H^+)
-0.726 = log(H^+)
antilog -0.726 = (H^+) = 0.1879
Do this on your calculator.
Punch in 0.726, change sign to - (or punch in -0.726 initially) and punch the button that says 10x. That should give you 0.1879.
Let me know if you still are confused.

Woops, that wasn't what I was confused about. That was just my stupidity, forgetting to change the sign, first. I just don't really know where to go after I have all of these things plugged in and such..

Ka = (H^+)(IO3^-)/(HIO3)
Ka = (0.1879)(0.1879)/(0.4-0.1879) = ??

Ka = 0.166

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asked by Trish
  1. 2. The pH of a .150 M solution of HClO is found to be 4.55 at 25 degrees C. What is KA for HClO at this temp?
    The answer is Ka=5.0*10^-9

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    posted by Ryann
  2. why did you subtract .1879 from .4? on this part:
    Ka = (0.1879)(0.1879)/(0.4-0.1879) = ??

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    posted by theresa

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