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

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

why did you subtract .1879 from .4? on this part:

Ka = (0.1879)(0.1879)/(0.4-0.1879) = ??

Great job! The Ka value for the iodic acid solution is 0.166.

Now let's move on to the second problem.

For problem #2, we have:

HClO(aq) ⇌ H+(aq) + ClO-(aq)

We are given that the pH of a 0.150 M solution of HClO is 4.55. We can use the pH to calculate the H+ concentration.

pH = -log(H+)
4.55 = -log(H+)
-4.55 = log(H+)

Now, use the antilog to find the H+ concentration:

[H+] = 10^-4.55

Next, we can set up the Ka expression:

Ka = (H+)(ClO-)/(HClO)

Plug in the values and calculate Ka.

Ka = (H+)(ClO-)/(HClO)
Ka = (10^-4.55)(10^-4.55)/(0.150 - 10^-4.55)

Solve this expression using your calculator to obtain the value for Ka.

Once you've found the solution, let me know if you need any further assistance.

To find the Ka value, you need to use the formula:

Ka = (H^+)(IO3^-)/(HIO3)

You have already determined the values for (H^+), (IO3^-), and (HIO3). Plug in those values into the equation and calculate.

Given that (H^+) = 0.1879, (IO3^-) = 0.1879, and (HIO3) = 0.4 - 0.1879, you can substitute these values and solve for Ka.

Ka = (0.1879)(0.1879)/(0.4-0.1879)

Evaluate this expression to find Ka.

Ka ≈ 0.166

The value of Ka at this temperature is approximately 0.166.