The problem reads as follows:
Calculate the pH of each of the following solutions.
a.) 0.100
M propanoic acid (HC
3H
5O
2, K
a = 1.3 X 10
-5)
b.) 0.100
M sodium propanoate (NaC
3H
5O
2)
c.) Pure H
2O
d.) A mixture containing 0.100
M HC
3H
5O
2 and 0.100
M NaC
3H
5O
2I've gotten a & b already (the first was easy, just plug and chug basically, then part b was slightly more difficult, having to find the K
b of the reaction then going through the regular motions). However, I can't figure out how to get part C. I have the values here if anyone wants to use them:
For part a.) [H
+] = 1.13X10
-3 and the pH = 2.95
For part b.) [OH
-] = 8.77X10
-6 and the pH = 8.94
The only thing I could think of for part C was to take the two equations:
HC
3H
5O
2 <-----------> H
+ + C
3H
5O
2-C
3H
5O
2- <-----------> HC
3H
5O
2 + OH
-Then cancel out the like terms on each side to give you
H
2O <---------> H
+ + OH
-But I'm stuck from there
Any help would be appreciated!
Thanks,
Lucas