Asked by Mandy
The Ksp of calcium hydroxide, Ca(OH)2, is 4.68 x 10-6 at 289.15K. Calculate the pH of a saturated solution.
Here's my work:
4.68 x 10-6= 4s^3
s= 0.0105
-log(0.0105)= 1.98
14-1.98=12.02
The answer is supposed to be 12.324, but I'm not sure what I'm doing wrong. Is it because it is not at 25C? Please help!
Here's my work:
4.68 x 10-6= 4s^3
s= 0.0105
-log(0.0105)= 1.98
14-1.98=12.02
The answer is supposed to be 12.324, but I'm not sure what I'm doing wrong. Is it because it is not at 25C? Please help!
Answers
Answered by
DrBob222
No, not the T. It could be EXCEPT Ksp is given at the T in the problem so that is ok. The problem here, I think, is that you failed to write the equation and because of that you slipped up on a tiny bit of the problem. Here is the dope.
.......Ca(OH)2 --> Ca^2+ + 2OH^-
I......solid........0........0
C......solid........s........2s
E......solid........s........2s
So you are exactly right and your work is great and s is your answer of 0.0105 M to the digit. BUT (OH^-) is 2s so if you will take the -log(2s) you will get the right answer for OH^- and convert that to pH.
Thanks for showing your work. It really helps us figure what's wrong.
.......Ca(OH)2 --> Ca^2+ + 2OH^-
I......solid........0........0
C......solid........s........2s
E......solid........s........2s
So you are exactly right and your work is great and s is your answer of 0.0105 M to the digit. BUT (OH^-) is 2s so if you will take the -log(2s) you will get the right answer for OH^- and convert that to pH.
Thanks for showing your work. It really helps us figure what's wrong.
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