Posted by Lena on Friday, July 24, 2009 at 1:07am.
Calculate the pH of 0.20 M NaCN solution.
NaCN ---> Na+ + CN-
CN- + H20+ ---> HCN+ + OH-
Initial conc. of CN- = 0.20 mol/L
change = -x
equillibrium = 0.20-x
HCN equill. = +x OK to here.
OH equill. = 1x10^-7+x This is also +x.
I will let you redo the math to get pH but I solve for pH and obtained 11.25.
Ka= 6.2 x 10^-10
Kb = KW/Ka = 1x10^-14/6n2 x 10^-10 = 1.6 x 10^-5
1.6 x 10^-5 = [x][1x10^-7+x]/0.2
QUADRATIC FORMULA:
x= 1.8 x 10^-3
pOH = -log(1.8x10^-3)
= 2.73
Ph = 14 - POH
=11.26
DID I DO THIS CORRECTLY?
How would I determine whether or not the CN- was an acid or a base. I was thinking this. The HCN is an acid so when you take the H off of it it becomes the conjugate base. Is this also correct? Yes, HCN is an acid, you remove the H and the CN^- becomes the conjugate base (because it will accept a proton).
I redid it and I got 11.25 :) Thanks :)
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