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
OH equill. = 1x10^-7+x
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 =
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?