You just need to set what you have equal to Kb for CN^- which isn't in any table but can be calculated this way.
Kb for CN^- = (Kw/Ka for HCN)
so (Kw/Ka for HCN) = (x)(x)/(0.8-x)
and solve for x.
what is the pH of 0.80 M NaCN?
what is the pH of 0.80 M NaCN? What is the concentration of HCN in the solution?
I did this so far:
I 0.8..... 0....0
C -x.......x....x
E 0.8-x....x....x
(x)(x)/0.8-x
How do I solve for x?
2 answers
what is the ph of 0.15 m NaCN
1 answer