Calculate the percent ionization of cyanic acid, Ka=2.0x10^-4, in a buffer soln that is .50M HCNO and .10M NaCNO.
Answer is .20%.
So,I know that NaCNO-->Na+ + CNO-
so [CNO-]=.10M from ICF chart.
and then HCNO <=>H+ + CNO-
where: [HCNO]= .5
and [CNO-]=.1
With an ice table , i keep on getting x=.001 which means the percent ionization would be 1% instead of .2%
Help!
2 answers
The ICE chart is not applicable in this case (actually you MAY use it but it's simpler to go another way). Use the Henderson-Hasselbalch equation, since this is a buffered solution, to solve for pH of the solution and convert that to H^+, then solve the percent ionization. I worked the problem and I did obtain 0.2%.
1. Find H+
=Ka(HA/A-)
= 2.0 x 10^-4 x (0.50/0.10)
H+ = 0.001
2. Find percent ionization
= H+/conc. of weak acid(HCNO)
= 0.001/0.50
= 0.002 x 100
= 0.2%
=Ka(HA/A-)
= 2.0 x 10^-4 x (0.50/0.10)
H+ = 0.001
2. Find percent ionization
= H+/conc. of weak acid(HCNO)
= 0.001/0.50
= 0.002 x 100
= 0.2%
5 answers