Calulate the percent ionization of cyanic acid Ka=2.0x10^-4 in a buffer soln. that is .50M HCNO and .10M NaCNO

Answer 1

Let x = moles/L HCNO that dissociate

[ HCNO ] = 0.50 -x

[ CNO- ] = 0.10 + x

[ H+ ] = x

for the equilibrium

HCNO <> H+ + CNO-

2.0 x 10^-4 = ( x ) (0.10 + x ) / 0.50 - x

x = 0.001 M

0.001 : 0.50 = x : 100

x = 0.2 %

Answer 2

O.K First, establish the chemical equilibrium HCNO <---> H+ + CNO- Then, write the concentrations. Since HCNO is a weak acid, we are yet to find the final concentrations of the H+ and CNO- ions. NaCNO is an ionic compound, so it dissociates completely into Na+ and CNO- ions. Therefore, the equilibrium (with concentrations at the bottom of the compounds) is: HCNO <--> H+ + CNO- (Since the Ka value is very small, we Initial 0.76M 0 0.4M will approximate 0.4 + x into x) Final 0.76-x x 0.4 + x Ka = [H+] [CNO-]/[HCNO] 2.0 * 10^-4 = [x][0.4]/[0.76+x] 1.52 * 10^-4 + 2.0 * 10^-4x = 0.4x Further approximation leads to 1.52 * 10^-4 = 0.4x x= 3.8 * 10^-4 M % dissociation = final concentration/initial concentration * 100% = 3.8 * 10^-4M/0.76M = 0.0005% I dunno, that's the answer I got. Maybe, I made a calculation error somewhere. So, if I were you, I would choose C) as the answer.

Answer 3

The formula would be:

(if HA acid)
% ionization = 100 / 1 + anti-log (pKa-pH)

or

(if BH acid)
% ionization = 100 / 1 + anti-log (pKa-pH)

to get the pKa, just multiply Ka with -log
to get the pH, since this is a buffer sol'n, then use the Henderson-Hasselbalch equation. (I can't type it well, so try googling it or something)

I hope this helps

Answer 4

good luck