The pH of a 1 .00 x 10-2 M solution of cyanic acid(HOCN) is 2.77 at 25 degrees. Calculate Ka and pKa for HOCN from this result.
...........HOCN ==> H^+ + OCN^-
I..........0.01.....0.......0
C..........-x.......x.......x
E........0.01-x.....x.......x
The problem tells you pH = 2.77; therefore, 2.77 = -log(H^+); therefore, (H^+) = 0.00170 = x in he above./
Evaluate H^+< OCN^-, and HOCN and calculate Ka.
pKa = -log K.
To calculate Ka and pKa for HOCN using the given pH value, we need to use the acid dissociation equation for HOCN:
HOCN ⇌ H+ + OCN-
The Ka expression for this equilibrium is:
Ka = [H+][OCN-] / [HOCN]
The pH value can be related to the concentration of H+ using the equation:
pH = -log[H+]
Given that the pH of the solution is 2.77, we can convert it to [H+] concentration:
[H+] = 10^(-pH)
[H+] = 10^(-2.77)
Now, we can assume that at equilibrium, the concentration of HOCN is equal to the initial concentration (1.00 x 10^(-2) M), and the concentration of OCN- is equal to the concentration of H+:
[HOCN] = 1.00 x 10^(-2) M
[OCN-] = [H+] = 10^(-2.77) M
Substituting these concentrations into the Ka expression, we get:
Ka = (10^(-2.77))(10^(-2.77)) / (1.00 x 10^(-2))
Now we can calculate Ka:
Ka = 10^(-2.77 + -2.77 - (-2))
Finally, we can calculate pKa using the equation:
pKa = -log(Ka)
pKa = -log(10^(-2.77 + -2.77 - (-2)))
Therefore, pKa = -(-2.77 + 2.77 - (-2)) = 7.77
So, Ka = 10^(7.77) and pKa = 7.77.