To find Ka and pKa for HOCN, we need to use the pH and concentration information provided.
The pH of a solution is related to the concentration of H+ ions present. The relationship between pH and concentration can be described by the equation:
pH = -log[H+]
In this case, we are given the pH value, which is 2.77. To find the concentration of H+ ions, we need to take the antilog of the negative pH value:
[H+] = 10^(-pH)
[H+] = 10^(-2.77)
[H+] = 1.995 x 10^(-3) M
Since HOCN is a weak acid, it will dissociate in water to form H+ and OCN- ions. The dissociation equation is:
HOCN โ H+ + OCN-
Let's define the initial concentration of HOCN as [HOCN]0 and the change in concentration of H+ and OCN- as x. At equilibrium, the concentrations will be:
[HOCN] = [HOCN]0 - x
[H+] = x
[OCN-] = x
Now, we can write the equilibrium constant expression for the dissociation of HOCN:
Ka = [H+][OCN-] / [HOCN]
Since the initial concentration of HOCN is given as 1.00 x 10^(-2) M, we substitute the known values:
Ka = (x)(x) / (1.00 x 10^(-2) - x)
Since the concentration of H+ is equal to x, we can simplify the expression:
Ka = x^2 / (1.00 x 10^(-2) - x)
Typically, in weak acid/base calculations, we assume that the "x" value is small compared to the initial concentration of the weak acid/base. Therefore, we can approximate (1.00 x 10^(-2) - x) as 1.00 x 10^(-2):
Ka โ x^2 / (1.00 x 10^(-2))
Now, we substitute the value of [H+] (1.995 x 10^(-3) M) into the expression to find Ka:
Ka โ (1.995 x 10^(-3))^2 / (1.00 x 10^(-2))
Ka โ 3.98 x 10^(-6)
To find pKa, we can take the negative logarithm of Ka:
pKa = -log(Ka)
pKa โ -log(3.98 x 10^(-6))
pKa โ 5.40
Therefore, the values of Ka and pKa for HOCN are approximately 3.98 x 10^(-6) and 5.40, respectively.