What is the Ka and a degree of ionization of HCl when the pH

What is the Ka and a (degree of ionization) of HCl when the pHs/concentrations are as follows:

Solution

to find Ka and a we have the formula,
pH = 1/2(pka - logC)

pH = 1

C = concentration = 1 M

1 = 1/2(pKa - log1)

pKa = 2

-log(Ka) = 2

Ka = 10^-2

and a = square root of (Ka/C)

a = square root of (10^-2/1)

a = 0.1

pH = 1.4

C = 0.1

pH = 1/2(pKa - logC)

1.4 = 1/2(pKa - log(0.1))

pKa = 1.8

Ka = 10^-1.8 = 0.01585

a = square root of (0.01585/0.1)

a = 0.3981

pH = 2.2

C = 0.01

pH = 1/2(pKa - logC)

2.2 = 1/2(pKa - log(0.02))

pKa = 2.701

Ka = 10^-2.701 = 1.991 * 10^-3

a = square root of (1.991 * 10^-3/0.01)

a = 0.4462