0.6nm = 6A

Forza ionica = I = 1/2*(0.0078 + 0.078) = 0.0078M

-logf(Ca++) = (0.51*2^2*(0.0078)^1/2) / (1 + 0.33*6*(0.0078)^1/2) = 0.153

f(Ca++) = 10^-0.153 = 0.703

a(Ca++) = [Ca++]*0.703