To find the pH of a 4.5 x 10^-3 M HI solution, we need to know the dissociation constant of HI (also known as the acid ionization constant, Ka). The Ka of HI is 1.3 x 10^-10.
HI ⇌ H+ + I-
At equilibrium, let x be the concentration of H+ ions, then [H+] = x and [I-] = x. The initial concentration of HI is 4.5 x 10^-3 M. Using the Ka expression, we can set up the following equation:
Ka = [H+][I-]/[HI]
1.3 x 10^-10 = (x)(x)/(4.5 x 10^-3)
Solving for x, we get:
x = 7.07 x 10^-6 M
Since pH = -log[H+], we can calculate the pH of the solution:
pH = -log(7.07 x 10^-6) = 5.15
Therefore, the pH of a 4.5 x 10^-3 M HI solution is 5.15.