In: Chemistry
The acid ionization constant for Pb(H2O)62+(aq) is 6.3×10-7. Calculate the pH of a 0.0442 M solution of Pb(NO3)2. pH =
Pb(H2O)62+ --------------------------------> [ Pb(H2O)5 OH ]+ + H+
0.0442 0 0
-x +x +x
0.0442-x x x
Ka = x^2 / 0.0442-x = 6.3×10^-7
x^2 + 6.3×10^-7x - 2.785 x 10^-8 = 0
x = 1.67 x 10^-4
[H+] = 1.67 x 10^-4 M
pH = -log [H+]
pH = -log (1.67 x 10^-4)
pH = 3.78