The acid ionization constant for Pb(H2O)62+(aq) is 6.3×10-7. Calculate the pH of a 0.0442 M solution...

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 =

Expert Solution

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

queen_honey_blossom answered 1 year ago

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