The acid ionization constant for Ga(H2O)63+ (aq) is 2.5×10-3 . Calculate the pH of a 0.1500...

The acid ionization constant for Ga(H2O)63+ (aq) is 2.5×10-3 . Calculate the pH of a 0.1500 M solution of Ga(NO3)3 .

Expert Solution

Ka for Ga(H2O)63+ (aq) = 2.5×10^-3 .

Ga(H2O)63+ (aq) ---------------> Ga(H2O)5.(OH)2+ + H+

0.1500 0 0

0.1500 - x x x

Ka = x^2 / 0.15 - x

2.5 x 10^-3 = x^2 / 0.15 - x

x = 0.0182

[H+] = 0.0182 M

pH = -log (0.0182)

pH = 1.74

queen_honey_blossom answered 1 year ago

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