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The acid ionization constant for Fe(H2O)62+(aq) is 3.0×10-6. Calculate the pH of a 0.0248 M solution...

The acid ionization constant for Fe(H2O)62+(aq) is 3.0×10-6. Calculate the pH of a 0.0248 M solution of Fe(NO3)2.

pH =

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Answer #1

Fe(NO3)2 salt of weakbase,strong acid.

kb of weak base = kw/ka = 10^-14/(3*10^-6) = 3.33*10^-9

pkb = -log(3.33*10^-9) = 8.47

pH of solution of Fe(NO3)2 = 7-1/2(pkb+logC)

                           = 7-1/2(8.47+log0.0248)

                           = 3.57

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