Question

The acid ionization constant for Fe(H2O)62+(aq) is 3.0×10-6. Calculate the pH of a 0.0394 M solution of Fe(NO3)2.

Answer 1

Solution :-

Ka= 3.6*10^-6

Using the given ka value we can find the equilibrium concentration of the hydronium ion

Fe(H2O)6^2+ + H2O ---- > [Fe(H2O)5 OH^-]^+ + H3O^+

0.0394 M                                                        0                                          0

-x                                                                       +x                                        +x

0.0394 M-x                                                       x                                            x

Ka=[[ Fe(H2O)5 OH^-]^+][H3O+] /[ Fe(H2O)6^2+]

3.6*10^-6 = [x][x]/[0.0394-x]

Ka is very small therefore we can neglect the x from denominator

3.6*10^-6 = [x][x]/[0.0394]

3.6*10^-6 * 0.0394 = x^2

1.42*10^-7 = x^2

Taking square root of both sides we get

3.77*10^-4 M= x =[H3O+]

Now lets calculate the pH

pH= -log [H3O^+]

pH= -log[3.77*10^-4]

pH= 3.42

Therefore the pH is 3.42