5.- The equilibrium constant, Ka, for the reaction below is 6.0*10^-3. Fe(H2O)6

Question

5.- The equilibrium constant, Ka, for the reaction below is 6.0*10^-3.

Fe(H2O)6 ^3+ (aq) + H2O(l) <--> Fe(H2O)5 (OH)^2+(aq) + H3O^+ (aq)

a)Calculate the pH of the 0.10M solution of Fe(H2O)6^3+.

b)Will a 1.0 M solution of iron (II) nitrate have a higher or lower pH than a 1.0 M solution of iron (III) nitrate?

6.- Calculate the pH of the following solutions:

a)2.0 M morphine hydrochloride, C17H19NO3 ?HCl(pKb=6.13)

b)0.50M AlCl3 (Ka of [Al(H2O)6] 3+ is 1.4 *10^-5)

7.-A 0.20M sodium shlorobenzoate (NaC7H4ClO2) solution has a pH of 8.65. Calculate the pH of a 0.20M Chlorobenzoic acid (HC7H4ClO2) solution.

Explanation / Answer

6.0 x 10^-3 = x^2 / 0.178-x
x = [H3O+]=0.033 M
pH = 1.5

sodium chlorobenzoate is a basic salt
[OH-]^2= Kb*Cb
[OH-] = 4.47*10^-6 M (at pH=8.65)
Kb= [OH-]^2/Cb = 9.97*10^-11
Ka of conjugate acid is Kw/Kb= 1.00*10^-4
so
[H+] = square root(Ka*Ca) =
= square root(1.00*10^-4 * 0.29) = 5.39*10^-3 M
pH = 2.27

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