Question

Please answer both questions 1&2 and circle the answers

1. + 0/0.1 points Previous Answers 1/4 Submissions Used Calculate the molar solubility of Fe(OH)2 in a buffer solution where the pH has been fixed at the indicated values. Kan = 7.9 x 10°. (8) pH 70 1.99e-9 XM (b) pH 10.8 . Хм (c) pH 12.1 Хм HopHelpCh17N6 2. + -10.1 points 0/4 Submissions Used Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 8.00 X 102 mL of solution and then titrate the solution with 0.168 M NaOH. C6H5CO2H(aq) + OH" (aq) + C6H5CO2 (aq) + H2O(l) What are the concentrations of the following ions at the equivalence point? Na+, H307, OH C6H5CO2 M Nat MH30+ MOH MC6H5CO2 What is the pH of the solution?

Answer 1

Fe(OH)2(s) --------> Fe^2+ (aq) + 2OH^-(aq)

                                 s                     2s

Ksp   = [Fe^2+][OH^-]^2

7.9*10^-16   = s*(2s)^2

s   = 5.82*10^-6

The solubility of Fe(OH)2 = 5.82*10^-6M

[OH^-]   = 2s = 2*5.82*10^-6   = 1.164*10^-5M

POH = -log[OH^-]

          = -log(1.164*10^-5)

         = 4.934

PH   = 14-POH

         = 14-4.934

         = 9.066

2.

no of moles of C6H5COOH   = W/G.M.Wt

                                             = 0.235/122.12

                                               = 0.00192moles

molarity of C6H5COOH = no of moles/volume in L

                                       = 0.00192/0.8 = 0.0024M

------- C6H5COOH(aq) + NaOH(aq) -------------> C6H5COONa(aq)    + H2O(l)

the exces molarity of NaOH = 0.168-0.0024   = 0.1656M

NaOH(aq) -------------> Na^+ (aq) + OH^-(aq)

0.1656M                    0.1656M        0.1656M

[Na^+]   = 0.1656M

[OH^-] = 0.1656M

[H3O^+] = Kw/[OH^-]

              = 1*10^-14/(0.1656)

               = 6.04*10^-14M

PH = -log[H3O^+]

        = -log(6.04*10^-14)

        = 13.22