Question

this is part 1. What am i doing wrong? Attached my previous answers

Part B Calculate the molar solubility of Ni(OH), when buffered at pH = 10.3. Express your answer using one significant figure. O A 0 2 ? 1.0 • 10-8 Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining

part 2

Answer 1

PH   = 10.3

POH   = 14-POH

          = 14-10.3

          = 3.7

POH   = -log[OH^-]

-log[OH^-]   = 3.7

    [OH^-]    = 2*10^-4M

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

                                        1*10^-4       2*10^-4

       Ksp    = [Ni^2+][OH^-]^2

                  = 1*10^-4 *(2*10^-4)^2

                  = 4*10^-12

part2

PH   = 11.9

POH   = 14-POH

          = 14-11.9

          = 2.1

POH   = -log[OH^-]

-log[OH^-]   = 2.1

    [OH^-]    = 8*10^-3M

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

                                        4*10^-3       8*10^-3

       Ksp    = [Ni^2+][OH^-]^2

                  = 4*10^-3 *(8*10^-3)^2

                  = 3*10^-7