Question

3) A solution is prepared by adding 40.0 mL of 0.060 M NaOH and 40.0 mL of 0050 M Sr(OH) a) Calculate the concentrations ofall ions in solution. b) Calculate the pOH of this solution.

Answer 1

NaOH

M1   = 0.06M                  M2 =

V1   = 40ml                     V2    = 40 + 40 = 80ml

M1V1 = M2V2

      M2    = M1V1/V2

             = 0.06*40/80    = 0.03M

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

0.03M                        0.03M         0.03M

[Na^+]   = 0.03M

[OH^-]   = 0.03M

Sr(OH)2

M1 = 0.05M                             M2 =

V1    = 40ml                               V2 = 40+40 = 80ml

           M1V1     =   M2V2

               M2      = M1V1/V2

                           = 0.05*40/80    = 0.025M

   Sr(OH)2 (aq) -----------------> Sr^2+ (aq) + 2OH^-(aq)

   0.025M                            0.025M        2*0.025M

    [Sr^2+]      = 0.025M

    [OH^-]        = 2*0.025M     = 0.05M

NaOH                                                           Sr(OH)2

M1   = 0.06M                                                 M2   = 0.05M

V1    = 40ml                                                    V2 = 40ml

     M         = M1V1   + M2V2/V1 + V2

                 = 0.06*40 + 0.05*40/40+40

                = 4.4/80   = 0.055M

POH   = -log[OH^-]

           = -log0.055

           = 1.2596