Question

1) Calculate pH for strong base solution: 13.0mL of 1.60

Answer 1

1 )     We have , 13 ml of 1.60 * 10^-2 Ca(OH)₂ is diluted to 440 ml

    i.e M1 = 1.60*10^-2 , V1 = 13ml

         M2 = ?    ,         V2 = 440ml

M1V1 = M2V2

13 * (1.60*10^-2 )= M2 * 440

                M2 = 13 * (1.60*10^-2 ) / 440

[OH^-] = M2   = 4.73 *10^-4 M

pOH = -log(OH^-) = -log ( 4.73 * 10^-4)

pOH = 3.3

We know that , pH + pOH = 14

                                  pH = 14- pOH

                                   pH = 14- 3.3

                                   pH = 10.7

2 ) 13 ml of 1.50 * 10^-2 M Ba(OH)₂ mixed with 49 ml of 5.9 * 10^-3M NaOH

          [OH^-] = { (13 * 1.5*10^-2) + (49 * 5.9*10^-3) }

                  =      0.195 + 0.289

          [OH^-] = 0.484M

3 )     FROM above calculation

             we have , [OH^-] = 0.484M

                             pOH = -log (OH^-)

                             pOH = -log (0.484) = 0.32

we know that , pH + pOH = 14

                               pH     = 14 - pOH

                              pH       = 14 - 0.32

                              pH        = 13.7