Question

1.

a) A solution of 1 x 10^-8 M KOH (aq) is prepared. What is the pH of this solution?

b) A phthalate buffer is prepared in which [hydrogen phthalate]= 0.047 M and [phthalic acid] = 0.083 M. If the Ka1 of phthalic acid acid is 1.12 x 10^-3, and the Ka2 = 3.91 x 10^-6, what is the expected pH of the buffer?

c) The pH of a solution formed from a weak monoprotic acid (HA) is 3.1. If the formal concentration of the acid is 0.1M, what is the acids Ka value?

d) The pH of a saturated solution of a metal hydroxide (with formula M(OH)₂) is 9.85. What is the Ksp of M(OH)₂?

Answer 1

1*10^-8 M KOH

KOH is strong base PH value range is 10 to 14.

1*10^-8 M KOH is very diluted solution.

The conc of OH^- is equal to conc of OH^- from base and conc of OH^- from water.

[OH^-]   = 10^-8 + 10^-7

            = 0.1*10^-7 + 10^-7

           = (0.1+1) *10^-7

           = 1.1*10^-7 M

POH = -log[OH^-]

           = -log1.1*10^-7

           = 6.9586

PH    = 14-POH

         = 14-6.9586 = 7.0414

b. PKa1 = -logka1

               = -log1.12*10^-3 = 2.9507

   PH = Pka1 + log[hydrogen phthalate]/[phthalic acid]

         = 2.9507 + log0.047/0.083

       = 2.9507 -0.2469   = 2.7038

c. PH = 1/2Pka -1/2logc

   3.1   = 1/2Pka-1/2log0.1

3.1    = 1/2PKa -1/2*-1

3.1+ 1   = 1/2PKa

1/2PKa = 4.1

PKa   = 8.2

-logKa = 8.2

Ka       = 10^-8.2   = 6.3*10^-9

d. PH = 9.85

   POH = 14-PH

             = 14-9.85   = 4.15

-log[OH^-] = 4.15

[OH^-]    = 10^-4.15 = 7.079*10^-5M

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

                                           7.079*10^-5       2*7.079*10^-5

   Ksp = [M^+]{OH^-]^2

Ksp = 7.079*10^-5 *(14.158*10^-5)^2   = 1.42*10^-12