Question

For a weak acid (HA), a 0.100 M solution was made. Students obtained the following pH value for the solution. Fill in the table.

0.100 M HA pH 5.03

[H3O+]=

[A-]=

[HA] (at equilibrium)=

Ka=

For the same solution fill out the following information:

0.100 M HA

pOH (Do not use scientific notation for this answer)=

[OH-]=

Kb=

Answer 1

PH   = 5.03

-log[H^+] = 5.03

   [H^+]   = 9.33*10^-6M

at equilibrium [A^-]   = [H^+]   = 9.33*10^-6M

[HA] = 0.1-9.33810^-6   = 0.1M

HA(aq) -------------> H^+ (aq) + A^- (aq)

Ka   = [H^+][A^-]/[HA]

        = 9.33*10^-6*9.33*10^-6/0.1   = 8.7*10^-10

POH = 14-PH

           = 14-5.03

           = 8.97

-log[OH^-]   = 8.97

   [OH^-]   = 1.07*10^-9M

Kb   = Kw/Ka

          = 1*10^-14/(8.7*10^-10)

           = 1.15*10^-5 >>>>>answer