Question

Please use an ice table and lable which answer is Ka for NH4+ and pka NH4+

balanced, ionic equation for the reaction of ammonium ion [NH4+] with water.

NH4+(aq) + H2O(l) ⇄ NH3(aq) + H3O+(aq)

equilibrium constant expression, Ka, for aqueous NH4+ Ka = [NH3][H3O+]/[NH4+]

measured pH values for each solution (A-B) to calculate [H3O+] .

Make an ICE table for each solution to find EQUILIBRIUM concentrations.

Calculate Ka for NH4+ and pka NH4+

Solution A - pH = 4.56 NH3 M = 0 NH4+ = 0.10

Solution B - pH = 4.21 NH3 = 0 NH4+ = 1.0

Answer 1

A)

pH =4.56

so -log H+ = 4.56

        so H+ ion conc = 10^-4.56

Now H+ ion conc = (Ka*c)^1/2   = (Ka* 0.10)^1/2   = 10^-4.56

so Ka*0.10 = ( 10^-4.56)^2

Ka = ( 10^-4.56)^2 /0.10 = 7.586*10^-9

B)

  

pH =4.21

so -log H+ = 4.21

        so H+ ion conc = 10^-4.21

Now H+ ion conc = (Ka*c)^1/2   = (Ka* 1.0)^1/2   = 10^-4.21

so Ka*1 = ( 10^-4.21)^2

Ka = ( 10^-4.21)^2 /1.0= 3.802 *10^-9