This question has multiple parts Work all the parts to get t

This question has multiple parts. Work all the parts to get the most points. Calculate the pH of a solution that is 0.0190 M in HPO4 (Kal =: 7.11x 10-3) pH = 2.08 Correct Correct O HpC2O4 (Kal = 5.60 × 10-2). pH = H3PO3 (Kal = 3 × 10-2). pH =

Solution

part B

         H2C2O4 + H2O <----> HC2O4^-(aq) + H3O+(aq)

initial 0.019                   0          0

change    -x                    +x          +x

equil   0.019-x                  x          x

Ka1 = [H3O+][HC2O4^-]/[H2C2O4]

5.6*10^-2 = (x^2/(0.019-x))

x = 0.015

[H3O+] = X = 0.015 M

pH = -log(H3O+) = -log(0.015)

   = 1.824

Part C

         H3po3 + H2O <----> H2po3-(aq) + H3O+(aq)

initial 0.019                   0          0

change    -x                    +x          +x

equil   0.019-x                  x          x

Ka1 = [H3O+][H2po3^-]/[H3po3]

3*10^-2 = (x^2/(0.019-x))

x = 0.0132

[H3O+] = X = 0.0132 M

pH = -log(H3O+) = -log(0.0132)

   = 1.88

part D

         H2so3 + H2O <----> Hso3-(aq) + H3O+(aq)

initial 0.019                   0          0

change    -x                    +x          +x

equil   0.019-x                  x          x

Ka1 = [H3O+][Hso3^-]/[H2so3]

1.23*10^-2 = (x^2/(0.019-x))

x = 0.0103

[H3O+] = X = 0.0103 M

pH = -log(H3O+) = -log(0.0103)

   = 1.99

part e

         H2s + H2O <----> Hs-(aq) + H3O+(aq)

initial 0.019                   0          0

change    -x                    +x          +x

equil   0.019-x                  x          x

Ka1 = [H3O+][Hs^-]/[H2s]

9.6*10^-8 = (x^2/(0.019-x))

x = 0.0103

[H3O+] = X = 4.266*10^-5 M

pH = -log(H3O+) = -log(4.266*10^-5)

   = 4.37