Question

Calculate the [H3O+] of the following polyprotic acid solution: 0.380 M H3PO4. Express your answer using two significant figures. [H3O+] =

Part B Calculate the pH of this solution. Express your answer using one decimal place. pH =

Part C Calculate the [H3O+] and pH of the following polyprotic acid solution: 0.330 M H2C2O4. Express your answer using two significant figures. [H3O+] =

Part D Calculate the pH of this solution. pH =

Answer 1

H3PO4 is triproic acid having three Ka values.

The ionization of H3PO4 is    H3PO4+ H2O<--------> H2PO4- + H3O+, Ka1= 7.5*10-3

Ka1= [H2PO4-[ [H3O+]/[H3PO4]

preparing the ICE table

component                                   initial                       change                 equilibrium

H3PO4                                          0.38                      -x                            0.38-x

H2PO4-                                        0                            x                                 x

H3O?+                                         0                             x                                x

ka1= x²/(0.38-x)= 7.5*10-3, when solved using excel, x= [H3O+]=0.04976   (1)

H2PO4- further undergoes ionization as H2PO4- + H2O ------>H3O+ + HPO4-2

Ka2= [H3O+] [HPO4-2]/ [H2PO4-] =6.2*10^-8

preparing the ICE Table again

component                                      initial                               change                    equilibrium

H2PO4-                                        0.04976                                 -x                              0.04976-x

HPO4-2                                             0                                        x                                      x

PO4-2                                               0                                         x                                     x

Ka2= x²/(0.04976-x)= 6.2*10^-8, when sovled using excel, [H3O+] =x=5.55*10^-5 (2)

the HPO4-2 formed undergoes further ionization as

HPO4-2 + H2O ---------->H3O++ PO4-3, Ka3= [H3O+] [PO4-3]/ [HPO4-2] = 1.7*10^-12

preparing ICE Table again

Component                                      iniitial                                 change                  equilibrium

HPO4-2                                            5.55*10-5                             -x                               5.55*10-5

PO4-3    0    x x

H3O+    0    x    x

Ka3= x²/(5.55*10^-5-x)= 1.7*10^-12 when solved using excel, x= [H3O+] =9.75*10^{-9                (3)}

total of H3O+ = 0.04976+5.55*10^-5 +9.75*10^-9=0.049816, pH= -log [H3O+]= 1.3

2. COOH-COOH is oxalic acid its Ka= 5.37*10-2

let HX= oxalic aicd.

HX + H2O --------->H3O+X-

Ka= [H3O+][ [X-]/[HX]

preparing the ICE TAble

component                                       initial                                  change                               equilibrium

HX                                                      0.33                                   -x                                           0.33-x

H3O+                                                   0                                       x                                               x

X-                                                        0                                       x                                                x

Ka= x²/(0.33-x)= 5.37*10^-2. when solved using excel, x= 0.109, pH= -log (0.109)= 0.96