Question

Calculate the pH of a mixture that contains 0.13 M of HNO3 and 0.20 M of HC6H5O.

Answer 1

Answer – We are given, [HNO₃] = 0.13 M , [HC₆H₅O] = 0.20 M

We know HNO₃ is strong acid and HC₆H₅O is weak acid,

So, [HNO₃] = [H⁺] = 0.13 M

Now we need to calculate the H⁺ in the HC₆H₅O

We know, Ka value for the HC₆H₅O = 1.6*10^-10

Now we need to put ICE chart

    HC₆H₅O + H₂O ------> H₃O⁺ + C₆H₅O^-

I    0.20                                0             0

C     -x                                  +x          +x

E 0.20-x                            +x           +x

Ka = [H₃O⁺][ C₆H₅O^-] / [HC₆H₅O]

1.6*10^-10 = x*x /(0.20-x)

We can neglect the x in the 0.20-x , since Ka value is too small

1.6*10^-10 *0.20 = x²

So, x = 5.66*10^-6

x = [H₃O⁺] = 5.66*10^-6 M

So, total [H₃O⁺] = 0.13 M + 5.66*10^-6 M

                          = 0.13 M

So, pH = -log [H₃O⁺]

            = - log 0.1300

             = 0.89