Question

Find the pH of each of the following solutions

A. 0.125M in HBr and 0.135M in HCHO2

B. 0.155M in HNO2 and 9.0�10^-2M in HNO3

C. 0.180M in HCHO2 and 0.23M in HC2H3O2

D. 6.0�10^-2M in acetic acid and 6.0�10^-2M in hydrocyanic acid

E. 7.00�10^-2M HNO3

F. 0.160M HNO2

G. 2.10�10^-2M KOH

H. 0.250M CH3NH3I

I. 0.320M KC6H5O

Answer 1

1) 0.125M in HBr and 0.135M in HCHO2

HBr is a strong acid. It dissociates completely.

[H+]initial = 0.125 M = [H3O+]

HCHO2 (formic acid) is a weak acid.

                                 HCHO2 + H2O <---> H3O+ + CHO2-        Ka = 1.77*10-4

                      initial      0.135         -           0.125          -

                    change        -x                           +x          +x

                       end      0.135-x                   0.125+x      x

Ka = [H3O+][CHO2-]/[HCHO2]

(0.125+x)(x)/(0.135-x) = 1.77*10-4

- assume x is very very small, almost zero.

(0.125)(x)/0.135 = 1.77*10-4

x = 1.912*10-4

Add this x back to your original [H3O+]

0.125+1.912E-4 = 0.1251912 = 0.125

pH=-log[H+]

ph= -log(0.125)

pH = 0.903

2) 0.155M in HNO2 and 9.0