Question

Calculate the change in pH when 0.32 mol H+

is added to 1.00 L of each of the following buffer

(a) a 0.56 M solution of pyridine (py) containing 0.50 MpyH+

Number....

(b) a 0.58 M solution of aniline (an) containing 0.88 M anH+

Number......

Answer 1

a)                                     py + H^+    pyH^+

No.of moles of each species:
                                                        0.56moles    0.32moles            0.5moles    
After the reaction:no.of moles

remainig                                           0.24moles                  0                       0.82moles

Hence pH = pKa +log [base]/[acid]             Handerson's equation

pKa for pyridine = 4.6

[base] = 0.24moles/1.0 L              pyridine is the base

[acid] = 0.82 moles/ 1.0 L             pyH+ is the conjugate acid of pyridine

pH = 4.6 + log [0.24moles/1.0 L]/[0.82 moles/ 1.0 L]

   = 4.06

b)                                        an + H^+    pyH^+

No.of moles of each species:
                                                        0.58moles    0.32moles            0.88moles    
After the reaction:no.of moles

remainig                                           0.26moles                  0                       1.20moles

Hence pH = pKa +log [base]/[acid]             Handerson's equation

pKa for aniline = 5.23

[base] = 0.26moles/1.0 L              aniline is the base

[acid] = 1.20 moles/ 1.0 L             anH+ is the conjugate acid of aniline

pH = 5.23 + log [0.26moles/1.0 L]/[1.20 moles/ 1.0 L]

   =4.57