Question

A buffer that contains 0.592 M base, B, and 0.286 M of its conjugate acid, BH+, has a pH of 8.87. What is the pH after 0.0020 mol of HCl is added to 0.250 L of the solution?

Answer 1

B +                 H+ <---> BH+

0.592                                         0.286

Before HCl is added:
pH=8.87
pOH=14-pH
      = 14-8.87
      =5.13

pOH = pKb + log ( [BH+] / [B] )
5.13 = pKb+ log (0.286/0.592)
5.13 = pKb - 0.316
pKb = 5.446

After HCl is added:

B and H+ from HCl willcombine to form BH+

Total number moles of BH+ = initial number of moles of BH+ + number of moles of acid added

                                                  = M*V + 0.002

                                                  = 0.286*0.25 + 0.002

                                                  = 0.0735 mol

Concentration of BH+ = number of moles / volume = 0.0735/0.25 = 0.294 M

Total number moles of B = initial number of moles of B - number of moles of acid added

                                                  = M*V - 0.002

                                                  = 0.592*0.25 - 0.002

                                                  = 0.146 mol

Concentration of B= number of moles / volume = 0.146/0.25 = 0.584 M

Use:

pOH = pKb + log ( [BH+] / [B] )
= 5.446+ log (0.294 /0.584)
= 5.446 - 0.298
= 5.148

so, pH = 14 -pOH = 14 - 5.148 = 8.85

Answer: 8.85