Question

24.172 mL of a 2.08 M HCl (aq) solution is combinded with 79.479 mL of a 0.465 M solution of a weak base with a pKb = 4.94. What is the pH of the resulting solution?

Answer 1

no of moles HCl   = molarity * volume in L

                             = 2.08*0.024172 = 0.05027 moles

no of moles of weak base = molarity * volume in L

                                          = 0.465*0.079479 = 0.03695 moles

   weak base + HCl ------------------> salt

I              0.03695         0.05027

ramining no of moles of HCl = 0.05027-0.03695 = 0.01332 moles

[H+]   = no of moles of HCl/total volume

           = 0.01332/0.024172+0.079479   = 0.01332/0.103651 = 0.1285M

PH   = -log[H+]

       = -log0.1285   = 0.891 >>>>>answer