Question

What is the pH when 125 mL of 0.1 M NaOH is added to 175 mL of 0.2 M CH3COOH (acetic acid) if pKa for acetic acid = 4.76?

Answer 1

CH3COOH + NaOH --------> CH3COONa + H2O

no of millimoles of NaOH = molarity * volume in ml

                                    = 0.1*125 = 12.5millimoles

no of moles of CH3COOH = 0.2*175 = 35 millimoles

                  CH3COOH + NaOH --------> CH3COONa + H2O

initial moles    35                  12.5                 0

Change moles -12.5               -12.5                +12.5

at equilibrium     22.5                0                     12.5

concentration of CH3COONa = no of millimoles/total volume = 12.5/300

concentration of CH3COOH   = no of millimoles/ total volume   = 22.5/300

P^H     = P^Ka + log[salt]/[acid]

         = 4.76 + log 12.5/300/22.5/300

         = 4.76 + log12.5/22.5

         = 4.76-0.2552

         = 4.5048 >>>> answer