Question

What is the pH of a solution in which 43 mL of 0.16 M NaOH is added to 16 mL of 0.140 HCl M ?

pH =

Answer 1

HCl + NaOH ----> NaCl + H2O

moles of HCl = M*Volume

                     = 0.140*0.016

                     = 0.00224 moles

moles of NaOH = M*Volume

                         = 0.16 *0.043

                         = 0.00688 moles

Moles of NaOH is more than that of HCl.

moles of NaOH left = 0.00688 - 0.00224

                               = 0.00464

pOH = -log(0.00464)

        = 2.33

pH = 14 - 2.33

     = 11.67