Question

Calculate the pH of a solution in which exactly 65 ml of 0.130 M NaOH(aq) is mixed with exactly 21 ml of 0.120 M HCl(aq). Enter the pH with 2 decimal places.

NaOH(aq)   +   HCl(aq)    →    H₂O   +   NaCl(aq)

Answer 1

NaOH(aq)   +   HCl(aq)    →    H₂O   +   NaCl(aq)

HCl                                                                             NaOH

MA   = 0.12M                                                            MB   = 0.13M

VA    = 21ml                                                                VB = 65ml

                      M    = MBVB-MAVA/VA+VB                     [MBVB    >   MAVA ]

                              = 0.13*65-0.12*21/21+65

                              = 5.93/86   = 0.069M

              M    =   [OH^-]   = 0.069M

            POH   = -log[OH^-]

                        = -log0.069   = 1.1611

        PH   = 14-POH

                = 14-1.1611    = 12.84>>>>>answer