Calculate the ph of the resulting solution if 19.0 ml of 0.190 m hcl(aq) is added to

a) when adding to 24 mL of 0.19 M NaOH:

first, we need to get moles of HCl = molarity * volume 

                                                        = 0.19 M * 0.019 L 

                                                        = 0.00361 moles

then, moles of NaOH = molarity * volume

                                    = 0.19 M * 0.024 L

                                    = 0.00456 moles

NaOH remaining = 0.00456 - 0.00361

                             = 0.00095 moles

when the total volume = 24mL+19 mL = 43 mL

∴ molarity of NaOH = moles / total volume 

                                 = 0.00095 / 0.043L

                                 = 0.0221 M

when POH = -㏒[OH-]

                   = -㏒0.0221 M

                   = 1.66

∴PH = 14 - POH

        = 14- 1.66

        = 12.34

b) when adding to 29 mL of 0.24 M NaOH:

when moles of HCl = 0.00361 moles 

then, moles of NaOH = molarity * volume

                                    = 0.24 M * 0.029 L

                                   = 0.00696 moles

∴ NaOH remaining = 0.00696 - 0.00361

                                 = 0.00335 moles

the total volume = 29 mL + 19 mL = 48 mL

molarity of NaOH = moles / total volume

                             = 0.00335 / 0.048L

                            = 0.0698 M

∴POH = -㏒[OH-]

           = -㏒ 0.0698

           = 1.16

∴ PH = 14- POH 

         = 14- 1.16

         = 12.84