Question

A 25.00 mL sample of 0.280 M NaOH analyte was titrated with 0.750 M HBr at 25 °C. Calculate the initial pH before any titrant was added. Calculate the pH after the 5.00ml titrant was added.

Answer 1

1) no. of moles of [OH-] = 0.280 M

Ph = 14 - poh

Ph = 14 - {-log[0.280]}

Ph = 14 - 0.55

Ph = 13.45

..........................................................................

no. of moles remaining = no.of moles of NaOH - no.of moles of HBr

                                 = (molarity of NaOH* vol. of NaOH in lt) - (molarity of HBr*vol.of HBr in lt.)

                                 = (0.280*0.025) - (0.750*0.005)

                                 = 0.007 - 0.00375

                                 = 0.00325 moles of NaOH

so Concentration OH- = no.of moles remaining/volume of solution in lt = 0.00325/(0.025 +0.005)

                                  = 0.108 M

Ph = 14- Poh

     = 14 - {-log[OH-]}

     = 14 - {-log(0.108)}

    = 14 - 0.96

    = 13.04