A 1.500 gram sample of HA (MM = 150 g) is dissolved to make 100.0 mL of solution. A 50.0 mL portion is neutralized with NaOH. This neutralized solution (volume of 80.3 mL) is mixed with 50.0 mL of unneutralized solution. The recombined solution has a pH of 6.35.Calculate:a.moles total of the HA sample in 100.0 mLb.moles HA in 50.0 mL portion not titratedc.moles HA in 50.0 mL portion that will be titrated (but before titration)d.moles A– formed in the titratione.[HA] and [A– ] in the combined mixture that has a pH of 6.35. (moles essentially do not change when the solutions are combined)f.Ka using the values for [HA] and [A–] together with the measured pH
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