Can the italicized portion of the problem be explained? Iunderstand up until the 0.015 moles of protonated acetic acidremained and I understand the HH portion of the problem after but Iwould like some clarity on how this number was concluded.
2. The following question has two parts.
a) What is the final pH of a solution obtained by mixing 250 ml of0.3 M acetic acid with 300 ml
of 0.2 M KOH? (pKb of acetate = 9.24). (Show your work!)
Moles of acetic acid = 0.25 l X 0.3 M = 0.075moles
Adding 0.3 l X 0.2 M = 0.06 moles of OH- to this solution willconvert 0.06 moles of acetic
acid to 0.06 moles of acetate, 0.015 moles ofprotonated acetic acid will remain.pKa = 14 -pKb = 14 - 9.24 =4.76
pH = pKa +log[CH3CO2-][CH3CO2H]
= 4.76 + log 0.06 moles / 0.55 l 0.015moles / 0.55 l
= 4.76 + log 0.06moles0.015 moles
= 5.36
pH = 5.36
