In: Chemistry
What is the ph of a solution made from 500 mL of 1.0 M HC2H3O2 and 1000 mL of .5 M C2H3O- after 80 mL 0.9 M NaOh has been added?
no of moles of HC2H3O2 = molarity * volume in L
= 1*0.5 = 0.5moles
no of moles of C2H3O2^- = molarity * volume in L
= 0.5*1 = 0.5 moles
PKa of HC2H3O2 = 4.75
PH = PKa + log[C2H3O2^-]/[HC2H3O2]
= 4.75+ log0.5/0.5
= 4.75+ log1
= 4.75 + 0 = 4.75
By the addition of NaOH
no of moles of NaoH = molarity * volume in L
= 0.9*0.08 = 0.072 moles of NaOH
no of moles of HC2H3O2 by the addition of NaOH = 0.5-0.072 = 0.428 moles
no of moles of C2H3O2^- by the addition of NaOH = 0.5+0.072 = 0.572 moles
PH = PKa + log[C2H3O2^-]/[HC2H3O2]
= 4.75 + log0.572/0.428
= 4.75+ 0.126 = 4.876 >>>>answer