8. no of moles of HC2H3O2 = molarity * volume in L = 0.15*1 = 0.15 moles no of m
ID: 699608 • Letter: 8
Question
8. no of moles of HC2H3O2 = molarity * volume in L
= 0.15*1 = 0.15 moles
no of moles of C2H3O2^- = molarity * volume in L
= 0.25*1 = 0.25 moles
PKa = -logKa
= -log1.8*10^-5
= 4.75
PH = PKa + log[C2H3O2^-]/[HC2H3O2]
= 4.75 + log0.25/0.15
= 4.75 + 0.2218 = 4.97
By the addition of 0.025 moles of HCl
no of moles of HC2H3O2 after addition of HCl = 0.15+0.025 = 0.175 moles
no of moles of C2H3O2^- after addition of HCl = 0.25-0.025 = 0.225 moles
PH = Pka + log[C2H3O2^-]/[HC2H3O2]
= 4.75 + log0.225/0.175
= 4.75 +0.1 = 4.85
Change in pH = 4.85-4.97 = -0.12 >>>>answer
Explanation / Answer
8. What is the pH of a solution that contains 0.15 M HC2H 02 and 0.25 M C2H302? Use K 1.8 x10% for HC2H302. By how much will the pH change if 0.025 mol of HCl is added to 1.00 L of the buffer. Answer: pH = 4.97; pH of the solution changes-0.12 When 50.0 mL of 0.050 M formic acid, HCHO2, is titrated with 0.050 M sodium hydroxide, what is the pH at the equivalence point? (Be sure to take into account the change in volume during the titration.) Answer: 8.07 9.
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