We’re going to titrate formic acid with the strong base, NaOH. There is initiall
ID: 1062217 • Letter: W
Question
We’re going to titrate formic acid with the strong base, NaOH. There is initially 100. mL of 0.50 M formic acid and the concentration of NaOH is 1.0 M.
A. What is the initial pH of the formic acid solution?
B. What is the percent ionization under initial conditions?
C. After the addition of 10 mL of NaOH, what is the pH?
D. After the addition of 25 mL of NaOH, what is the pH? Think about where in the titration this brings you.
E. What volume of NaOH is required to reach the equivalence point?
F. What is the pH at the equivalence point?
G. What is the pOH at the equivalence point?
H. If, instead of NaOH being added, 0.05 moles of HCl is added by bubbling the gas through the solution. Assume that the volume has not changed. What is the percent dissociation of formic acid?
Sorry! i know its alot
Explanation / Answer
formic acid pKa = 3.75
a)
initially
HA <->H + + A-
Ka= [H+][A-]/[HA]
10^-3.65 = x*x/(0.5-x)
x = 0.00934
pH = -log(0.00934) = -log(0.00934
b)
% ion = [H+]/M* 100% = 0.00934/0.5 * 100 = 1.868 %
c)
mmol of acid = MV = 100*0.5 = 50 mmol of acid
mmol of base = MV = 10*1 = 10
mmol of acid left after reaction = 50-10 = 40
mmol of ocnjugate formed = 0 + 10 = 10
this is buffer so
pH = pKa + log(A-/HA)
pH = 3.75 + log(10/40) = 3.147
d)
mmol of acid = MV = 100*0.5 = 50 mmol of acid
mmol of base = MV = 25*1 = 25
mmol of acid left after reaction = 50-25= 25
mmol of ocnjugate formed = 0 + 25= 25
this is buffer so
pH = pKa + log(A-/HA)
pH = 3.75 + log(25/25) = 3.75
this is half euqivalence point
e)
equivlance point
mmol of acid = 50
mmol of base required = 50
so...
V = mmol/V = 50/1 = 50 mL of NaOH
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