Consider the titration of 0.967 L of 0.893 M ascorbic acid (H2C6H6O6) with 1.72
ID: 882235 • Letter: C
Question
Consider the titration of 0.967 L of 0.893 M ascorbic acid (H2C6H6O6) with 1.72 M NaOH. What is the pH at the second equivalence point of the titration?
At this point, we have only C6H6O62– (weak base) left. And this would be in a weak base equilibrium. Let’s put this in ICE chart form (in molarity, so we should divide by the total volume: 0.967 L + 0.502 L (1st eq. pt.) + 0.502 L (2nd eq. pt) = 1.97 L
C6H6O62–
+ H2O
HC6H6O6– + OH–
I
0.4381
—
0
0
C
-x
—
+x
+x
E
0.4381 - x
—
x
x
Write out Kb and solve for x ([OH–], pOH, and pH:
K =[HC6H6O6-][OH-] b [C H O2- ]
0.00625 » x2 0.4381
x = 0.0523M = [OH- ] pOH = -log(0.0523)
= 1.28 pH = 12.72
I dont understand how they got .4381or the Kb value???
C6H6O62–
+ H2O
HC6H6O6– + OH–
I
0.4381
—
0
0
C
-x
—
+x
+x
E
0.4381 - x
—
x
x
Explanation / Answer
Moles of ascorbic acid = volume in L * molarity
= 0.967 L * 0.893 M
= 0.8635 mol
There are two equivalence points.
Reaction between ascorbic acid and NaOH for first equivalence point,
H2C6H6O6 (aq) + OH- (l) ---- > HC6H6O6- (aq) + H2O
This suggests that one mol of ascorbic acid needs one mole NaOH
Lets calculate number of moles of NaOH
Number of moles of NaOH = number of moles of ascorbic acid
Number of moles of NaOH = 0.8635 mol
Now we calculate volume of NaOH needed for first equivalence point
Volume of NaOH = mol of NaOH / molarity of NaOH
=0.8635 mol NaOH / 1.72 M NaOH
= 0.5020 L NaOH
Reaction for second equivalence point.
HC6H6O6- (aq) + OH- (l) ---- > C6H6O62- (aq) + H2O
In this reaction also volume of NaOH needed = 0.5020 L
Total volume of NaOH needed to reach second equivalence point = 0.5020 L + 0.5020 L = 1.0041 L
Finally in the solution there is only moles of C6H6O62- (aq) are present
Moles of C6H6O62- (aq) = moles of H2 C6H6O6 (aq)= 0.8635 mol
We find concentration of C6H6O62- (aq).
Total volume = volume of ascorbic acid + volume of NaOH
= 0.967 L + 1.0041 L
= 1.971 L
[C6H6O62- ]= 0.8635 mol / 1.9671 L
= 0.4381 M
We know C6H6O62- acts as a base and abstracts a proton from water to form HC6H6O6-
So we use ka2 value of HC6H6O6- to calculate kb
Kb = 1.0 E-14 / ka2
Ka2 = 1.6E-12
Kb = 1.0 E-14 / 1.6E-12
= 0.00625
Lets set up ICE table of the reaction of C6H6O62- (aq) with water
C6H6O62- (aq) + H2O (l) ----- > HC6H6O6- (aq) + OH- (aq)
I 0.4381 0 0
C -x +x +x
E (0.4381-x) x x
Kb = x2/ ( 0.4381 –x)
0.00625 = x2/ ( 0.4381 –x)
0.00625 * ( 0.4381 –x) = x2
0.002738 – 0.00625 x = x2
X2 + 0.00625 x – 0.002738 = 0
By solving quadratic equation we get
x = 0.04929
[OH-]= x = 0.04929 M
pOH = -log [OH-]= -log ( 0.04939 )
= 1.3072
pH = 14 –pOH
= 14- 1.3072 = 12.69
pH = 12.70