Consider the titration of 80.0 mL of 0.0200 M C5H5N (a weak base; Kb = 1.70e-09)
ID: 971511 • Letter: C
Question
Consider the titration of 80.0 mL of 0.0200 M C5H5N (a weak base; Kb = 1.70e-09) with 0.100 M HNO3.
Calculate the pH after the following volumes of titrant have been added:
(a) 0.0 mL pH =
(b) 4.0 mL pH =
(c) 8.0 mL pH =
(d) 12.0 mL pH =
(e) 16.0 mL pH =
(f) 25.6 mL pH =
PART TWO:
A buffer solution contains 0.14 mol of ascorbic acid (HC6H7O6) and 0.84 mol of sodium ascorbate (NaC6H7O6) in 2.80 L.
The Ka of ascorbic acid (HC6H7O6) is Ka = 8e-05.
(a) What is the pH of this buffer?
pH =
(b) What is the pH of the buffer after the addition of 0.07 mol of NaOH? (assume no volume change)
pH =
(c) What is the pH of the original buffer after the addition of 0.21 mol of HI? (assume no volume change)
pH =
Explanation / Answer
I will answer Part 1. I highly reccomend you to post your second question in another post to get answered.
First, let's calculate the equivalence point volume,:
Va = 0.02 * 80 / 0.1 = 16 mL
This means that in 16 mL we reached the equivalence point. Beyond this point, we have excess of acid.
a) At 0 mL, no acid is added, only base is present:
r: C5H5N + H2O ---------> C5H5NH+ + OH-
i: 0.02 0 0
e: 0.02-x x x
1.7x10-9 = x2 / 0.02-x ---> Kb is small so the value of x will be small too, and we can neglect 0.02-x to 0.02 only:
1.7x10-9 * 0.02 = x2
x = [OH] = 5.83x10-6 M
pOH = -log(5.83x10-6) = 5.23
pH = 14-5.23 = 8.77
b) 4 mL of acid added:
moles acid = 0.1 * 0.004 = 0.0004 moles
moles base = 0.02 * 0.080 = 0.0016 moles
reaction general: C5H5N + HNO3 ---------> C5H5NH+ + NO3-
the base is in excess, so, the remaining moles would be:
moles base remaining = 0.0016-0.0004 = 0.0012 moles
using the HH equation:
pOH = pKb + log[BH/B] ---> pKb = -log(1.7x10-9) = 8.77
pOH = 8.77 + log(0.0004/0.0012)
pOH = 8.29
pH = 14-8.29 = 5.71
c) at 8 mL, is the half equivalence point, and pH = pKa so:
pKa = 14-8.77 = 5.23 = pH
d) 12 mL of acid added:
moles of acid = 0.1 * 0.012 = 0.0012 moles
moles of base remaining = 0.0016 - 0.0004
pOH = 8.77 + log(0.0012/0.0004)
pOH = 9.25
pH = 4.75
e) at the equivalence point, all the base is consumed and now the acid is in excess, the reaction in general is the following:
moles of conjugate acid = moles acid added = 0.0016 moles
[Concentration] = 0.0016 / 0.096 = 0.0167 M
r: C5H5NH+ <---------< C5H5N + H+ ---> Ka = 10-pKa = 10-5.23 = 5.89x10-6
i: 0.0167 0 0
e: 0.0167-x x x
5.89x10-6 = x2 / 0.0167-x
5.89x10-6 * 0.0167 = x2
x = [H+] = 3.14x10-4 M
pH = -log(3.14x10-4) = 3.5
f) finally at 25.6 mL
moles base = 0.0016 moles
moles acid added = 0.1 * 0.0256 = 0.00256 moles
moles acid remaining = 0.00256 - 0.0016 = 0.00096 moles
Now calculate pH with the HH equation
Hope this helps