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 K_a of ascorbic acid (HC₆H₇O₆) is K_a = 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 =  

Answer 1

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