Question

If I have an unknown acid that has a pH of 6.87 and I add 200ml water to make is a solution with a pH of 5.33 and then I titrate it with 11.1 ml of 0.100M NaOH and at my equivalence point i have added 2.089 ml NaOH and the pH is 7.10 and my half equivalence point is after adding only 1.0445 ml of NaOH with a pH of 6.04.

1. What is the pH of the unknown solution that will be used to determine the ionization constant (ka)?

2. what are the ionization constants determined?

3. What is the molarity of the unknown solution used?

Answer 1

Q1 & Q2

we should use the solution of the HALF equivalence point, in order to facilitate the pKa value from the pH equation (Henderson Hasselbach equation)

From the previous procedure,

note that for a weak acid, which this is the case:

the pH at the HALF equivalence point will be calculated as follows

pH = pKa + log(A-/HA)

note that in the half point, A- = HA; the anion concentration = molecular acid concentration (i.e. half point)

so

pH = pKa is true

we know pH at half point, which is 7.10

so

safely assume pH = pKa = 6.04

for our acid

pKa = 6.04

so

Ka = 10^-pKa = 10^-6.04 =9.120*10^-7

now...

Q3.

Molaity of unkown solution used...

now, use first equation:

Ka = [H+][A-]/[HA]

is the constant of the acid, by definition

so

we know:

Ka = 9.120*10^-7

[H+] = [A-] = 10^-pH (in a pure acid, due to stichiometry)

[H+] = 10^-6.87 = 1.348*10^-7

Get [HA] as follows

[HA] = M - x ( account fot that in solution/equilibrium)

9.120*10^-7 = ( 1.348*10^-7)/ (1.348*10^-7) / (M- 1.348*10^-7)

solve for "M"

M = (( 1.348*10^-7)/ (1.348*10^-7)) / (9.120*10^-7) + (1.348*10^-7)

Which makes sense,

since it is impossible to have a pH = 6.87 initially, then add 200 mL of water and decrease the pH to 5.33 (if we decrease [H+] concentration, then we increase pH)

so this is most likely:

pH initially is 4.87 and NOT 6.87

proof:

[H+] = [A-] = 10^-pH (in a pure acid, due to stichiometry)

[H+] = 10^-4.87= 0.00001348962

Get [HA] as follows

[HA] = M - x ( account fot that in solution/equilibrium)

9.120*10^-7 = (0.00001348962*0.00001348962) / (M- 0.00001348962)

solve for "M"

M = ((0.00001348962*0.00001348962) /( 9.120*10^-7) + 0.00001348962

M = 0.000213 mol of acid /L

which makes much more sense now