Question

I am having a difficult time solving the problems below from a practice quiz which was assigned to help with my upcoming Chemistry quiz. Please help!

3. Suppose a buffer (pKa = 4) contained 5 moles of HA and 5 moles of A- , calculate the pH when:

A. 1 mole of NaOH is added.

B. 1 mole of HCl is added.

4. A 1M aqueous solution of a weak acid (HA) was made, its pH at equilibrium was 1, calculate the Ka and the pKa of this weak acid. Hint: Make an ice table.

Answer 1

Solution:

3) According to Hendersen -Hasselbatch equation,

pH = pka + log [A-] /[HA]

Case 1) 1 mol NaOH is added:

The addition of base increases the concentration of A- and decreases the concentration of HA.

Thus,

[A-] = 5 +1 = 6 mol

[HA] = 5 - 1 = 4 mol

Hence,

pH = pka + log [A-] / [HA]

pH = 4 + log 6/4 = 4 + log 1.5

pH = 4 + 0.18 = 4.18

Case 2) 1 mol HCl is added:

The addition of HCl increases HA concentration while decreases A- concentration.

Thus,

[A-] = 5 -1 = 4 mol

[HA] = 5 +1 = 6 mol

pH = 4 + log 4/6

pH = 4 + log 0.666

pH = 4 - 0.18 = 3.82

4) The ICE table for weak acid dissociation is given as,

HA + H2O. = A- + H3O+

1 M -----------------0 ----------0 (Initial)

-X ---------------- + X -------- +X (change)

1 - X -------------- X ---------- X (equilibrium)

Thus,

ka = X . X / (1 - X)

Since, HA is weak acid, hence X can be neglected from denominator. Thus,

ka = X^2 / (1)

Given, pH = 1

Then, [H3O+] = X = 10^-pH = 10^-1 M = 0.1 M

Therefore,

ka = X^2 = (0.1)^2 = 0.01 = 1 x 10^-2

pka = - log ka = - log 1 x 10^-2 = 2