Suppose that you wanted to prepare a acetate ion/acetic acid buffer solution with a pH of 4.35. What is the value 34 Marks: 2 off [A l/IHA) for the correct buffer solution for this event? The Ka for acetic acid is 1.8 x 10 5. a. 0.39
b. 0.407
c. 0.372
d. 4.74

b. 0.407

Explanation:

Acetic acid / Acetate ion is a buffer (Mixture of a weak acid, acetic acid, with its conjugate base, acetate ion) with pKa = -log Ka = 4.74.

The simplest way to determine the pH of a buffer is using Henderson-Hasselbalch formula:

pH = pKa + log [Conjugate base] / [Weak acid]

For acetic buffer with pH = 4.35:

4.35 = 4.74 + log [A⁻] / [HA]

-0.39 = log [A⁻] / [HA]

0.407 = [A⁻] / [HA]

Thus, right option is:

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).