Your boss is studying a pepsin homolog present in fruit and ask you to make 1.00

Question

Your boss is studying a pepsin homolog present in fruit and ask you to make 1.00 L of a 0.100 m acetate buffer at pH 4.50. You have powdered acetate and glacial acetic acid (also distilled water and all the usual lab equipment) available. Note the the pKa of acetic acid is 4.79.

a. What do you mix in what quantity to make the buffer?

b. Upon adding 0.010 millimols of a strong acid to a 1.0 ml sample of your buffer, what is the new pH?

c. For comparison, what is the pH of a 0.010 M aqueous solution of aceitc acid? Of 0.010 M per-chloric acid (a strong acid)?

Explanation / Answer

a. 0.1 m of 1 L buffer preparation

density of water = 1g/ml

So 0.1 m = 0.1 M

pH = pKa + log([base]/[acid])

4.50 = 4.79 + log([acetate]/[acetic acid])

[aceate] = 0.513[acetic acid]

[acetic acid] + [acetate] = 0.1 M x 1 L = 0.1 mol

[acetic acid] + 0.513[acetic acid] = 0.1

So amount of

[acetic acid] = 0.066 mols

and,

[acetate] = 0.1 - 0.066 = 0.034 mols

or preparation of buffer

b. moles of HCl added = 0.02 mmol

new [acetic acid] = 0.1 x 1 + 0.01/1 = 0.11 M

new [acetate] = 0.1 M x 1 ml - 0.01 mmol/1 = 0.09 M

new pH = 4.79 + log(0.09/0.11) = 4.70

c. pH of 0.01 M acetic acid (aq)

Ka = 1.62 x 10^-5 = x^2/0.01

x = [H+] = 4.02 x 10^-4 M

pH = 3.39

pH of 0.01 M HCl

[H+] = 0.01 M

pH = 2

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