Question

Calculate the change in pH of a 1.00 L of a buffered solution preparing by mixing 0.50 M acetic acid (Ka = 1.8 x 10^-5) and 0.50 M sodium acetate when 0.010 mole of NaOH is added. Compare this pH change with that when the same amount 0.010 mole of NaOH was added to 1.00 L of water. (Remember what we did in class a week ago, the pH will jump from 7 to 12 when that tiny amount of NaOH was added).

Answer 1

The change in pH of a 1.00 L of a buffered solution preparing by mixing 0.50 M acetic acid (Ka = 1.8 x 10^-5) and 0.50 M sodium acetate when 0.010 mole of NaOH is added is 4.75

when the same amount 0.010 mole of NaOH was added to 1.00 L of water the pH = 12

Explanation:

given that:

concentration of acetic acid = 0.50 M

Concentration of base sodium acetate = 0.50 M

ka = 1.8 x 10^-5)

pka = -log [ka]

pka = 4.74

From Henderson-Hasselbalch Equation:

pH = pKa + log $\frac{[base]}{[acid]}$

pH = 4.74 + Log $\frac{[0.5]}{[0.5]}$

pH = 4.74 + 0

pH = 4.74

Number of moles of NaOH = 0.010 moles

volume 1 litre

molarity = 0.010 M

Moles of acetic acid and sodium acetate before addition of NaOH

FORMULA USED:

molarity = $\frac{number of moles}{volume in litres}$

acetic acid,

0.5 = number of moles

0.5 is the number of moles of sodium acetate.

number of moles of NaOH  0.010 moles

NaOH reacts in 1:1 molar ratio with acetic acid so

number of moles in acetic acid = 0.5 - 0.010 = 0.49

number of moles in sodium acetate = 0.5 +0.010 = 0.51

new pH

pH = pKa + log $\frac{[base]}{[acid]}$

pH= 4.74 + log[0.51] - log[0.49]

pH= 4.75

PH of NaOH of 0.01 M (BASE)

pOH = -Log[0.01]

pOH         = 2

pH can be calculated as

14= pH +pOH

pH= 14-2

pH = 12