Respuesta :
Using henderson-hasselbach equation, the mass of sodium acetate(dry) is 0.2g and volumes of glacial acetic acid and dH₂O needed to prepare 50 mL of 0.2 M acetate buffer of pH 4.5 are 9.125 mL and 15. 875 mL
The Henderson-Hasselbach equation is :
pH = pKₐ + log [A ⁻]
[HA]
We have a pH 4.5 solution of acetic acid and sodium acetate. pKₐ of acetic acid is 4.75. Taking ratio of weak acid to conjugate base:
4.5 = 4.75 + log [sodium acetate]
[acetic acid]
[Sodium acetate] = 0.5623
[acetic acid]
Now, number of moles of each component to be calculated.
If total concentration is- 0.10 M, then:
[HA] + 0.5754[HA] = 0 .10 M
[HA] = 0.10
1.0000 + 0.5754
[A⁻] = 0.0365 M
Total volume given is - 50 mL(50 mL= 0.050 L), so the moles of each component :
nₐ₋ = 0.0365 mol x 0.050 L = 0.001825 mol
L
nhₐ = 0.0635 mol x 0.050 L = 0.003175 mol
L
Let the starting concentration were 0.20M, then we can find volume they both start with:
Vₐ₋ = 1L x 0.001825 mols A⁻ = 0.009125 L
0.20 mols A⁻
= 9.125 mL
Vhₐ = 1L x 0.003175 mols HA = 0.015875 L
0.20 mols HA
= 15.875 mL
To calculate mass of sodium acetate , using the equation:
pKₐ = -log Kₐ = -log 1.8 x 10⁻⁵ = 4.74 (kₐ of acetic acid is 1.8 x 10⁻⁵)
4.5 = 4.74 + log [acetate] / [acetic acid]
log[acetate] / [acetic acid] = -0.24
[acetate] / [acetic acid] = 0.182
x/0.2=0.182
x = [acetate] = 0.0364 M
0.0364 mol/L x 0.050 L x 136.08 g/mol = 0.247 g
Since the molarity is given to only one significant figure, answer can be reported as 0.2 g
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