Answer:
Molarity of sodium acetate you will need to add is 0.0324M
Explanation:
<em>Assuming volume of the buffer is 1L.</em>
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The pH of a buffer can be determined using Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pKa is pKa of the weak acid, [A⁻] molar concentration of conjugate base and [HA] molar concentration of weak acid</em>
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Replacing for the acetic buffer (pKa = 4.76):
pH = 4.76 + log [Sodium Acetate] / [Acetic Acid]
As you have 0.010 moles of acetic acid in 1L:
[Acetic Acid] = 0.010mol / 1L = 0.010M
And you require a pH of 5.27:
5.27 = 4.76 + log [Sodium Acetate] / [0.010M]
0.51 = log [Sodium Acetate] / [0.010M]
10^0.51 = [Sodium Acetate] / [0.010M]
3.236 = [Sodium Acetate] / [0.010M]
3.236 [0.010M] = [Sodium Acetate]
0.0324M = [Sodium Acetate]
<h3>Molarity of sodium acetate you will need to add is 0.0324M</h3>
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