Asolution of acetic acid is prepared in water by adding 11.1 g of sodium acetate to 200 ml of water. the final ph is measured to be 6.5. what are the concentrations of acetate and acetic acid in solution?

Given buffer contains:

Acetic acid: CH3COOH.

Sodium Acetate : CH3COONa.

Mass of CH3COONa taken = 11.1 g

Molar mass of CH3COONa = 82.03 g/mol

# moles of CH3COONa = 11.1g/82.03 g.mol-1 = 0.1353 moles

Volume of water = 200 ml = 0.2 L

Therefore, the initial concentration of CH3COONa = 0.1353/0.2 = 0.677 M

Buffer pH = 6.5

pKa (CH3COOH) = 4.76

Based on Henderson-Hasselbalch equation:

pH = pKa + log [CH3COONa]/[CH3COOH]

Now, initially there is only 0.677 M CH3COONa . If 'x' is the concentration of the protonated species (i.e CH3COOH) formed, then the amount of CH3cooNa remaining = 0.677-x

6.5 = 4.76 + log [0.677-x]/[x]

x = 0.011 M

i.e. [CH3COOH] = 0.011 m

[CH3COONa] = 0.677-0.011 = 0.666 M

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the answer is minerals.