Question

# In lab, you will be provided with a 1.8M acetic acid solution and a 1.0M NaOH...

In lab, you will be provided with a 1.8M acetic acid solution and a 1.0M NaOH solution. You want to make 100.0mL of a 0.5M acetic acid buffer at pH 5.00. What volume (mL) of the 1.8M acetic acid solution, NaOH solution, and deionized water should you combine? Show all work. (pKa of acetic acid= 4.75).

total mole of acetic acid and sodium acetate = 0.100 L * 0.5 mole / L = 0.05 mole.

mole of NaOH = mole of CH3COONa = x mole

and

mole of CH3COOH in buffer solution = (0.05 - x) mole.

Using Henderson equation,

pH = pKa + log [salt] / [acid]

or

5.00 = 4.75 + log [x / (0.05 - x)]

or

x = 0.032 mole.

thus

mole of NaOH = 0.032 mole.

volume of NaOH = 1000 ml * 0.032 mole / 1.0 mole = 32 ml

and

mole of acetic acid = (0.05 - 0.032) = 0.018 mole.

volume of 1.8M acetic acid added = 1000 ml * 0.018 mole / 1.8 mole = 10.0 ml

and

volume of water added = 100 - (32 + 10.0) = 58.0 ml

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