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You need to prepare an acetate buffer of pH 5.33 from a 0.631 M acetic acid...

You need to prepare an acetate buffer of pH 5.33 from a 0.631 M acetic acid solution and a 2.20 M KOH solution. If you have 480 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.33 ? The p K a of acetic acid is 4.76.

Expert Solution

First calculate initial moles of acetic acid .

We have, [ CH₃COOH ] = No. of moles of CH₃COOH / Volume of solution in L

No. of moles of CH₃COOH = [ CH₃COOH ] Volume of solution in L

No. of moles of CH₃COOH = 0.631 mol / L 0.480 L

No. of moles of CH₃COOH = 0.30288 mol

Now, consider reaction of NaOH with acetic acid.

CH₃COOH + KOH CH₃COOK + H₂O

Let's use ICE table.

moles	CH₃COOH + KOH CH₃COOK
I	0.30288	X
C	-X	-X	+X
E	0.30288 - X	0	+X

After addition of X moles of KOH , solution contain weak acid Acetic acid and its salt potassium acetate.

This solution acts as a buffer solution and its pH is calculated by using Henderson's equation.

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

pH = pKa + log [ CH₃COOK ] / [ CH₃COOH ]

We have to prepare buffer solution of pH 5.33 .

5.33 = 4.76 + log X / 0.30288 - X

log X / 0.30288 - X = 5.33 -4.76

log X / 0.30288 - X = 0.57

X / 0.30288 - X = 10 ^0.57 = 3.72

X = 3.72 ( 0.30288 - X )

X = 1.127 - 3.72 X

X +3.72 X = 1.127

4.72 X =1.127

X = 1.127 / 4.72

X = 0.2388 moles

If we add 0.2388 moles of KOH to 480 ml of 0.631 M acetic acid we will get buffer solution of pH 5.33.

Now, calculate volume of KOH required to add to acetic solution to make buffer.

We have, [ KOH ] = No. of moles of KOH / Volume of solution in L

Volume of solution in L = No. of moles of KOH / [ KOH ]

Volume of solution in L = 0.2388 moles / 2.20 mol / L

Volume of solution in L = 0.1085 L

Volume of solution in L = 108.5 ml

ANSWER : 108.5 ml

queen_honey_blossom answered 1 year ago

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