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In: Chemistry

You need to prepare an acetate buffer of pH 5.50 from a 0.899 M acetic acid...

You need to prepare an acetate buffer of pH 5.50 from a 0.899 M acetic acid solution and a 2.73 M KOH solution. If you have 775 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.50? The pKa of acetic acid is 4.76.

Expert Solution

no of moles of CH3COOH = molarity * volume in L

= 0.899*0.775 = 0.696 moles

let no of moles of KOH is x

CH3COOH + KOH ---------------> CH3COOK + H2O

I 0.696 x 0

C -x -x +x

E 0.696-x 0 +x

PH = PKa + log[CH3COOK]/[CH3COOH]

5.5 = 4.76 + logx/0.696-x

logx/0.696-x = 5.5-4.76

logx/0.696-x = 0.74

x/0.696-x = 10^0.74

x/0.696-x = 5.4954

x = 5.4954*(0.696-x)

x = 0.588

no of moles of KOH = x = 0.588 moles

no of moles of KOH = molarity * volume in L

0.588 = 2.73 * volume in L

volume in L = 0.588/2.73 = 0.2154L = 215.4ml >>>>answer

volume of KOH = 215.4ml

queen_honey_blossom answered 2 years ago

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