You need to prepare an acetate buffer of pH 5.85 from a 0.649 M acetic acid...

You need to prepare an acetate buffer of pH 5.85 from a 0.649 M acetic acid solution and a 2.22 M KOH solution. If you have 925 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.85? The pKa of acetic acid is 4.76.

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Expert Solution

moles of acetic acid in 0.649M of 925ml= Molarity (Moles/Liter)* Volume (liters)= 0.649*0.925 =0.600

let x= volume of KOH in L, moles of KOH added= 2.2*x= 2.2x

volume of solution = 2.2x +0.925

assuming KOH is limiting reactant and all the KOH is consumed, moles of potassium acetate formed due to reaction

CH3COOH+KOH------>CH3COOK( potassium acetate)+ H2O

=2.2x, potasium acetate suppliments acetate ions

moles of Acetic acid remaining = 0.6-2.2x

concentrations : potassium acetate (CH3COO-)= 2.2x/(0.925+x), acetic aicd = (0.6-2.2x)/ (0.925+x)

since pH= pKa+ log (CH3COO-)/ [CH3COOH]

5.85= 4.76 + log (2.2x/(0.6-2x)

12.3= 2.2x/(0.6-2x)

12.3*0.6-12.3*2x= 2.2x

12.3*0.6= x*(2.2+24.6)= 26.8

x= 12.3*0.6/26.8= 0.275 L= 275 ml

queen_honey_blossom answered 1 year ago

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