You need to prepare an acetate buffer of pH 5.87 from a 0.841 M acetic acid...

You need to prepare an acetate buffer of pH 5.87 from a 0.841 M acetic acid solution and a 2.92 M KOH solution. If you have 975 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a bufer of pH 5.87? The pKa of acetic acid is 4.76.

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Expert Solution

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

no of moles of CH3COOH = molarity *volume in L

= 0.841*0.975 = 0.82 moles

PH = PKa + log[CH3COONa]/[CH3COOH]

5.87 = 4.76 + logx/0.82-x

logx/0.82-x = 5.87-4.76 = 1.11

logx/0.82-x = 1.11

x/0.82-x = 10^1.11

x/0.82-x = 12.88

x = 12.88*(0.82-x)

x = 0.76moles = no of moles of KOH

no of moles of KOH = molarity * volume in L

0.76 = 2.92*volume in L

volume in L = 0.76/2.92 = 0.26L = 260ml

volume of KOH = 260ml >>>> answer

queen_honey_blossom answered 1 year ago

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