You need to prepare an acetate buffer of pH 6.43 from a 0.743 M acetic acid...

You need to prepare an acetate buffer of pH 6.43 from a 0.743 M acetic acid solution and a 2.20 M KOH solution. If you have 830 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.43? The pKa of acetic acid is 4.76.

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Expert Solution

SOLUTION:

pH = pKa + log[CH3COO-] / [CH3COOH]

6.43 = 4.76 + log[CH3COO-] / [CH3COOH]

log[CH3COO-] / [CH3COOH] = 6.43 - 4.76 = 1.67

[CH3COO-] / [CH3COOH] = antilog(1.67) = 46.77

[CH3COO-] = [CH3COOH] 46.77 =====1)

[CH3COO-] + [CH3COOH] = 0.743M=====2)

Inserting above equation 2) in 1)

[CH3COOH] 47.77 = 0.743M

[CH3COOH] = 0.743M / 47.77 = 0.0155M

Hence from equation2) [CH3COO-] = 0.743M - [CH3COOH] = 0.743M -0.0155 = 0.727M

Hence Molarity of acetate ion must be 0.727M.

moles of CH3COO- = Molarity X Volume = 0.727 X 830 = 603.41mmol = 0.603 moles

CH3COO- is formed by the reaction of KOH with CH3COOH. One mole of KOH produces one mole of CH3COO- as

CH3COOH + KOH ----> CH3COO- + K+ + H2O

Hence moles of KOH required to produce 0.603 moles of CH3COO- = 0.603 Moles

Number of moles of KOH = Molarityof KOH X Volume

Volume = 0.603 Moles / 2.2 = 0.274L = 274mL

Hence volume of KOH needed = 274mL

queen_honey_blossom answered 1 year ago

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