Describe the preparation of 1 liter of a 0.20 M acetate buffer pH 4.47 starting

Question

Describe the preparation of 1 liter of a 0.20 M acetate buffer pH 4.47 starting from a 1 M solution of acetic acid and a 1.5 M solution of KOH.

Explanation / Answer

Acetic acid and its salt with strong baseforms the acidic buffer system. According to Hendersonequation , pH = pKa + log [salt] /[acid] Ka of aceticacid = 1.8 x 10-5 ? pKa = - log 1.8 x 10-5 = 4.744 Given that 1 Lof 0.2M acetate buffer has to be prepared. ?Total number of moles of acetic acid and potassium acetate =0.2 M *1L = 0.2 mol let the number of moles of acetic acid = n1 Number of moles of potassium acetate = n2 Let V1 volune(L) of acetic acid and V2 Volume (L) of KOH aremixed to form the requied buffer. CH3COOH + KOH -------> CH3COOK+ H2O ? n1 = 1M *V1 n2 = 1.5 M*V2 Given pH of the buffer = 4.47. ? 4.47 = 4.744 + log n2/n1 Bysolving n2/n1 = 0.532 we have n2 + n1 = 0.2 mol by solving , n2 = 0.069 mol , n1 = 0.131mol ? 0.069 moles of KOH reacts with 0.069 moles of aceticacid to form 0.069 moles of potassiumacetate CH3COOH + KOH -------> CH3COOK+ H2O 0.069 mol 0.069 mol 0.069 mol ? 0.131 + 0.069 = 0.2 mol of acetic acid has to betaken. Volume of acetic acid = 0.2 mol / 1M =0.2 L Volumeof KOH = 0.069 mol / 1.5M =0.046 L ? The required buffer can be prepared by adding 0.2 L of 1M acetic acidto 0.046 L of 1.5 M KOH.

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