Using the Henderson Hasselbach equation, determine the amount (in grams of mL) o

Question

Using the Henderson Hasselbach equation, determine the amount (in grams of mL) of the acid and base components needed to prepare 500 mL of both of the following two buffer systemsL 0.1000 M phosphate buffer (pH 7.2) and 0.1000 M acetate buffer (pH 4.7). For the phosphate buffer you willcombine sodium phosphate, monobasic (NaH2PO4*H2O) and sodium phosphate, dibasic (Na2HPO4*7H2O). For the acetate buffer you will combine glacial acetic acid (CH3COOH, 17.4 M) and anhydrous sodium acetate (CH3COO- Na+).

Explanation / Answer

For 500 mL phosphate buffer :

monobasic sodium phosphate is the acid( represented by 'a') and dibasic sodium phosphate is the conjugate base( represented by 'cb')

For this acid base pair, pKa = 6.86

pH = 7.2

Using equation : pH = pKa + log10 ( [cb]/[a])

7.2 = 6.86 + log10 ( [cb]/[a])

([cb]/[a]) = 2.18---(1)

Also, molarity of buffer = 0.1

Thus, [cb]+[a] = 0.1 ---(2)

Solving (1) and (2), we get : [cb] = 0.068 M , [a] = 0.0314 M

Since volume is 0.5 L , thus

Moles of hydrated dibasic sodium phosphate(conjugate base) required = 0.068/2 = 0.034 moles

Thus, mass of hydrated dibasic sodium phosphate required = 0.034*molecular mass = 0.034*268.07 = 9.11 g

Moles of hydrated monobasic sodium phosphate(acid) required = 0.0314/2 = 0.0157 moles

Thus, mass of hydrated monobasic sodium phosphate required = 0.0157*molecular mass = 0.0157*138.01 = 2.16 g

For 500 mL acetate buffer :

Glacial acetic acid is the acid and anhydrous sodium acetate is the conjugate base

For this acid base pair, pKa = 4.75

pH = 4.7

Using equation : pH = pKa + log10 ( [cb]/[a])

4.7 = 4.75 + log10 ( [cb]/[a])

([cb]/[a]) = 0.891---(1)

Also, molarity of buffer = 0.1

Thus, [cb]+[a] = 0.1 ---(2)

Solving (1) and (2), we get : [cb] = 0.046 M , [a] = 0.052 M

Since volume is 0.5 L , thus

Moles of anhydrous sodium acetate(conjugate base) required = 0.046/2 = 0.023 moles

Thus, mass of anhydrous sodium acetate required = 0.023*molecular mass = 0.023*82.03 = 1.88 g

Moles of glacial acetic acid(acid component) required = 0.052/2 = 0.026 moles

Conc. of given acetic acid = 17.4 M

Thus, volume required = 0.026/17.4 = 0.00143 L = 1.43 ml

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