1) Prepare 250 mL of a 0.1 M phosphate buffer of pH approximately 7.2 . Thus, mo

Question

1) Prepare 250 mL of a 0.1 M phosphate buffer of pH approximately 7.2. Thus, monosodium phosphate and its conjugate base, disodium phosphate will be used to prepare the buffer solution. Use the Henderson-Hasselbalch equation to determine how much acid and conjugate base to use. Use a pKa value of 6.86 for a buffer of pH 7.2. Be sure to show ALL calculations.

2) Prepare a fresh sodium dithionite solution (Na2S2O4) with a concentration of 0.003 M in a 50 mL volumetric flask. How much sodium dithionite is needed? Show ALL calculations in answering this.

Explanation / Answer

1) Molarity of the buffer is the sum of the molarities of acid and conjucate base

[acid] + [conjucate base] = 0.1 M

pH = pKa + log ( base / acid)

7.2 = 6.8 + log (base/acid)

(base/acid) = 2.5118

[base] = 2.5118 [acid] ratio

[acid] = 0.02847 M

[base] = 0.0715 M

M = moles / L

But we need to prepare for 250 mL

So moles of acid = 250 mL * 0.02847 mol / 1000 mL => 0.007117 mol

Molar mass of monosodium phosphate ( acid, NaH2PO4) = 119.98 g/mol

So mass of NaH2PO4 required = 119.98 g/mol *0.007117 mol => 0.85 g

Moles of base = 250 mL * 0.0715 mol / 1000 mL => 0.017875 mol

Molar mass of disodium phosphate => 141.96 g/mol

Mass of base ( Na2HPO4) required = 141.96 g/mol * 0.017875 mol => 2.54 g

So to prepare 250 mL of 0.1 M phosphate buffer add 0.85 g of monosodium phosphate and 2.54 g of disodium phosphate in 250 mL of distilled water , and the resulting solution will have pH of 7.2.

2) Na2S2O4 = 50 mL * ( 0.003 mol / 1000 mL) => 0.00015 mol

Molar mass of Na2S2O4 => 174.1 g/mol

Mass of Na2S2O4 required = 0.00015 mol * 174.1 g/mol => 0.026 g

0.026 g or 26 mg of Na2S2O4 has to be added in 50 mL of distilled water to get 0.003 M of Na2S2O4

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