Question

1. ______ / (10 pts): Theoretical Buffer Solution: Preparing a 0.050 M buffered solution using a conjugate weak acid/weak base pair. You will prepare 100 mL of buffered solution with pH = ____7___. The formal concentration of this buffer should be 0.050 F (i.e., the total concentration of the weak acid and its conjugate base is 0.050 M). (You will lose 5 points if your first attempt at these calculations in incorrect) • Record your calculations below, clearly showing all your work with proper labels (e.g., mass of HCO3-). • Find the total amount (g or mL) of each species you will have to add to 100 mL of water to get your assigned pH.

2. ______ / (10 pts) Practical Buffer Solution: Describe how you would prepare 100 mL of a 0.050 M _sodium phosphate monobasic and .5 NaOH_buffer to a pH of ___7____ .This method is accomplished by preparing a solution of a weak species (acid/base), and then adding the appropriate strong species (NaOH) to get to the desired pH. (You will lose 5 points if your first attempt at these calculations in incorrect) Outline in bullet form in the space provide below and make sure to include what amounts and volumes you will use based on your calculations

FOR BOTH QUESTIONS USE:

SODIUM PHOSPHATE MONOBASIC molecular W=137.99

.5M NaOH

phosphoric acid pka2=7.198

Answer 1

1) The buffer system consists of sodium phosphate monobasic, NaH₂PO₄ and sodium phosphate dibasic, Na₂HPO₄. The acid dissociation reaction is

H₂PO₄^- (aq) <======> HPO₄^2- (aq) + H⁺ (aq)

The pH of the buffer solution is 7.000 while the pK_a2 of phosphoric acid is 7.198. Use the Henderson-Hasslebach equation as below.

pH = pK_a2 + log [Na₂HPO₄]/[NaH₂PO₄]

===> 7.000 = 7.198 + log [Na₂HPO₄]/[NaH₂PO₄]

===> -0.198 = log [Na₂HPO₄]/[NaH₂PO₄]

===> [Na₂HPO₄]/[NaH₂PO₄] = antilog (-0.198) = 0.6339

===> [Na₂HPO₄] = 0.6339*[NaH₂PO₄] ……(1)

Again, it is given that [NaH₂PO₄] + [Na₂HPO₄] = 0.050 M

===> [NaH₂PO₄] + 0.6339*[Na₂HPO₄] = 0.050 M

===> 1.6339*[NaH₂PO₄] = 0.050 M

===> [NaH₂PO₄] = (0.050 M)/1.6339 = 0.0306 M.

Therefore, [Na₂HPO₄] = 0.6339*[NaH₂PO₄] = 0.6339*0.0306 M = 0.01939 M ≈ 0.0194 M.

The formal concentrations of sodium phosphate monobasic and sodium phosphate dibasic required are 0.0306 M and 0.0194 M.

We wish to prepare 100 mL of the buffer. The M.W. of sodium phosphate monobasic is 137.99 g/mol; the M.W. of sodium phosphate dibasic is 141.96 g/mol.

Mass of sodium phosphate monobasic required = (100 mL)*(1 L/1000 mL)*(0.0306 mol/L)*(137.99 g/mol) = 0.4222 g (ans).

Mass of sodium phosphate dibasic required = (100 mL)*(1 L/1000 mL)*(0.0194 mol/L)*(141.96 g/mol) = 0.2754 g/mol (ans).

2) We prepare the buffer by adding only sodium phosphate monobasic and sodium hydroxide (NaOH, 0.5 M).

We have already calculated the moles and hence, mass of NaH₂PO₄ present in the buffer. Calculate the moles of sodium phosphate dibasic as (100 mL)*(1 L/1000 mL)*(0.0194 mol/L) = 0.00194 mole.

Write down the neutralization reaction as below.

NaH₂PO₄ (aq) + NaOH (aq) --------> Na₂HPO₄ (aq) + H₂O (l)

As per the stoichiometric equation,

1 mole Na₂HPO₄ = 1 mole NaH₂PO₄ = 1 mole NaOH

Hence, 0.00194 mole Na₂HPO₄ is produced by the neutralization of 0.00194 mole NaH₂PO₄ with 0.00194 mole NaOH.

Mass of NaH₂PO₄ corresponding to 0.0194 mole = (0.00194 mole)*(137.99 g/mol) = 0.2677 g.

Add this mass of NaH₂PO₄ to the mass present in the buffer above; the total mass of NaH₂PO₄ required is (0.4222 + 0.2677) g = 0.6899 g.

Volume of NaOH required = (moles of NaOH required)/(molar concentration of NaOH) = (0.00194 mole)/(0.5 mol/L) = 0.00388 L = (0.00388 L)*(1000 mL/1 L) = 3.88 mL.

Therefore, in order to prepare the buffer, add 0.6899 g NaH₂PO₄ in a 100 mL volumetric flask, add 3.88 mL of 0.5 M NaOH and make up the volume to the 100 mL mark with deionized water. Shake well until a homogenous solution is obtained. The prepared solution is the NaH₂PO₄/Na₂HPO₄ buffer at pH 7.000 (ans).