You wish to prepare 50.0 milliliters of 0.0500 M phosphate buffer, pH 6.90, from

Question

You wish to prepare 50.0 milliliters of 0.0500 M phosphate buffer, pH 6.90, from phosphoric acid [a liquid; specific gravity = 1.69, assay by weight = 85.696; Mw = 97.999] and potassium hydroxide [a solid; purity 8696; Mw = 56.11] Use pKa (not pKa') from Table 2-1 in the coursepack. Calculate your answers to three (3) significant figures, and be sure to include the correct units in your answer. (6 points each) What is the molar concentration of the phosphoric acid starting material? Submit Answer Tries 0/2 What volume (in milliliters) of phosphoric acid would you need to make the buffer? Submit Answer Tries 0/2 What mass (in grams) of KOH would you need to prepare 25.0 mL of 1.00 M KOH? Submit Answer Tries 0/2 What volume (in milliliters) of the 1.00 M KOH would be needed to adjust the pH to 6.90? Submit Answer Tries 0/2 At pH 6.90 choose whether each species below is present to a significant degree KH2PO4 H3PO4 K3PO4 H2PO4 HPO4 K2HP4 PO4 Submit Answer Tries 0/2

Explanation / Answer

Solution

1)

a) Concentration of phosphoric acid in starting material:

Assay by weight = 85.6 %

So, 85.6 g of phosphoric acid is present in 100 g of solution.

Density = 1.69 g/ml

Volume of solution = mass of solution/density of solution = 50/1.69

= 29.58 ml = 0.02958 L

Mass of phosphoric acid = 85.6 g

molar mass of phosphoric acid = 97.999 g/mol

Moles of phosphoric acid = mass/molar mass = 85.6/97.999 = 0.873 mol

Concentration of phosphoric acid = moles/volume of solution = 0.873/0.02958 = 29.5 M

b)

Volume of phosphoric acid:

Volume of phosphate buffer = 50 ml

Concentration of phosphate buffer = 0.0500 M

By applying M1V1 = M2V2

0.0500*50 = 29.5 *V2 =

V2 = 0.085 ml

c)

concentration of KOH = 1.00 M

Volume = 25.0 ml = 0.025 L

concentration = moles of KOH/volume

1.00 = moles/0.025

Moles of KOH = 0.025 mol

molar mass of KOH = 56.11 g/mol

Mass of KOH = moles*molar mass = 0.025*56.11 = 1.40 g

Purity is 86 %

So, 86 % of x g of KOH = 1.40 g

x g = 1.40*100/86 = 1.63 g

Mass of KOH = 1.63 g

d)

pH = 6.90

pOH = 14-pH = 14-6.90 = 7.10

pOH = -log[KOH]

7.10 = -log[KOH]

[KOH] = 10-7.1 = 7.94 x 10-8 M

KOH available = 1.00 M

volume = 100 ml

By applying M1V1 = M2V2

7.94 x 10-8 x 100 = 1.000 x V2

V2 = 7.94 x 10-6 ml

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