You wish to prepare 100. milliliters of 0.0400 M phosphate buffer, pH 6.50, from

Question

You wish to prepare 100. milliliters of 0.0400 M phosphate buffer, pH 6.50, from phosphoric acid [a liquid; specific gravity = 1.69, assay by weight = 85.6%; MW = 97.999] and potassium hydroxide [a solid; purity 86%; MW = 56.11].

What is the molar concentration of the phosphoric acid starting material?

=14.8 M

What volume (in milliliters) of phosphoric acid would you need to make the buffer?

=??

What mass (in grams) of KOH would you need to prepare 25.0 mL of 1.00 M KOH?

=1.63 g

What volume (in milliliters) of the 1.00 M KOH would be needed to adjust the pH to 6.50?

=??

At pH 6.50, what is the concentration (in millimolar) of the conjugate base?

=??

Explanation / Answer

1) 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 = 100/1.69 = 59.17 ml = 0.059 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 = 0.873/0.059 = 14.80 M

2) Volume of phosphoric acid:

Volume of phosphate buffer = 100 ml

Concentration of phosphate buffer = 0.0400 M

By applying M1V1 = M2V2

0.0400*100 = 14.80*V2

V2 = 0.27 ml

Volume of phosphoric acid = 0.27 ml

3) 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

4) pH = 6.50

pOH = 14-pH = 14-6.50 = 7.50

pOH = -log[KOH]

7.50 = -log[KOH]

[KOH] = 3.16*10-8 M

KOH available = 1.00 M

volume = 100 ml

By applying M1V1 = M2V2

3.16*10-8*100 = 1.000*V2

V2 = 3.16*10-6 ml

Volume of KOH used = 3.16*10-6 ml

5) pH = 6.50

pKa of phosphoric acid = 2.12

pH = pKa + log [A-]/[HA]

6.50 = 12.67 + log [A-]/14.8

-6.17 = log[A-]/14.8

[A-]/14.8 = 6.76*10-7

[A-] = 1.00*10-5 M

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