5 Steps to a 5 AP Chemistry 2019

Experimental Investigations ❮ 293

and water vapor: 2 NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(g) + CO 2 (g). The loss of mass is the

loss in mass of CO 2 + H 2 O. Examining the equation for the decomposition reaction, you

can see that there is a 1:1 ratio of moles of water and carbon dioxide.

Calculations

If you let z = moles CO 2 = moles H 2 O, then the total grams of mass lost can be shown as the

sum of the moles of each (which will be the same) times the molar mass of each substance:

Mass lost (grams) = (z × 18.02 g H 2 O/mole) + (z × 44.01g CO 2 /mole)

You can then solve for z, the number of moles. As you can see from the balanced equation,

the moles of NaHCO 3 solid that decomposed is 2z. The mass of NaHCO 3 that decom-

posed will be:

2 z × 84.02g NaHCO 3 /mole

The percent of NaHCO 3 in the mixture will be the mass of the sodium bicarbonate divided

by the mass of the mixture sample times 100%:

(grams NaHCO 3 /grams mixture) × 100%

Comments

In order to increase the precision (and hopefully the accuracy) of the determination, several

runs should be made and an average taken. This same procedure may be applied to many

other reactions and mixtures. These samples could also be analyzed by a titration procedure.

Experiment 8: Redox Titration

Synopsis

In this experiment, the concentration of a substance will be determined by using a redox

titration. The titrant will need to be standardized before it can be used in the titration.

Commonly, the redox titration involves the titration of hydrogen peroxide (H 2 O 2 ) with

potassium permanganate (KMnO 4 ), with the goal of analyzing the commercial hydrogen

peroxide that can be found in a pharmacy. The KMnO 4 solution can be standardized

against a Fe(NH 4 ) 2 (SO 4 ) 2 ·6H 2 O solution. You will prepare a standard (known concentra-

tion) solution of the Fe(NH 4 ) 2 (SO 4 ) 2 ·6H 2 O, a sulfuric acid solution, and a solution of

potassium permanganate. The redox half-reactions involved in the standardization are:

Fe^2 +(aq) → Fe^3 +(aq) + 1 e- and

MnO 4 - (aq) + 8 H+(aq) + 5 e- → Mn^2 +(aq) + 4 H 2 O(l),

giving an overall redox reaction of:

5 Fe^2 +(aq) + MnO 4 - (aq) + 8 H+(aq) → 5 Fe^3 +(aq) + Mn^2 +(aq) + 4 H 2 O(l)

The half-reactions involved in the titration of the hydrogen peroxide are:

H 2 O 2 (aq) → O 2 (g) + 2 H+(aq) + 2 e- and

MnO 4 - (aq) + 8 H+(aq) + 5 e- → Mn^2 +(aq) + 4 H 2 O(l),

giving the overall redox-reaction:

5 H 2 O 2 (aq) + 2 MnO 4 - (aq) + 6 H+(aq) → 2 Mn^2 +(aq) + 8 H 2 O(l) + 5 O 2 (g)