Illustrated Guide to Home Chemistry Experiments

204 DIY Science: Illustrated Guide to Home Chemistry Experiments

LABORATORY 11.4:

STANdARdIzE A HydRoCHLoRIC ACId SoLUTIoN By TITRATIoN

The process used to determine the

concentration of a solution with very high

accuracy is called standardizing a solution.

To standardize an unknown solution, you react

that solution with another solution whose

concentration is already known very accurately.

RIREEqU d EqUIpmENT ANd SUppLIES

£ goggles, gloves, and protective clothing

£ balance and weighing papers

£ beaker, 150 mL (2)

£ volumetric flask, 100 mL

£ funnel

£ pipette, 10 mL

£ burette, 50 mL

£ ring stand

£ burette clamp

£ storage bottle, 100 mL (labeled “sodium

carbonate, 1.500 m”)

£ hydrochloric acid, 1 m bench solution (~10 mL)

£ sodium carbonate, anhydrous (15.90 g)

£ phenolphthalein indicator solution (a few drops)

£ distilled or deionized water

For example, to standardize the hydrochloric acid solution that
we made up in a preceding lab, we might very carefully measure
a known quantity of that solution (called an aliquot) and
neutralize that aliquot with a solution of sodium carbonate whose
concentration is already known very accurately. Adding a few
drops of an indicator, such as phenolphthalein or methyl orange,
to the solution provides a visual indication (a color change)
when an equivalence point is reached, when just enough of the
standard solution has been added to the unknown solution to
neutralize it exactly. By determining how much of the sodium
carbonate solution is required to neutralize the hydrochloric acid,
we can calculate a very accurate value for the concentration of
the hydrochloric acid. This procedure is called titration.

Titration uses an apparatus called a burette (or buret), which
is a very accurately graduated glass cylinder with a stopcock or
pinchcock that allows the solution it contains to be delivered in
anything from a rapid stream to drop-by-drop. Because titration
is a volumetric procedure, the accuracy of the results depends
on the concentration of the reagent used to do the titration. For
example, if 5.00 mL of 1.0000 M sodium carbonate is required
to neutralize a specific amount of the unknown acid, that same
amount of acid would be neutralized by 50.00 mL of 0.10000 M
sodium carbonate. If our titration apparatus is accurate to 0.1 mL,
using the more dilute sodium carbonate reduces our level of error
by a factor of 10, because 0.1 mL of 0.10000 M sodium carbonate
contains only one tenth as much sodium carbonate as 0.1 mL of
1.0000 M sodium carbonate. For that reason, the most accurate
titrations are those performed with a relatively large amount of a
relatively dilute standard solution.

The obvious question is how to obtain an accurate reference
solution. For work that requires extreme accuracy, the best
answer is often to buy premade standard solutions, which are
made to extremely high accuracy (and the more accurate, the
more expensive). None of the work done in a home lab requires
that level of accuracy, so the easiest and least expensive method
is to make up your own standard solutions. (In fact, to illustrate
the principles of standardization and titration, you don’t even

need a truly accurate reference solution; you can simply

pretend that a 1 M solution is actually 1.0000 M and proceed

on that basis. Your results won’t be accurate, but the principles

and calculations are the same.)

When you make up a standard solution, take advantage

of the difference between absolute errors and relative errors.

For example, if your balance is accurate to 0.01 g, that means

that any sample you weigh may have an absolute error of as

much as 0.01 g. But that absolute error remains the same

regardless of the mass of the sample. If you weigh a 1.00 g

sample, the absolute error is 1% (0.01 g/1.00 g · 100). If you weigh

a 100.00 g sample, the absolute error is 0.01% (0.01 g/100.00 g

· 100). Similarly, volumetric errors are absolute, regardless of the

volume you measure. For example, a 10.00 mL pipette may have

an absolute error of 0.05 mL. If you use that pipette to measure

10.00 mL, the relative error is (0.05 mL/10.00 mL · 100) or 0.5%.

If you measure only 1.00 mL, the relative error is ten times as

large, (0.05 mL/1.00 mL · 100) or 5.0%.