Standardization of Solution and Titration Lab Report, Preparing a Dilute HCl Solution from a Concentrated One Titrating NAOH Solution with HCl Solution (of Known Concentration)

Chemistry

International College

Objective

preparing a dilute HCl solution from a concentrated one

titrating NAOH solution with HCl solution (of known concentration)

Procedure

Section A: Preparation of 100.0cm3 0.480 mol/dm3 HCl solution

a)Determine the volume of the concentrated acid needed.

b)Introduce a small amount of the concentrated acid solution into a 50cm3 beaker.

c)Add some distilled water into the appropriate volumetric flask from the wash bottle.

d)Using a pipette, take from the beaker the necessary volume of the concentrated acid and admit it into the volumetric flask.

e)Add distilled water to obtain the required volume of the needed solution.

Section B: Preparing the basic solution to be standardized

a)Weigh approximately 2g of NaOH pellets on the electronic balance, and introduce them into 150cm3 beaker.

b)Dissolve the pellets in 100cm3 of distilled water.

c)Cover the beaker with watch glass.

How solution containing NaOH was prepared:

a)Weigh approximately 2g of NaOH pellets on the electronic balance, and introduce them into 150cm3 beaker.

b)Dissolve the pellets in 100cm3 of distilled water.

c)Cover the beaker with watch glass.

d) Insert the solution into burette using a funnel

How solution containing dilute HCl was prepared (preparation of dilute HCL is explained in data collection):

After preparing dilute HCl by determining the volume of the concentrated acid needed, and introducing 4.07 dm3 of the concentrated acid solution into a 100cm3 beaker.

We add some distilled water into the beaker from the wash bottle until the 100 cm3 mark is reached.

Data collection

Table 1: Weight of NaOH pellets and volume of distilled water in which they are introduced

Weight of NaOH pellets(2.00±0.02)g

Volume of water in which they are introduced(100.00±0.02)cm3

Table 2: Volume and concentration of HCl solution to be diluted in order to obtain HCl solution of 0.480mol/dm3 concentration

Volume of HCl solution4.07±0.02dm3

HCl concentration11cm3

Quantitative data:

Table 3: Readings from the burette

Trial*Vi ± 0.02cm3**Vf ± 0.02cm3

Rough0.008.61

1st titration8.6116.95

2nd titration16.9525.29

3d titration25.2933.62

*Vi= initial burette reading

**Vf= final reading of burette

Qualitative data:

Table 4: Aspects of acid/base mixture

Trialcolor

RoughVery dark purple pink/purple

1st titration20 % less darker than rough

2nd titrationExtremely faint pink

3d titrationFaint pink, darker than 2nd

Some notable observations:

In Erlenmeyer one drop causes the solution's color to go from colorless to dark pink

concentrated HCl solution possesses a smell that irritates the nose

when HCl solution is mixed with water for dilution, mixture becomes "wavy" in nature for a while

HCl, NAOH and phenolphthalein are colorless

NaOH pellets when exposed to air become rapidly covered with a liquid like layer

Data processing and presentation

Table 5: Volumes of the base needed to reach end point

Trial*VbE ± 0.04 cm3

1st8.34

2nd 8.34

3 d8.33

*VbE= volume of base needed to reach equivalence point

Concentration of dilute HCL solution:

n1=n2

c1xv1=c2xv2

v=(0.1x0.480)/11.8= 4.07 dm3

in order to get 100 cm3 of dilute HCL acid with concentration of 0.480mol/ dm3 we extract 4.07 dm3 from the concentrated acid, and insert the amount into an erlemeyer, then we add distilled water until the volume of the solution in Erlenmeyer reaches 100 cm3.

Sample calculation (trial 1)

VbE = Vf-Vi

= (16.95± 0.02) - (8.61± 0.02)

= 8.34 ± 0.04 cm3

Calculating concentration of the base:

HCl(aq)+NaOH(aq) NaCl(aq)+H2O(l)

At equivalence point:

n(NaOH)=n(HCl)

Cb x ΔVbE= Ca x Va

Cb= (Ca x Va)/ ΔVbE

ΔVbE= (VbE1+ VbE2 VbE3)/3

= (25.01±0.12)/3

= 8.34 ± 0.04cm3

Ca= 0.000408mol/cm3

Va= 100 ± 0.02 cm3

Cb= (Ca x Va)/ ΔVbE

= (0.000408 x 100.02± 0.02)/ 8.34± 0.02

≈ (0.005±0.029)mol/cm3

Evaluation and conclusion

Our results show that for the 1st and 2nd trial the volume of the base added are identical, but the solution's color differs. A primary standard solution should:

Be of high purity

Remain unreacted in air while weighing and remain stable (no reaction or decomposition) during storage

Have a high molar mass. The higher the molar mass, the lower the possibility of weighing errors.

React with the solution to be standardized in a direct, well-defined reaction

A solution of NaOH (in water) is one of the worst standard solutions. This is because during the preparation of the solution, some unwanted side reactions might occur. The pellets of sodium hydroxide, when exposed to the atmosphere during the weighing process, readily react with carbon dioxide (an acid) present in the air in the following manner: 2NaOH + CO2 → Na2CO3 + H2O. Furthermore, NaOH absorbs moisture rapidly from the air. This was evident because of the slippery feel experienced while handling the pellets for weighing. In addition, it is almost impossible to obtain NaOH of sufficient purity in order to use it as a primary standard*. Sodium hydroxide also reacts with glass, producing sodium silicate. This allows us to conclude that a solution of sodium hydroxide is a poor primary standard. All these factors might cause a decrease in the concentration of the base to be titrated thereby affecting the experiment's results, which explains why a larger volume of sodium hydroxide was needed as time passed by. Experimental errors also may have caused .Since the burette was not covered, a small volume of the base might have reacted with the CO2 in the air. This might explain why an identical volume of the base added to the same amount of the acid has different effects each time.

Another source of error might come from the drops of the acid on the inner sides of the Erlenmeyer and the drops of the base on the tip of the burette. By merely rinsing the tip of the burette we viewed a change in the color in the solution, hence we were adding to the acid a smaller volume of base than we thought. When we rinced the Erlenmeyer with water when the acid/base mixture was purple, the solution's color disappeard and the solution became colorless again, hence we weren't reacting 10 ml of acid with the base, but a lower volume since the drops even make a difference.