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UMSC 1124 CHEMISTRY LABORATORY I
Experiment No
:6
Title : Acid-Base Titration: Determination Of The Percentages (%) Sodium Carbonate (Na2CO3) and Sodium Hydroxide (NaOH) in a Mixture By Using pH Meter.
Name Name : Ong Ong Boon Boon Hooi Hooi
ID No : 09A 09ALB LB034 03481 81
Nam Name of of partner
: Saw Ai Ling & Wan Sh Sher Teen
Date Date of Expe Experi rime ment nt
: 24/0 24/06/ 6/20 2009 09
Semester
: June 2009
Nam Name of of lecturer
: Ms. Chan hang Chew Cheen
Title Acid-Base Titration: Determination Of The Percentages (%) Sodium Carbonate (Na2CO3) and Sodium Hydroxide (NaOH) in a Mixture By Using pH Meter.
Objective To determine the respective weight percent of sodium carbonate (Na2CO3) and sodium hydroxide (NaOH) in a mixture by using pH meter to detect the equivalent points.
Introduction A pH meter is used in the acid-base titration to determine the total alkali (Na2CO3 and NaOH) present in the sample solution. Using this technique, unknown concentration of sample solution is place into the conical flask and titrates with HCl which place in buret. The change of pH from the pH meter will be recorded and the curve of the titration can be drawn. From the observation of the graph it will shows that there will be a drastic change of Ph when it reach its equivalent points.
There is an amount of acid needed for first stage and second stage for the reaction with the sample solution. Hence, the percentage of the bases in the mixture can be determined.\
Materials
0.1M hydrochloric acid Sample solution which contains of 5g in total weight of Na2CO3 and NaOH in 1000cm3 solution. 0.1M barium chloride solution pH meter
Methodology Part A Titration 1
25cm3 of the sample solution is pipetted into conical flask.
The electrode of the pH meter is inserted to the conical flask to determine the pH of the solution.
0.50cm³ of 0.1M HCl was titrated into the conical flask.
The electrode is then been inserted again to determine the pH.
The pH reading is recorded.
A titration curve of pH versus amount of acid added into the sample is plotted.
Equivalent point can be found from the graph plotted.
Result
Initial reading (cm3)
pH
Amount of HCl added (cm3)
8.5
12.18
0.0
9.0
12.05
0.5
9.5
11.96
1.0
10.0
12.09
1.5
10.5
12.15
2.0
11.0
12.03
2.5
11.5
12.00
3.0
12.0
11.96
3.5
12.5
11.88
4.0
13.0
11.95
4.5
13.5
11.89
5.0
14.0
11.74
5.5
14.5
11.83
6.0
15.0
11.82
6.5
15.5
11.84
7.0
16.0
11.73
7.5
16.5
11.58
8.0
17.0
11.39
8.5
17.5
11.18
9.0
18.0
10.99
9.5
18.5
10.78
10.0
19.0
10.61
10.5
19.5
10.45
11.0
20.0
10.29
11.5
20.5
10.14
12.0
21.0
9.98
12.5
21.5
9.91
13.0
22.0
9.62
13.5
22.5
9.36
14.0
23.0
8.98
14.5
23.5
7.90
15.0
24.0
7.29
15.5
24.5
7.10
16.0
25.0
6.89
16.5
25.5
6.72
17.0
26.0
6.55
17.5
26.5
6.39
18.0
27.0
6.20
18.5
27.5
6.08
19.0
28.0
5.87
19.5
28.5
5.18
20.0
29.0
4.93
20.5
29.5
3.32
21.0
30.0
3.02
21.5
30.5
2.82
22.0
31.0
2.72
22.5
31.5
2.64
23.0
32.0
2.56
24.0
32.5
2.51
24.5
Calculation
15cm3 of HCl is used to reach the first equivalent point where sodium hydroxide is fully reacted. This reaction produced sodium ch loride and sodium hydrogen carbonate which is converted from sodium carbonate. The chemical equations involved are shown below, NaOH + HCl Na2CO3 + HCl
NaCl + H2O NaHCO3 + NaCl
20.5cm3 of HCl is used to reach the second equivalent point where sodium hydrogen carbonate is converted into sodium chloride.
The chemical equation involved is shown below, NaHCO3 + HCl
NaCl + H2O + CO2
\
For the chemical equation shown below, Na2CO3 + 2HCl
2NaCl + H2O + CO2
The volume of HCl required to complete the following reaction is (20.5 – 15)
2 cm3 = 11 cm3
=
=
=
2.2 10-2 mol dm-3
The concentration of Na2CO3 is 2.2x10-2 mol dm-3 The molecular weight of Na2CO3 is 106 g/mol The concentration of Na2CO3 in the sample solution in the unit of g/dm3 = (2.2x10-2) mol/dm3 * (106) g/mol
= 2.332 g/dm3
Given that the weight of the sample solution is 5 g / dm3 The weight percentage (wt%) of Na2CO3 in the sample solution
=
= 46.64%
100%
For the chemical equation shown below, NaOH + HCl
NaCl + H2O
The volume of HCl required to complete the following reaction is (15 – 5.50) cm3 = 9.5 cm3
Using:
=
= = 3.8
10-2 mol dm-3
The concentration of NaOH in the sample solution is 3.8 10-2 mol dm-3. Given that the molecular weight of NaOH is 40g/mol. The concentration of NaOH in the sample solution in the unit of g/dm3 = (3.8 10-2) mol/dm3
(40.0) g/mol
= 1.52 g/dm3 Given that the weight of the sample solution is 5 g/dm3. The weight percentage (wt %) of NaOH in the sample solution
=
= 30.40%
100%
Question s
1. Calculate the respective concentrations of Na2CO3 and NaOH in the sample solution. A. The concentration of Na2CO3 in sample solution is 2.2 x 10-2 mol dm-3 while the concentration of NaOH in sample solution is 3.8 x 10-2 mol dm-3.
2. Hence calculate the weight percentage (wt%) of each alkali in the sampl solution. A. The weight percentage of Na2CO3 in sample solution is 46.64% while the weight percentage of NaOH in sample solution is 30.40%.
Discussion This is the experiment using pH meter to determine the concentration and weight percentage of Na2CO3 and NaOH. A pH meter is an electronic instrument used to measure the pH (acidity or alkalinity) of a liquid (though special probes are sometimes used to measure the pH of semi-solid substances). A typical pH meter consists of a special measuring probe (a glass electrode) connected to an electronic meter that measures and displays the pH reading. Using pH meter to do this experiment is not an easy one. First of all, both end points are not very well visible (curve is not too steep), so accurate judgement of color is very important. Second, during titration (especially before first end point) it is possible to lose some carbon dioxide from the solution. If that happens we will overestimate amount of sodium hydroxide and underestimate amount of carbonate. Other important point is that both calculations of amount of carbonate and hydroxide are based on two measurements, which means final error can be twice as large as in the case of simple titration with one end point. The titration curve of pH versus the amount of HCl added into the sample was plotted and it showed a curve with two buffer regions and two equivalence points. This is because there are two different compounds in the sample solution, which is strong base and another one is weak base. It is also show that the equivalence points and doesn’t always occur at the pH7 due to it has to depend on how much of the OH- or H+ moles is exhibited in the solution being titrated. From the calculation we get, the weight percentage of sodium hydroxide is 30.40% while the weight percentage of sodium carbonate is 46.64%. Precaution steps •
In this experiment, it is advise that we must wear glove when doing this experiment. This is because the chemical is strong acid.
•
The pH metre has to be calibrate before using it with 2-3 buffer solution which have the pH 4 , pH 7 and pH 10.
Conclusion
• •
The weight percentage of Na2CO3 is 46.64%. The weight percentage of NaOH is 30.40%. .
Reference •
Reference Book
Graham Hill and John Holman (2000). Fifth Edition Chemistry: Chemistry In Context. Nelson Edition.
