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Department of Pure and Applied Chemistry: Analysis of An Antacid

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DEPARTMENT OF PURE AND APPLIED CHEMISTRY

College of Arts and Sciences


VISAYAS STATE UNIVERSITY
Visca, Baybay City, Leyte
Chem 140 Laboratory

Name: Honey Fher I. Casido Date Performed: October 17, 2018


Lab Schedule: Wednesday (R196) 7:00-10:00 Date Submitted: October 24, 2018
Group No.: 2 Rating:____________________

EXERCISE NO.6
Analysis of an Antacid

I. Objectives

Materials

Acid buret 0.1 N HCl


Base buret 0.1 N NaOH
Buret Clamp phenolphthalein
250mL Erlenmeyer flask Antacid sample
Iron stand

II. Results

HCl
Trial # Initial reading (mL) Final reading (mL) Titrant used(mL)
1 28.7 29.4 0.7
2 29.4 30 0.6
3 30 30.7 0.7

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NaOH
Trial # Initial reading (mL) Final reading (mL) Titrant used (mL)
1 31.4 31.7 0.3
2 31.7 32 0.3
3 32 32.3 0.3

Computation Flow
0.103mol
Mole of HCl n= x volume of HCl
L

0.103mol
Trial 1 n= x 0.0007 L of HCl n = 7.25 x 10-5 mol.
L

0.103 mol
Trial 2 n= x 0.0006 L of HCl n = 6.18 x 10-5 mol.
L

0.103mol
Trial 3 n= x 0.0007 L of HCl n = 7.25 x 10-5 mol.
L

0.106
Mole of NaOH n= x volume of NaOH
L

0.106 mol
Trial 1 n= x 0.0003 L of NaOH n = 3.18 x 10-5 mol.
L

0.106 mol .
Trial 2 n= L x 0.0003 L of NaOH n = 3.18 x 10-5 mol.

0.106 mol .
Trial 3 n= L x 0.0003 L of NaOH n = 3.18 x 10-5 mol.

Compute mole of HCl reacted with antacid


Mol= mol. Of HCl – mol. Of NaOH
Trial 1 = (7.25 x 10-5 mol HCl) – (3.18 x 10-5 mol. NaOH) T1 mol = 4.07 x 10-5 mol NaOH
Trial 2 = (6.18 x10-5 mol HCl) – (3.18 x 10-5 mol NaOH) T2 mol = 3.10 x 10-5 mol NaOH
Trial 3 = (7.25 x 10-5 mol HCl) – (3.18 x 10-5 mol. NaOH) T1 mol = 4.07 x 10-5 mol NaOH

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III. Discussion

A.Preparation of the acid/base burets, preparing the two buret for 0.1 N NaOH and 0.1 N
HCl by rinsing each buret with each solution allowing the liquid to run out through the tip
of the buret

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IV. Conclusion

V. Answers to Questions

1. Write the chemical equations representing the chemical reactions involved in this
Experiment.
Answer: reaction of antacid in human stomach
HCl (aq) > H+(aq) + Cl- (aq)
H+ (aq) + OH- (aq) > H2O(i)
2 H+ (aq) + CO32- (aq) > H2O(l) + CO2(g)
Distillation of antacid in water and titration using acids and basis
Mg (OH)2 + CO32-(aq) Mg 2+ + 2Cl- + 2H2O
CaCO3 + 2HCl Ca2+ + 2Cl- + 2H2O

Tablet [mg(OH)2 | CaCO3] + HCl > neutralized tablet + excess acid > acidic
solution
Excess HCl + NaOH > neutral solution

2. Briefly explain why (a) each buret must be rinsed with the solution to be used, (b) any
air space at the alkali buret nozzle must be removed before titration.
Answer:
(a) Each buret must be rinsed with the solution because when cleaning the glassware
particularly buret, rinse it off with the solution to be used in the experiment. If the
buret is not completely dry the time to use it for the experiment, the remaining traces

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of the water on the inside will make your titrant more dilute and change its
concentration.
(b) Any air space at the alkali buret nozzle must be removed before titration because a
bubble in the nozzle of buret will produce an inaccurate volume reading if the bubble
escapes during titration. During a titration such small bubbles begin to move in the
direction of the nozzle but may remain in place even though there is a moderate flow
of titrant (above right). Even when the buret valve is wide open some bubbles remain
in place until yountake your eyes off them. Then they sneak through the nozzle and
ruin the titration.

3. Calculate the theoritical number of moles HCl equivalent to 1 mole of (a) NaHCO 3
(b) CaCO3 (c) Al2O3.

4. How many miligrams of NaHCO 3 is in a 500-mg tablet if 40.00 mL of 0.12N HCl is


required for neutralization of this sample?

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