Experiment-4


FE 211 25.10.2004

EXPERIMENT 4

LAB REPORT

Submitted By : Mutlu DEMİREL Group: 3

MATERIALS:

HCl , HAc , Phenolphthalein , Methyl Orange , NaOh , Dilute Water ,Buret ,

Erlenmayer Flask ,

PURPOSE:

In this experiment we titrated HCI and HAc mixture with NaOH to determine

amounth of HCI , HAc and NaOH and the molarity of NaOH.

METHOD:

Gravimetric Analysis

THEORY:

The capacity of water to accept protons is called alkalinity. Alkalinity is important in water treatment and in the chemistry and biology of natural waters. Alkalinity serves as a pH buffer and reservoir for inorganic carbon. It helps to determine the ability of water to support algal growth and other aquatic life; thus, it can be used as a measure of water fertility. In natural waters, the species responsible for alkalinity are OH-, HCO3-, and CO32-. Alkalinity is equal to [OH-] + [HCO3-] + 2[CO32-]. The coefficient of “2” before [CO32-] is necessary because carbonate accepts two protons. However, in natural waters of pH from 7 – 9, the predominant species is bicarbonate.

Alkalinity can be measure by titration with a standard solution of HCl. The titration reaction for the bicarbonate ion is: HCO3- + H+ _ H2CO3 The endpoint can be determined with the methyl orange indicator or by measuring the pH of the solution throughout the titration. In the later case, analysis of the titration curve allows determination of the end point. Alkalinity is expressed as the number of moles of H+ required to titrate one liter of water sample or as mg CaCO3/L of water.

PROCEDURE:

5 ml HCI and 5 ml HAc was poured into a erlenmayer flask. Then 2-3 drops of methyl orange indicator was added to this solution and titrated with standart solution until the color of the solution change. Then 2-3 drops of phenolphthalein was added to the solution and again titrated with NaOH until the end point was observed.

RESULTS AND CALCULATION :

HCI = n NaOH fw of HCI = 36.5 g volume of NaOH = 8 mL

NaOH =

NaOH = n HCI = 0.0008 mol

fw of

volume of NaOH = 11 mL

NaOH = n

NaOH =

NaOH = n = 0.0011 mol

DISCUSSION:

In this titration reaction we have a mixture of strong acid , HCI , and weak acid HAc , as primary solution. The mixture of acids were titrated with base , NaOh because the Ka for weak acid was abouth or less so two end point was observed. The first end point was for HCI , the second was for HAc . To observe the end points of this two different acids, two different indicators were used. Methyl orange for HCI and Phenolphthalein for HAc.

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