Standardization of Can Alkality Titrimetric Analysis

Published: 2021-06-29 06:53:53
essay essay

Category: Biographies

Type of paper: Essay

This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.

Hey! We can write a custom essay for you.

All possible types of assignments. Written by academics

Standardization of NaOH by PHP.
Determination of strength of 0.1 HCl by 0.1N NaOH.

Experiment No. 01 Standardization of NaOH by PHP.
Standardization of can alkality titrimetric analysis.
PHP reacts with NaoH giving rise to the generation of

Colour change of the end point:
Colour change runs from the colourless solution of faint pink color at the end point.
Glass ware and apparatus:
Conical flask (150 ml)
Pipette ( 20ml)
Volumetric flask ( 1000ml)
Reagents and chemicals:
PHP( )
Phenoptheline indicator.
Preparation of 0.1 N NaOH:
4 gram of NAOH was weight by balance and it was transferred to 1000ml volumetric flask containing about 1000ml of water. The flask was shaken gently to mixed. The solution was diluted to mark and mixed well.

Preparation of 0.1PHP (C_8 H_5 〖KO〗_4):
Exactly 2.04 gm PHP (phenoptheline indicator) weight by balance and transferred to 100ml volumetric flask containing about 100 ml of water. The flask was shaken gently to get substance dissolved. The solution was dilution to the mark and mixed.
Titration of 0.1N NaOH with PHP:
10ml of 0.1N NaOH was taken in a conical flask 2-3 drops of phenoptheline added. The solution was colourless and tritation was carried out with 0.1N PHP fill the end point is reached. The end point is indicated by the disappearance of the pink color get solution and three times to get the main values of the tritant
We know, Here,
(burette reading ×molecular weight of PHP)/(portence of PHP ×molecular weight of PHP) Burette point = 10.4
= (10.4×204.22/(1000×10))/(0.203 ×99.5/100) Portence of PHP = 99.5%
= (10.4×0.020422)/(0.203×0.995) Molecular weight of PHP = 204.22g
= 0.2123/0.201985
= 1.0 Factor
Normality of NaOH = 1.005 x 0.1 = 0.1005
Table: Showing titant values and normality of NaOH.
Initial burette rate (ml) Final burette rate (ml) Difference (ml) Average (ml) Normality of NaOH

Warning! This essay is not original. Get 100% unique essay within 45 seconds!


We can write your paper just for 11.99$

i want to copy...

This essay has been submitted by a student and contain not unique content

People also read