Wednesday, February 27, 2013

Determination of the concentration of sulphuric acid by titration

Determination of the concentration of sulphuric tart by titration

begin
To determine the concentration of a solution of sulphuric caustic by titration with sodium hydroxide

Apparatus
250mL E-flask, stirrer with magnet, burette +0.1 ml, 10.0ml pipette

Chemical
H2SO4(aq), 0.100M NaOH(aq), BTB (pH-indicator)

Experimental
The burette is rinsed by and filled up by NaOH-solution. It is required that no transfer bubbles present in the burette. 10.0ml of the H2SO4 is filled into the E-flask and diluted by distilled water supply to about 100 cm3. wherefore 10 drops of BTB and a magnetized stirring bar are added to it. After dropping 5 drops of BTB, the H2SO4 starts to change colour to pale yellow. Furthur dropping 5 more than drops, it becomes yellow. The E-flask is placed on the magnetic stirrer. The liquid is stirred. H2SO4 is titrated with NaOH. The H2SO4 is observed to create a drop of ghastly colour liquid when adding a drop of NaOH. However, it gradually disappear. The titration is considered to be end until the colour turns to green. Then the above procedure is repeated twice. The collected data is shown in table 1


Table 1. Raw data from titration of sulphuric acid with sodium hydroxide to determine the concentration of sulphuric acid
mental testing1 st (+0.1ml)2nd (+0.1ml)3rd (+0.1ml)
Vol. Of NaOH used when a blue drop starts to appear but gradually disappear12.2 ml11.7 ml12 ml
Vol.

Ordercustompaper.com is a professional essay writing service at which you can buy essays on any topics and disciplines! All custom essays are written by professional writers!

Of NaOH used when the colour is green20.0ml19.3 ml19.4 ml




Calculations
The equation of titration: 2NaOH + H2SO4  2H2O + Na2SO4
Average of volume of NaOH used for titration: (20.0+19.3+19.4)/3=19.57ml
 come up of NaOH = Concentration of NaOH
Volume of NaOH
fare of NaOH =0.1
0.1957
Amount of NaOH =1.957 x 10-3 mol

 mill ratio of NaOH and Na2SO4 : 2:1
NaOH =0.5 Na2SO4
Amount of Na2SO4: 1.957 x 10-3 /2 =9.785x 10-4 mol
Amount of Na2SO4 = Concentration of Na2SO4
Volume of Na2SO4
9.785x 10-4 =...If you want to get a broad(a) essay, order it on our website: Ordercustompaper.com



If you want to get a full essay, wisit our page: write my paper

No comments:

Post a Comment