Topic : Volumetric analysis – Stoichiometry
Purpose : To determine the exact concentration of a monobasic acid, HX.
Material : i) Monobasic acid, HX.
ii) Solid sodium hydroxide,NAOH.
Apparatus : i) One volumetric flask 250ml and stopper.
ii) Electronic balance ±0.01g
iii) Three tiltration flask.
iv) One 25 cm³ pipette and pipette filler
v) One 50 cm³ burette.
vi) One retort stand and clamp.
vii) One white tile.
viii) One wash bottle filled with distilled water.
Theory : Neutralization process occurred between acid and base. Then H+ ion from acid react with OH־ ion from the base to form H2O.
Procedure : a) i) A piece of filter paper is weighted using the electronic weight.
ii) A spoon of solid sodium hydroxide,NAOH is taken from the container
using spatula and weighted.
iii) The reading of the measurement is recorded.
iv) The sodium hydroxide,NAOH is then put into the beaker .
v) A little amount of distilled water is filled into the beaker and stirred using a glass rod to dissolve the solid sodium hydroxide , NaOH .
vi) The solution is then transferred into 250cm³ volumetric flask
vii) Using the 25 cm³ pipette to pipette 25 cm³ of NAOH solution into the
viii) Monobasic acid, HX solution is poured into the 50ml burette.
ix) Titrate the solution as well.
x) Record your readings in the table below.
xi) Repeat the tiltration as many as you think necessary to achieve
Results : b) Record your tiltration reading in the table below :
Third I Average
11.40 I 11.36
0 I 0
Volume of KA1/cm
11.40 I 11.36
( ii ) Calculate your average tiltrate value showing the suitable tiltrate values.
Average volume : 21.2 + 20.3 + 20.0 cm³
3= 20.5 cm³
Average volume = 20.5 cm³
Calculation : Mass of filter paper : 0.0 g
Mass of filter paper and solid sodium hydroxide : 4.0 g
Mass of solid sodium hydroxide : 4.0 g – 0.0 g= 4.0 g
= 40.0 g mol־¹
Relative molecular mass of NaOH = 23.0 + 16.0 + 1.0
No. of mole of NaOH = ___4.0 g___ Molarity of NaOH = _0.1 mol_
40.0 g mol־ 0.25 dm³
HX + NaOH → NaOH + H2O
From the equation, 1 mole of HX react with 1 mole of NaOH.
Concentration of monobasic acid, HX 4 M1V1 = M2V2
M2 = 0.4 × 25
= 0.49 mol dm־³
Conclusion : 20.5 cm³ of HX react with 25.0 cm³ of 0.40 moldm־³ of NaOH to produce
water and salt. The concentration of HX obtained through calculation is
0.49 mol dm־³.
Comment : The value obtained is slightly different from the theory because there is
some parallax error when the experiment is carried on.