Estimation of
sodium hydroxide
To
estimate the amount of sodium hydroxide dissolved in 250 ml of the given
unknown solution volumetrically. For this you are given with a standard
solution of sodium carbonate solution of normality 0.0948 N and hydrochloric
acid solution as link solution.
Neutralization
of Sodium carbonate by HCl is given below. To indicate the end point, methyl
orange is used as an indicator.
Na2CO3 + 2HCl →
2NaCl +
CO2 + H2O
Neutralization
of Sodium hydroxide by HCl is given below. To indicate the end point,
phenolphthalein is used as an indicator.
NaOH +
HCl → NaCl + H2O
Burette
is washed with water, rinsed with HCl solution and filled with same HCl
solution up to the zero mark. Exactly 20 ml of standard Na2CO3solution
is pipetted out into the clean, washed conical flask. To This solution 2 to 3
drops of methyl orange indicator is added and titrated against HCl link
solution from the burette. HCl is added drop wise till the colour change from
straw yellow to pale pink. Burette reading is noted and the same procedure is
repeated to get concordant values.
Calculation
:
Volume
of HCl (link) solution V1 = ml
Normality
HCl (link) solution N1 = ?
N
Volume of
standard Na2CO3 solution V2 = 20
ml
Normality
of standard Na2CO3 solution N2 = 0.0948 N
According
to normality equation: V1× N1 = V2 × N2
N1
= V2 × N2 / V1
Normality
of HCl (link) solution (N1) = ------------X--------- N
Burette
is washed with water, rinsed with HCl solution and filled with same HCl
solution up to the zero mark. Exactly 20 ml of unknown NaOH solution is
pipetted out into the clean, washed conical flask. To This solution 2 to 3
drops of phenolphthalein indicator is added and titrated against HCl link
solution from the burette. HCl is added drop wise till the pink colour
disappears completely. Burette reading is noted and the same procedure is
repeated to get concordant values.
Volume of
Unknown NaOH solution V1 = 20
ml
Normality
of Unknown NaOH solution N1 = ?
N
Volume
of HCl (link) solution V2 = ml
Normality
HCl (link) solution N2 = N
According
to normality equation:
V1
x N1 = V2 x N2
N1 = V2 x N2 / V1
Normality
of Unknown HCl solution N1 =
__________Y_____ N
The
amount of NaOH dissolved in 1 lit of the solution = (Normality) x (equivalent
weight)
The
amount of NaOH dissolved in 250 ml of the solution = Normality x
equivalentweight x 250 / 1000
= Y × 40 × 250 / 1000
The
amount of NaOH dissolved in 750 ml of the solution = g
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