Question

A solution of MgCl2 is standardized by titrating 25.00 mL aliquots of the solution with a 0.01162 M EDTA solution, using eriochrome black T as the indicator. The volumes of titrant required for the titration of three aliquots are given in the table below. Determine the concentration of the MgCl2 solution and its standard deviation. (4 pts)

aliquot Volume EDTA (mL)

1 39.83

2 40.03

3 39.97

Answer 1

moles of Mg²⁺ is equivalent to moles of Edta.

so, M₁ V₁ = M₂ V₂

where, M₁ = molar concentration of MgCl₂

V₁ = volume of MgCl₂ solution ( ml)

M₂ = molar concentration of Edta solution

V₂ = volume of Edta solution ( ml)

For, trial 1

25*M₁ = 39.83*0.01162

or, M₁ = 0.01851 M

Trial 2

25*M₁ = 40.03*0.01162

M₁ = 0.001860 M

Trial 3

25* M₁ = 39.97*0.01162

or, M₁ = 0.01857 M

Now,

average ( ) =( 0.01851+0.01860 + 0.01857)/3 = 0.01856 M

Standard deviation ( S.D.) =✓

n is trial number

or, S.D. = ✓[ ( 0.01856 - 0.01851)² + ( 0.01856 - 0.01860)² + ( 0.01856-0.01857)²/ 2]

= ✓[ (0.00005)² + (0.00004)² + (0.00001)²/2]

⁼✓[ ( 4.2*10^-9)/2]

S.D. = 4.59*10^-5