Question

Volume Barium chlol Cond 0.0 5.0 10.0 15.0 20.0 25.0 27.0 29.0 31.0 32.0 33.0 34.0 35.0 40.0 42.0 44.0 48.0 50.0 2.07 2.12 2.16 2.25 2.30 2.35 2.38 2.42 2.50 2.55 2.61 2.67 2.72 3.01 3.11 3.21 3.41 3.50

Refer data from conductometric titration of a solution containing Mg2+ and SO42- with BaCl2.

(a) Determine the concentration of sulfate ion in the 40mL aliquot if it was titrated with 0.05M BaCl2 solution.

(b) Use these molar ionic conductances to explain the SHAPE of the titration curve.

Cl- = 76.4
Ba2+ = 127.3
SO4 2- = 160.0
Mg2+ = 106.0

Answer 1

This is a type of Conductometric Precipitation titration where precipitates of BaSO₄ are formed.

BaCl₂(aq) + MgSO₄(aq) → MgCl₂(aq) + BaSO₄(s)

OR

Ba2+ (aq) + SO42- (aq) → BaSO4 (s) White precipitates are formed. ans.a.)  As titration proceeds, each chloride is being replaced by a sulphate ion. From the given data ,we find that initialy there is very slow increase in the conductance. After reaching equivalent point when all the SO42- ion is consumed, further addition of simply add more Ba2+and Cl- in solution and hence conductance increase suddenly as we go from 35 mL to 40 mL. thus the Equivalance point must me 35mL. Given that, Concentration of BaCl2, M1 = 0.05M and Volume of BaCl2 from the above data is, V1 = 35mL Volume of MgSO4 , V2 = 40mL Thus we will use the equation, M1V1 = M2V2 Concentration of MgSO4 is,M2 0.05 × 35 = M2 × 40 M2 = (0.05 × 35) /40  = 0.04475 ≈ 0.045 M