Question

In: Chemistry

10.00 ml of seawater containing Mg2+ ions is determined by precipitation with an oxalate (C2O42-) solution....

10.00 ml of seawater containing Mg2+ ions is determined by precipitation with an oxalate (C2O42-) solution. Mg2+ + C2O42- <> MgC2O4 The precipitate was filtered, redissolved, and titrated to endpoint requiring 18.21 ml of a standard MnO4-. Beforehand, the MnO4- solution was standardized with 10.00 ml of a 2.00 * (10)-3 M oxalate solution and required 32.10 ml of MnO4- solution. Calculate the concentration of Mg2+ in the seawater sample. 2 MnO4- + 5 C2O42- + 6H+ ==> 10 CO2 + 2 Mn2+ + 8 H2O

Solutions

Expert Solution

Since it a a redox titration, it is better to use normality for all the concentrations than the molarity.

For converting molarity to normality conversion, we use the following formula.

normality (N) = n-factor x molarity; where n-factor = no. of electrons gained or lost by a species in redox reaction.

Standardizing MnO4- ​solution:

given redox reaction 2MnO4- + 5 C2O42- + 6H+ ​ 2 Mn2+ + 10 CO2 + 8H2O

MnO4- getting reduced for =7 oxidation state to +2 state. i.e. it is gaining 5 e-.

Hence n-factor for MnO4- = 5

given volume of MnO4- (V1) = 32.10 ml

Similarly C2O42- is getting oxidised from +6 to +8 oxidation state. i.e. loosing 2e-

hence n-factor of C2O42- = 2

givenmolarity of C2O42- (M2) = 2 x 10-3 M

Normality of C2O42- (N2) = 2x2x 10-3 N = 4 x 10-3 N

volume of C2O42- (V2) = 10 ml

from the law of equivalents

no. of milli equivalents of MnO4- = no. of milli equivalents of C2O42-

i.e. V1N1 = V2N2​N1 = V2N2/ V1

​ N1 = 10 x 4 x 10-3 / 32.10 = 1.2461 x 10 -3 N

Estimating concentration of Mg2+:

given Mg2+ + C2O42- MgC2O4

from the reaction no. of moles of Mg2+ = no. of moles of C2O42-

i.e. normality of Mg2+ in the solution = normality of C2O42-

normality of C2O42- can be calted as follows.

volume of MnO4- (V1) = 18.21 ml

normality of MnO4- (N1) = 1.2461 x 10-3 N

Volume of C2O42- (V2) = 10.0 ml

normality of C2O42- = N2

N1V1 = N2V2​ N2 = N1V1 / V2

hence N2 = 1.2461 x 10-3 X 18.21 / 10 = 2.2691 x 10-3 N

Hence concentration of Mg2+ in sea water = 2.2691 X 10-3 N


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