Question

1) A Na3PO4 standard solution is prepared by transferring 3.50 g of primary standard-grade sodium carbonate to a 500.0-mL volumetric flask, dissolving the sample in ~250 mL of distilled deionized water and diluted to the mark. A 50.00-mL aliquot is taken and titrated with 46.65 mL of HCl solution. Calculate the concentration of the HCl solution.

please show your work clear and net answer

2) If the measured potential of a half-cell is 0.584 V when measured against the S.C.E, what is the potential if the measurement is made using the saturated Ag-AgCl reference electrode? E = 0.197 V for the saturated Ag-AgCl electrode and E = 0.241 V for the S.C.E.

Answer 1

ans 1.

Molecular mass of Na3PO4 = 163.94 gmol^-1

Given mass = 3.5 g

Mole of Na3PO4 = (3.50/163.94)g g^-1mol = 0.0214mol

Given volume = 500mL = 0.500L

Concentration of Na3PO4 (M₁) = mole/volume = 0.0214mol/ 0.500L = 0.04270M

In titrations Volume of Na3PO4 (V₁) = 50mL

Let Concentration of HCl (M₂) and given HCl Volume (V₂) is = 46.65mL

Now using relation

M_1*V1₌M_2*V₂

Putting value

M₂ = 0.04270M* 50mL/46.65mL = 0.04576M

Hence concentration of HCl is 0.04576M

ans 2.

Given that the measured potential of a half-cell is 0.584 V is measured against the S.C.E

S.C.E. = 0.241 V , we know that this value is respect to S.H.E.( Standard Hydrogen Electrode)

So with respect to S.H.E. potential of a half-cell is given as= 0.584V - 0.241 V = 0.343V

We also know that saturated Ag-AgCl electrode reference potential E = 0.197 V with respect to S.H.E.

Hence potential of a half-cell with respect to Ag-AgCl electrode is E = 0.197 V + 0.343V = 0.540V

Hence potential of a half-cell with respect to Ag-AgCl electrode is E = 0.540V