Question

A aliquot of solution containing of in required of EDTA solution for titration. How many milligrams of will react with of this EDTA solution?

Answer 1

The answer is "
`\bold{1.39 \ mg}`".

Step-by-step explanation:

Please find the complete question in the attached file.

Total Moles of
`Mg_2^(+)` = moles of
`MgSO_4`

`= (mass)/(molar \ mass \ of MgSO_4)\\\\= (0.450)/(120.37)\\\\ = 0.0037385 \ mol`

`Mg_2^(+) + EDTA4^(-) \longrightarrow Mg(EDTA)2^(-)`

EDTA mol in 37.6 mL of solution = 50.0 mL of
`Mg2^(+)`

`= (50.0)/(500) *` total moles of
`Mg2^(+)`

`= (50.0)/(500) * 0.0037385\\\\= 3.7385 * 10^((-4)) \ mol\\`

`Ca2^(+) + EDTA4^(-) \longrightarrow Ca(EDTA)2^(-)`

`CaCO_3 Moles = Ca2^(+) Moles =` EDTA moles in a solution of 1.40 mL

`= (1.40)/(37.6) * 37.6 \ mL` the solution of EDTA moles.

`= (1.40)/(37.6) * 3.7385 * 10^((-4)) \\\\= 1.392 * 10^((-5))\ mol\\\\`

Mass of
`CaCO_3 =`mole
`*` the molar mass of
`CaCO_3`

`= 1.392 * 10^((-5)) * 100.09\\\\= 0.00139 \ g\\\\ = 1.39 \ mg`