Question

What is the conecntration of Fe3 and the concentration of No3- present in the solution that result when 30.0 ml of 1.75M Fe(No3)3 are mixed with 45.0 ml of 1.00M Hcl?

Answer 1

`[Fe^(+3)]=0.700 M`

`[NO_(3)^(-)]=2.10 M`

Step-by-step explanation:

Here, a solution of Fe(NO₃)₃ is diluted, as the total volume of the solution has increased. The formula for dilution of the compound is mathematically expressed as:

`C_(1). V_(1)= C_(2).V_(2)`

Here, C and V are the concentration and volume respectively. The numbers at the subscript denote the initial and final values. The concentration of Fe(NO₃)₃ is 1.75 M. As ferric nitrate dissociates completely in water, the initial concentration of ferric is also 1.75 M.

Solving for [Fe],

`[Fe^(+3)]=(C_(1).V_(1))/(V_(2) )`

`[Fe^(+3)]=((1.75).(30.0))/(45.0+30.0 )`

`[Fe^(+3)]=0.700 M`

For [NO₃⁻],

There are three moles of nitrate is 1 mole of Fe(NO₃)₃. This means that the initial concentration of nitrate ions will be three times the concentration of ferric nitrate i.e., it will be 5.25 M.

`[NO_(3)^(-)]=(C_(1).V_(1))/(V_(2) )`

`[NO_(3)^(-)]=((5.25)(30.0))/(30.0+45.0 )`

`[NO_(3)^(-)]=2.10 M`