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

Answer:

$[Fe^{+3}]=0.700 M$

$[NO_{3}^{-}]=2.10 M$

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],

$[Fe^{+3}]=\frac{C_{1}.V_{1}}{V_{2} }$

$[Fe^{+3}]=\frac{(1.75).(30.0)}{45.0+30.0 }$

$[Fe^{+3}]=0.700 M$

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.

$[NO_{3}^{-}]=\frac{C_{1}.V_{1}}{V_{2} }$

$[NO_{3}^{-}]=\frac{(5.25)(30.0)}{30.0+45.0 }$

$[NO_{3}^{-}]=2.10 M$