# What is the final concentration of the solution produced when 225.5 ml of a 0.09988 M solution of...

## Question:

What is the final concentration of the solution produced when {eq}225.5 \ ml {/eq} of a {eq}0.09988 \ M {/eq} solution of {eq}Na_2CO_3 {/eq} is allowed to evaporate until the solution volume is reduced to {eq}45.00 \ ml {/eq}?

## Dilution Equation:

The equation which is utilized for the determination or evaluation of an unknown concentration in mol/L is named as the dilution equation. The dilution equation consists of two volumes along with two concentrations of the given solution.

## Answer and Explanation: 1

Given Data:

• The volume of {eq}{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}} {/eq} is {eq}{V_1} = 225.5{\rm{ mL}} {/eq}.
• The concentration of {eq}{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}} {/eq} is {eq}{M_1} = 0.09988{\rm{ M}} {/eq} .
• The reduced volume of the solution is {eq}{V_2} = 45.00\;{\rm{mL}} {/eq}.

The final concentration that is {eq}{M_2} {/eq} of the solution can be calculated as given below.

{eq}{M_2} = \dfrac{{{M_1}{V_1}}}{{{V_2}}} {/eq}

Where,

• {eq}{M_2} {/eq} is the final concentration.
• {eq}{V_2} {/eq} is the reduced volume.
• {eq}{M_1} {/eq} is the concentration of {eq}{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}} {/eq}.
• {eq}{V_1} {/eq} is the volume of {eq}{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}} {/eq}.

Substitute the respective values in above equation.

{eq}\begin{align*} {M_2} &= \dfrac{{0.09988{\rm{ M}} \times 225.5{\rm{ mL}}}}{{45.00\;{\rm{mL}}}}\\ &= \dfrac{{22.5}}{{45}}{\rm{M}}\\ &= 0.5\;{\rm{M}} \end{align*} {/eq}

Therefore, the final concentration of the solution produced is 0.5 M.