# 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**.

#### Learn more about this topic:

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Chapter 8 / Lesson 5Want to know how to calculate dilution factor? See dilution equations, the dilution formula, and learn how to dilute acid and how to dilute a solution.