Question

Calculate the final concentration of each of the following:

a. $2.0Lofa6.0MHCl$solution is added to water so that the final volume is $6.0L.$

b. Water is added to $0.50Lofa12MNaOH$solution to make $3.0L$of a diluted $NaOH$solution.

c. A $10.0mL$sample of a $25\%(m/v)KOH$solution is diluted with water so that the final volume is $100.0mL.$

d. A $50.0mL$sample of a $15\%(m/v)H_{2}S{O}_4$solution is added to water to give a final volume of $250mL$

Short answer

(part a) The concentration of the given solution is $M_2=2.0M$

(part b) The final concentration of the solution provided is as$M_2=2.0M$

(part c) The concentration of$KOH$solution is as$2.5\%(m/v)KOH$

(part d) The final concentration of$H_{2}S{O}_4$solution is derived as$3.0\%(m/v)H_{2}S{O}_4$

Step-by-step solution

Introduction (part a).

A solution concentration is an indicator of the total of drop wise in a given amount of solvent or solution. A saturated solution is one which includes a relatively large amount of dissolved solute.

Given Information (part a).

The following data is displayed to determine the concentration of the solution:

Volume of $HCl,V_1=2.0L$

Concentration of $HCl,M_1=6.0M$

After adding $HCl$to water, Then the volume of solution is$V_2=6.0L$

Explanation (part a).

To find the final concentration $M_2,$

role="math" localid="1652680959610" $$\begin{gathered} M_1{V}_1=M_2{V}_2 \\ 6.0M\times 2.0\mathrm{L}=M_2\times 6.0\mathrm{L} \\ {\mathrm{M}}_2=\frac{6.0\mathrm{M}\times 2.0\mathrm{L}}{6.0\mathrm{L}} \\ =2.0\mathrm{M} \end{gathered}$$

Given Information (part b).

The following data is displayed to determine the concentration of the solution:

Concentration of $NaOH,M_1=12M$

Volume of $NaOH,V_1=0.5L$

After the mixing of water to$NaOH,$Volume is$V_2=3.0\hspace{0.17em}L$

Explanation (part b).

To know the final concentration as,

$$\begin{gathered} M_1{V}_1=M_2{V}_2 \\ 12M\times 0.5\mathrm{L}=M_2\times 3.0\mathrm{L} \\ {\mathrm{M}}_2=\frac{12\mathrm{M}\times 0.5\mathrm{L}}{3.0\mathrm{L}} \\ =2.0\mathrm{M} \end{gathered}$$

Given Information (part c).

The following data is displayed to determine the concentration of the solution:

Volume of $KOH$solution, $V_1=10.0mL$

Concentration of the solution is$25\%(m/v)KOH$

Explanation (part c).

Find the weight of the solution in $10.0mL$solution is,

localid="1653895266022" $$\begin{gathered} \text{Mass of}\mathrm{KOH}\text{in}10.0\mathrm{mL}\text{solution}=\frac{25\mathrm{g}}{100\mathrm{mL}}\times 10.0\mathrm{mL} \\ =2.5\mathrm{g} \end{gathered}$$

Then find out the concentration of $KOH$contains $2.5g$in the concentration solution is,

localid="1653895249172" $$\begin{gathered} \text{Final concentartion}(\mathrm{m}/\mathrm{v})=\frac{2.5\mathrm{g}}{100\mathrm{mL}}\times 100 \\ =2.5\%(m/v) \end{gathered}$$

Given Information (part d).

The following data is displayed to determine the concentration of the solution:

Volume of $H_{2}S{O}_4$solution is $V_1=50.0mL$

The concentration is provided as $15\%(m/v)H_{2}S{O}_4$

Explanation (part d).

Find the weight of the solution in $50.0mL$of solution is,

localid="1653895224802" $$\begin{gathered} \text{Mass of}{H}_{2}S{O}_4\text{in5}0.0\mathrm{mL}\text{solution}=\frac{15\mathrm{g}}{100\mathrm{mL}}\times 50.0\mathrm{mL} \\ =7.5\mathrm{g} \end{gathered}$$

Therefore find out the concentration of $H_{2}S{O}_4$containing $50.0mL$of concentration solution is ,

localid="1653895235372" $$\begin{gathered} \text{Final concentartion}(\mathrm{m}/\mathrm{v})=\frac{7.5\mathrm{g}}{250\mathrm{mL}}\times 100 \\ =3\%(m/v) \end{gathered}$$