Q. 9.55 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the final concentration of each of the following:

a. $2.0 L of a 6.0M HCl$ solution is added to water so that the final volume is $6.0 L.$

b. Water is added to $0.50 L of a 12M NaOH$ solution to make $3.0 L$ of a diluted $NaOH$ solution.

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

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

Step-by-Step Solution

Answer

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

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

(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 $3.0\% (m/v) H_{2}S{O}_4$

Step1: 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.

Step2: Given Information (part a).

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

Volume of $HCl, V_1=2.0 L$

Concentration of $HCl, M_1=6.0 M$

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

Step3: Explanation (part a).

To find the final concentration $M_2,$

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

Step4: Given Information (part b).

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

Concentration of $NaOH, M_1=12M$

Volume of $NaOH, V_1=0.5 L$

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

Step5: Explanation (part b).

To know the final concentration as,

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

Step6: Given Information (part c).

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

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

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

Step7: Explanation (part c).

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

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

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

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

Step8: 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 $V_1=50.0 mL$

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

Step9: Explanation (part d).

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

$$\begin{gathered} \text{ Mass of }{H}_{2}S{O}_4\text{ in 5}0.0\mathrm{mL}\text{ solution }=\frac{15 \mathrm{g}}{100 \mathrm{mL}}\times 50.0 \mathrm{mL} \\ =7.5 \mathrm{g} \end{gathered}$$

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

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