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College Chemistry : Reactions

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Example Questions

Example Question #1 : Acid Base Reactions

Screen shot 2016 03 24 at 9.10.57 pm

Determine the volume in mL of HCl solution it would take to neutralize $35.0\ mL$ of a $0.0025\ M$ NaOH with a $0.012\ M$ HCl solution during a titration.

Possible Answers:

$131.0\ mL$

$23.0\ mL$

$24.0\ mL$

$72.0\ mL$

Correct answer:

$72.0\ mL$

Explanation:

Determine the moles of NaOH used:

$0.035\ L\times \frac{0.025\ moles}{L}=8.75\times10^{-4}\ moles\ NaOH$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ HCl=8.75\times10^{-4}\ moles$

Convert the moles of HCl to liters:

$\frac{L}{0.012\ moles}\times8.75\times10^{-4}\ moles=0.072\ L\ of HCl$

Convert the liters to milliliters:

$\frac{1\ mL}{10^{-3}L}\times0.072\ L=72.0\ mL\ HCl$

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Example Question #1 : Titrations

Screen shot 2016 03 24 at 9.10.57 pm

Determine the volume in mL of HCl solution it would take to neutralize $42.0\ mL$ of a $0.0021\ M$ NaOH with a $0.014\ M$ HCl solution during a titration.

Possible Answers:

$6.3\ mL$

$40.5\ mL$

$1.2\ mL$

$4.2\ mL$

Correct answer:

$6.3\ mL$

Explanation:

Determine the moles of NaOH used:

$0.042\ L\times \frac{0.0021\ moles}{L}=8.82\times10^{-5}\ moles\ NaOH$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ HCl=8.82\times10^{-5}\ moles$

Convert the moles of HCl to liters:

$\frac{L}{0.014\ moles}\times8.82\times10^{-5}\ moles=0.0063\ L\ of HCl$

Convert the liters to milliliters:

$\frac{1\ mL}{10^{-3}L}\times0.0063\ L=6.3\ mL\ HCl$

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Example Question #11 : Acid Base Reactions

NaOH reacts with $CH_{3}COOH$ in aqueous solution according to the chemical equation provided. If it took $32.0\ mL$ of a $0.1\ M$ NaOH solution to titrate $25.0\ mL$ of an $CH_{3}COOH$ solution, what was the concentration of original $CH_{3}COOH$ solution?

Possible Answers:

$1.4\ M$

$0.13\ M$

$0.0032\ M$

$0.25\ M$

Correct answer:

$0.13\ M$

Explanation:

Using the concentration of NaOH as a conversion factor, we can convert the volume of NaOH to moles by dimensional analysis:

$\frac{0.1\ moles}{L}\times0.032\ L=0.0032\ moles\ NaOH$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.0032\ moles$

$Molarity=\frac{0.0032\ moles}{0.025\ L}=0.13\ M$

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Example Question #1 : Titrations

Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . Acetic acid reacts with NaOH in aqueous solution according to the chemical equation provided. Determine the percentage (by volume) of vinegar in a $40.0\ mL$ sample that reacted with $15.0\ mL$ of $0.1\ M$ NaOH during a titration to reach the equivalence. (Density of $CH_{3}COOH$ = $1.05\frac{g}{mL}$ )

Possible Answers:

$10\%$

$45\%$

$3.5\%$

$0.21\%$

Correct answer:

$0.21\%$

Explanation:

$\frac{0.1\ moles}{L}\times0.015L=0.0015\ moles$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.0015\ moles$

Convert the moles of CH_3COOH to grams:

$0.0015\ moles\ CH_{3}COOH\times \frac{60.05\ g}{moles}=0.090\ g\ CH_{3}COOH$

Using the density of CH_3COOH given, convert the grams to mL:

$0.09\ g \times \frac{mL}{1.05g}=0.086\ mL\ CH_{3}COOH$

Determine the percentage of CH_3COOH by dividing the volume of by the total volume of vinegar:

$\frac{0.086\ mL}{40.0\ mL}=0.21\%\ CH_{3}COOH\ in\ vinegar$

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Example Question #1 : Titrations

NaOH reacts with $CH_{3}COOH$ in aqueous solution according to the chemical equation provided. If it took $26.7\ mL$ of a $0.1\ M$ NaOH solution to titrate $20.8\ mL$ of an $CH_{3}COOH$ solution, what was the concentration of original $CH_{3}COOH$ solution?

Possible Answers:

$0.309\ M$

$1.4\ M$

$0.005\ M$

$0.154\ M$

Correct answer:

$0.154\ M$

Explanation:

Using the concentration of NaOH as a conversion factor, we can convert the volume of NaOH to moles by dimensional analysis:

$\frac{0.1\ moles}{L}\times0.0267\ L=0.00267\ moles\ NaOH$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.00267\ moles$

$Molarity=\frac{0.0032\ moles}{0.0208\ L}=0.154\ M$

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Example Question #2 : Titrations

Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . Acetic acid reacts with NaOH in aqueous solution according to the chemical equation provided. Determine the percentage (by volume) of acetic acid in a $100.0\ mL$ sample of vinegar that reacted with $30.0\ mL$ of $2.0\ M$ NaOH during a titration to reach the equivalence. (Density of $CH_{3}COOH$ = $1.05\frac{g}{mL}$ )

Possible Answers:

$7.0\%$

$24\%$

$3.4\%$

$1.5\%$

Correct answer:

$3.4\%$

Explanation:

Convert the liters of NaOH to moles:

$\frac{2.00\ moles}{L}\times0.030L=0.06\ moles\ NaOH$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.06\ moles$

Convert the moles of CH_3COOH to grams:

$0.06\ moles\ CH_{3}COOH\times \frac{60.05\ g}{moles}=3.6\ g\ CH_{3}COOH$

Using the density of CH_3COOH given, convert the grams to mL:

$3.6\ g \times \frac{mL}{1.05g}=3.4\ mL\ CH_{3}COOH$

Determine the percentage of CH_3COOH by dividing the volume of by the total volume of vinegar:

$\frac{3.4\ mL}{100.0\ mL}=3.4\%\ CH_{3}COOH\ in\ vinegar$

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Example Question #1 : P H

Determine the pH of a solution that is $0.40M \text{HCl}$ .

Possible Answers:

1.42

0.22

0.39

0.40

Correct answer:

0.40

Explanation:

Since $\text{HCl}$ is a strong acid, the concentration of H_3O^+ is equal to the concentration of the acid itself.

Thus, [H_3O^+]=0.40M .

Recall how to find the pH of a solution:

$\text{pH}=-\log([H_3O^+])$

Plug in the given hydronium ion concentration to find the pH of the given solution.

$\text{pH}=-\log(0.40)=0.40$

Remember to maintain the correct number of significant figures.

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Example Question #2 : P H

Find the pH for a solution that is $0.520\%$ HBr by mass. Assume a density of $1.01\frac{g}{mL}$ for the solution.

Possible Answers:

2.001

1.188

1.233

3.007

Correct answer:

1.188

Explanation:

Start by assuming that there is liter of the solution. From this, we can use the given density to find the mass of the solution.

$\text{Mass of solution}=1000\text{mL of Solution} \cdot \frac{1.01g}{mL}=1010\text{g of solution}$

Next, find the mass of $\text{HBr}$ that is present in the solution.

$\text{Mass of HBr}=0.00520(1010g)=5.252\text{g of HBr}$

Now, find the number of moles of $\text{HBr}$ that is present in the solution.

$5.252\text{g HBr}\frac{1\text{mol HBr}}{80.91\text{g HBr}}=0.06491\text{moles of HBr}$

Since we initially assumed that we had liter of the solution, we now also know the concentration of $\text{HBr}$ in this solution.

$[\text{HBr}]=0.06491M$

Since $\text{HBr}$ is a strong acid, the concentration of hydronium ions in the solution will be the same as the concentration of $\text{HBr}$ .

$[\text{HBr}]=[H_3O^+]=0.06491M$

Recall how to find the pH of a solution.

$\text{pH}=-\log([H_3O^+])$

$\text{pH}=-\log(0.06491)=1.188$

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Example Question #3 : P H

Find the pH of a solution that is $5.06\% \text{HI}$ by mass. Assume a density of $1.01\frac{g}{mL}$ for the solution.

Possible Answers:

0.398

0.451

0.211

0.336

Correct answer:

0.398

Explanation:

Start by assuming that there is liter of the solution. From this, we can use the given density to find the mass of the solution.

$\text{Mass of solution}=1000\text{mL of Solution} \cdot \frac{1.01g}{mL}=1010\text{g of solution}$

Next, find the mass of $\text{HI}$ that is present in the solution.

$\text{Mass of HI}=0.0506(1010g)=51.106\text{g of HI}$

Now, find the number of moles of $\text{HI}$ that is present in the solution.

$51.106\text{g HI}\frac{1\text{mol HI}}{127.91\text{g HI}}=0.3995\text{ moles of HI}$

Since we initially assumed that we had liter of the solution, we now also know the concentration of $\text{HI}$ in this solution.

$[\text{HI}]=0.3995M$

Since $\text{HI}$ is a strong acid, the concentration of hydronium ions in the solution will be the same as the concentration of $\text{HI}$ .

$[\text{HI}]=[H_3O^+]=0.3995M$

Recall how to find the pH of a solution.

$\text{pH}=-\log([H_3O^+])$

$\text{pH}=-\log(0.3995)=0.398$

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Example Question #61 : Reactions

Which of the following techniques will decrease the pH of a solution?

Possible Answers:

Increasing the amount of solvent

Increasing the amount of Hydroxide molecules

Increasing the concentration of protons

Adding more acid at the same Molarity of the solution

Decreasing the concentration of protons

Correct answer:

Increasing the concentration of protons

Explanation:

Increasing the concentration of protons of a solution will make the solution more acidic; therefore, it lowers the solution’s pH. Decreasing the concentration of protons will make the solution more basic, raising the pH. Adding more acid of the same molarity of the original solution will not increase the concentration of protons and will not increase acidity or lower the pH. Increasing the amount of hydroxide ions will make the solution more basic and raise the pH. Increasing the amount of solvent will lower the concentration, affecting molarity and not lowering the pH.

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College Chemistry : Reactions

Study concepts, example questions & explanations for College Chemistry

All College Chemistry Resources

Example Questions

Example Question #1 : Acid Base Reactions

Example Question #1 : Titrations

Example Question #11 : Acid Base Reactions

Example Question #1 : Titrations

Example Question #1 : Titrations

Example Question #2 : Titrations

Example Question #1 : P H

Example Question #2 : P H

Example Question #3 : P H

Example Question #61 : Reactions

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