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College Chemistry : Acid-Base Reactions

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Example Question #2 : Titrations

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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:

$40.5\ mL$

$1.2\ mL$

$6.3\ mL$

$4.2\ mL$

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 #3 : Titrations

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 #3 : 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:

$0.21\%$

$3.5\%$

$10\%$

$45\%$

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 #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 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:

$3.4\%$

$24\%$

$7.0\%$

$1.5\%$

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 #11 : Acid Base Reactions

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

Possible Answers:

0.39

0.40

0.22

1.42

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 #12 : Acid Base Reactions

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:

1.188

2.001

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 #13 : Acid Base Reactions

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:

0.451

0.398

0.336

0.211

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 #1 : P H

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

Possible Answers:

Decreasing the concentration of protons

Increasing the amount of Hydroxide molecules

Adding more acid at the same Molarity of the solution

Increasing the concentration of protons

Increasing the amount of solvent

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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Example Question #14 : Acid Base Reactions

What is the of a solution that contains $3.0\: M$ hydrofluoric acid and $2.5\:M$ fluoride?

Note: $K_{a}=6.76\cdot10^{-4}$ for hydrofluoric acid.

Possible Answers:

3.00

3.09

3.26

3.17

Correct answer:

3.09

Explanation:

In this question, we're given the concentrations of both a weak acid and its conjugate base in solution. We're also provided with the acid-dissociation constant for this acid, and we're asked to find the pH of the solution.

The easiest way to go about solving this problem is to first convert the acid-dissociation constant given into pKa. Then, we can use the Henderson-Hasselbalch equation to solve for the pH of this solution.

$pK_{a}=-log(K_{a})$

$pK_{a}=-log(6.76\cdot 10^{-4})$

$pK_{a}=3.17$

Now that we have the pKa, we can use it, along with the concentrations of the weak acid and its conjugate base, in order to solve for the pH of the solution.

$pH=pKa+log(\frac{base}{acid})$

$pH=3.17+log(\frac{2.5}{3.0})$

pH=3.09

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College Chemistry : Acid-Base Reactions

Study concepts, example questions & explanations for College Chemistry

All College Chemistry Resources

Example Questions

Example Question #2 : Titrations

Example Question #3 : Titrations

Example Question #3 : Titrations

Example Question #1 : Titrations

Example Question #1 : Titrations

Example Question #11 : Acid Base Reactions

Example Question #12 : Acid Base Reactions

Example Question #13 : Acid Base Reactions

Example Question #1 : P H

Example Question #14 : Acid Base Reactions

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