Titrations

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AP Chemistry › Titrations

Questions 1 - 8

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Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . Acetic acid reacts with 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 $0.1\ M$ during a titration to reach the equivalence. (Density of $CH_{3}COOH$ = $1.05\frac{g}{mL}$ )

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 to grams:

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

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

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

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

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

The titration of a $15.0\text{mL}$ sample of an $\text{HCl}$ solution at an unknown concentration requires $17.08\text{mL}$ of a $0.200\text{M NaOH}$ solution. What is the concentration of the unknown hydrochloric acid solution in ?

Explanation

Start by writing the chemical equation for this acid-base reaction.

$\text{HCl}(aq)+\text{NaOH}(aq)\rightarrow\text{H}_2\text{O}(l)+\text{NaCl}(aq)$

Next, find the number of moles of $\text{NaOH}$ that was added.

$17.08\text{mL NaOH}\cdot\frac{1\text{L}}{1000\text{mL}}\cdot\frac{0.200\text{ moles NaOH}}{\text{L}}=0.003416\text{ moles of NaOH}$

At the equivalence point of a titration, the moles of acid and the moles of base are the same.

$\text{Moles of HCl}=\text{Moles of NaOH}=0.003416$

Finally, find the molarity of the solution.

$\text{Molarity}=\frac{\text{moles of solute}}{\text{liters of solution}}=\frac{0.003416}{15.0\times10^{-3}}=0.228M$

Make sure to round the answer to three significant figures.

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

$0.13\ M$

$1.4\ M$

$0.0032\ M$

$0.25\ M$

Explanation

Using the concentration of as a conversion factor, we can convert the volume of 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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Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . Acetic acid reacts with 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 $2.0\ M$ during a titration to reach the equivalence. (Density of $CH_{3}COOH$ = $1.05\frac{g}{mL}$ )

Explanation

Convert the liters of 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 to grams:

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

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

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

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

$72.0\ mL$

$131.0\ mL$

$23.0\ mL$

$24.0\ mL$

Explanation

Determine the moles of 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 to liters:

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

Convert the liters to milliliters:

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

A $75.0\text{mL}$ sample of sodium hydroxide is titrated with hydrochloric acid. Find the pH of the solution after adding $45.0\text{mL}$ of hydrochloric acid.

Explanation

Start by finding the initial number of moles of sodium hydroxide by using the given volume and molarity. Since sodium hydroxide is a strong base, it will dissociate completely into $\text{OH}^-$ .

$\text{Initial moles of NaOH}=\text{Initial moles of OH}^-=0.075\text{L}(\frac{0.100\text{ mol}}{\text{L}})=0.0075\text{ moles}$

Next, find the number of moles of the hydrochloric acid added by using the given volume and molarity. Since hydrochloric acid is a strong acid, all of the $\text{H}_3\text{O}^+$ will come from the acid.

$\text{Moles of HCl added}=\text{Moles of H}_3\text{O}^+=0.045\text{L}(\frac{0.150\text{ moles}}{\text{L}})=0.00675\text{ moles}$

Now, as the acid is added, it will neutralize some of the base as shown by the following equation:

$\text{OH}^-(aq)+\text{H}_3\text{O}^+(aq)\rightarrow 2\text{H}_2\text{O}(l)$

Since the reactants are found in a ratio, subtract the number of moles of acid added from the number of initial moles of base to get the amount of remaining base.

$\text{Remaining moles of NaOH}=0.0075-0.00675=7.5\times10^{-4}\text{ moles of NaOH}$

Now, divide this number of moles by the total volume of the solution to find the concentration of the base.

$[\text{OH}]^-=\frac{7.5\times 10^{-4}\text{ moles}}{0.075\text{L}+0.045\text{L}}=0.00625\text{ M}$

Next, find the $\text{pOH}$ value.

$\text{pOH}=-\log(0.00625)=2.204$

Finally, find the $\text{pH}$ value.

$\text{pH}=14-\text{pOH}$

$\text{pH}=14-2.204=11.80$

Your answer should have significant figures.

$\text{pH}=11.8$

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