Titrations

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AP Chemistry › Titrations

Questions 1 - 10
1

A buffer using acetic acid (pKa=4.76) is titrated with NaOH. What is the pH at half the equivalence point?

2.38

4.76

7.00

9.52

12.36

Explanation

The pH at half the equivalence point is equal to the pKa of the acid.

2

Where does the flattest region of a titration curve of the titration of a weak acid with a strong base occur?

At the pKa of the acid

At the pKb of the base

At a pH greater than 7

At a pH of 7

Explanation

In this question, titration curve would graph the pH of acid solution versus the amount of base added. Since the base is strong and the acid is weak, we can conclude that the pH will be slightly greater than 7 at the equivalence point. The equivalence point is found in the steepest region of the curve.

The half-equivalence point is the flattest region of the titration curve and is most resistant to changes in pH. This corresponds to the pKa of the acid. Within this region, adding base (changing the x-value) results in very little deviation in the pH (the y-value). This region is also the buffer region for the given acid.

3

Where does the flattest region of a titration curve of the titration of a weak acid with a strong base occur?

At the pKa of the acid

At the pKb of the base

At a pH greater than 7

At a pH of 7

Explanation

In this question, titration curve would graph the pH of acid solution versus the amount of base added. Since the base is strong and the acid is weak, we can conclude that the pH will be slightly greater than 7 at the equivalence point. The equivalence point is found in the steepest region of the curve.

The half-equivalence point is the flattest region of the titration curve and is most resistant to changes in pH. This corresponds to the pKa of the acid. Within this region, adding base (changing the x-value) results in very little deviation in the pH (the y-value). This region is also the buffer region for the given acid.

4

A buffer using acetic acid (pKa=4.76) is titrated with NaOH. What is the pH at half the equivalence point?

2.38

4.76

7.00

9.52

12.36

Explanation

The pH at half the equivalence point is equal to the pKa of the acid.

5

39mL of a solution of is mixed with 24mL of an solution of unknown concentration produced a solution with a pH of 7. What is the concentration of the solution?

None of these

Explanation

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the acid:

Plug in known values and solve.

6

39mL of a solution of is mixed with 24mL of an solution of unknown concentration produced a solution with a pH of 7. What is the concentration of the solution?

None of these

Explanation

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the acid:

Plug in known values and solve.

7

A 100mL solution is composed of 25% ethanol by volume and water. What is the mass of the solution?

Explanation

First we determine the mass of the ethanol in solution using its density. Using the percent by volume of ethanol, we know that there are 25mL of ethanol in a 100mL solution. The remaining 75mL are water.

Since the density of water is 1g/mL, we know that the mass of 75mL of water is 75g. The total mass is the sum of the ethanol and the water.

8

A 100mL solution is composed of 25% ethanol by volume and water. What is the mass of the solution?

Explanation

First we determine the mass of the ethanol in solution using its density. Using the percent by volume of ethanol, we know that there are 25mL of ethanol in a 100mL solution. The remaining 75mL are water.

Since the density of water is 1g/mL, we know that the mass of 75mL of water is 75g. The total mass is the sum of the ethanol and the water.

9

You are given 500 mL of a HCl solution of unknown concentration and you titrate is with 0.0540 M NaOH. It takes 32.1 mL of the NaOH solution to reach your end point. What is [HCl] of your original solution?

3.47\times 10^{-3}\hspace{1 mm}M

1.73\times 10^{-3}\hspace{1 mm}M

8.41\times 10^{-1}\hspace{1 mm}M

5.41\times 10^{-1}\hspace{1 mm}M

None of the available answers.

Explanation

First, let us write out the reaction that occurs:

HCl_{(aq)}+NaOH_{(aq)}\rightarrow NaCl_{(aq)}+H_2O_{(l)}

32.1\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.0540\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}HCl}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=1.73\times 10^{-3}\hspace{1 mm}moles\hspace{1 mm}HCl

\frac{1.73\times 10^{-3}\hspace{1 mm}moles\hspace{1 mm}HCl}{500\hspace{1 mm}mL}\times\frac{1000\hspace{1 mm}mL}{1\hspace{1 mm}L}=3.47\times 10^{-3}\hspace{1 mm}M

10

You are given 500 mL of a HCl solution of unknown concentration and you titrate is with 0.0540 M NaOH. It takes 32.1 mL of the NaOH solution to reach your end point. What is [HCl] of your original solution?

3.47\times 10^{-3}\hspace{1 mm}M

1.73\times 10^{-3}\hspace{1 mm}M

8.41\times 10^{-1}\hspace{1 mm}M

5.41\times 10^{-1}\hspace{1 mm}M

None of the available answers.

Explanation

First, let us write out the reaction that occurs:

HCl_{(aq)}+NaOH_{(aq)}\rightarrow NaCl_{(aq)}+H_2O_{(l)}

32.1\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.0540\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}HCl}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=1.73\times 10^{-3}\hspace{1 mm}moles\hspace{1 mm}HCl

\frac{1.73\times 10^{-3}\hspace{1 mm}moles\hspace{1 mm}HCl}{500\hspace{1 mm}mL}\times\frac{1000\hspace{1 mm}mL}{1\hspace{1 mm}L}=3.47\times 10^{-3}\hspace{1 mm}M

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