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High School Chemistry : Titrations

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Example Questions

Example Question #1 : Identifying Unknown Volume

Starting with 75mL of a .030M HNO_3 solution, how many milliliters of .022M NaOH solution are needed to titrate it?

Possible Answers:

102mL

25mL

50mL

75mL

None of these

Correct answer:

102mL

Explanation:

For this question use the following formula:

N_A M_A V_A = N_B M_B V_B

N_A 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 volume of the base:

$\frac{N_A M_A V_A }{N_B M_B}= V_B$

Plug in known values and solve.

V_B=102mL

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Example Question #2 : Identifying Unknown Volume

Starting with 12mL of a .015M NaOH solution, how many milliliters of .000M HBr solution are needed to titrate it?

Possible Answers:

440mL

710mL

360mL

None of these

540mL

Correct answer:

540mL

Explanation:

For this question use the following formula:

N_A M_A V_A = N_B M_B V_B

Rearrange the equation for the volume of the base:

$\frac{N_A M_A V_A }{N_B M_B}= V_B$

Plug in known values and solve.

V_B=540mL

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Example Question #3 : Identifying Unknown Volume

Starting with 50mL of a .015M HBr solution, how many milliliters of .010M KOH solution are needed to titrate it?

Possible Answers:

50mL

75mL

150mL

275mL

None of these

Correct answer:

75mL

Explanation:

For this question use the following formula:

N_A M_A V_A = N_B M_B V_B

Rearrange the equation for the volume of the base:

$\frac{N_A M_A V_A }{N_B M_B}= V_B$

Plug in known values and solve.

V_B=75mL

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Example Question #41 : Reaction Types

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

Possible Answers:

4.76

12.36

2.38

9.52

7.00

Correct answer:

4.76

Explanation:

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

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Example Question #1 : Titrations And Indicators

At what pH does the equivalence point lie for a strong-acid / strong-base titration?

Possible Answers:

Slightly basic

Slightly acidic

It is impossible to determine

Correct answer:

Explanation:

The equivalence point for a strong-acid / strong-base titration will be at neutral pH, 7. This is because each equivalent of the acid will neutralize each equivalent of the base, and you will be left with a neutral solution.

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Example Question #1 : Titrations And Indicators

At what pH does the equivalence point lie for a strong-acid / weak-base titration?

Possible Answers:

Slightly acidic

It is impossible to determine

Slightly basic

Correct answer:

Slightly acidic

Explanation:

The equivalence point for a strong-acid / weak-base titration will be at a slightly acidic pH. This is because the acid is stronger and dissociates to a greater degree, while the base is not quite as strong, so doesn't dissociate to a large enough extent to neutralize each equivalent of the acid.

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Example Question #1 : Titrations And Indicators

At what pH does the equivalence point lie for a weak acid-strong base titration?

Possible Answers:

Slightly acidic

It is impossible to determine

Slightly basic

Correct answer:

Slightly basic

Explanation:

The equivalence point for a weak-acid / strong-base titration will be at a slightly basic pH. This is because the base is stronger and dissociates to a greater degree, while the acid is not quite and strong and doesn't dissociate to a large enough extent to neutralize each equivalent of the base.

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Example Question #2 : Help With Titration Curves

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?

Possible Answers:

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

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

None of the available answers.

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

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

Correct answer:

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

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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Example Question #2 : Titrations And Indicators

A titration is a drop-by-drop mixing of an acid and a base in order to determine the concentration of an unknown solution, via addition of a solution with known concentration. A titration curve can be graphed showing the relationship between the mixture pH and the amount of known solution added.

What would the titration curve look like for a strong base being titrated with a strong acid?

Possible Answers:

A positively sloped line

A decreasing sigmoidal curve

An increasing sigmoidal curve

A negatively sloped line

Correct answer:

A decreasing sigmoidal curve

Explanation:

There are two things to consider here.

1. Since the solution is originally a strong base, the pH will be originally elevated. As a strong acid is added to the solution, the pH will decrease. As a result, the titration curve will be decreasing as the volume of titrant increases.

2. The titration curve will never be a straight line. Eventually, the strong acid will be much larger in volume than the original base; however, the pH will eventually even out at the pH of the added titrant.

Since we are titrating a strong base with a strong acid, the titration curve will be represented by a decreasing sigmoidal curve.

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Example Question #4 : Help With Titration Curves

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

Possible Answers:

At a pH greater than 7

At the pKa of the acid

At the pKb of the base

At a pH of 7

Correct answer:

At the pKa of the acid

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.

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High School Chemistry : Titrations

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #1 : Identifying Unknown Volume

Example Question #2 : Identifying Unknown Volume

Example Question #3 : Identifying Unknown Volume

Example Question #41 : Reaction Types

Example Question #1 : Titrations And Indicators

Example Question #1 : Titrations And Indicators

Example Question #1 : Titrations And Indicators

Example Question #2 : Help With Titration Curves

Example Question #2 : Titrations And Indicators

Example Question #4 : Help With Titration Curves

All High School Chemistry Resources

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