MU #1 Titration is a laboratory process in which a volume of solution of known concentration is used to determine the concentration of another solution.

A titration reaction involves a complete neutralization reaction where the moles of H⁺ equals the moles of OH^-

An indicator previously added to the solution, such as litmus or phenolphthalein, indicates the change from acid to neutral or basic to neutral. See table M

The concentration of acids or molarity (M_A) is measured by the number of moles of H⁺ over the liters of solution or (V_A)

M_A = mol H⁺ / V_A   

Therefore, mol H⁺ = M_A V_A

The concentration of bases or (M_B) is measured by the number of moles of OH^- over the liters of solution or (V_B)

M_B = mol OH^- / V_B  

Therefore, mol OH^- = M_B V_B

Since in a neutralization reaction (mol H⁺ = mol OH^-)

Therefore M_A V_A = M_B V_B    See table T 

Where

M_A = molarity of H⁺

V_A = volume of acid

M_B = molarity of OH^-

V_B =  volume of base

The molarity of an acid can be calculated if a base of known concentration (standard base) is added, drop by drop, to a specific volume of the acid until the indicator changes color. At this exact point (the end point of titration) moles of H⁺ are equal to moles of OH^-

Example

What is the molarity of HCl (aq) if 10. milliliters of 4.0 M NaOH (aq) neutralizes exactly 20. milliliters of HCl (aq)?

M_A = ?                 

V_B = 10 ml      

M_B = 4.0 M

V_A= 20 ml   

From  M_A V_A = M_B V_B        M_A = M_B V_B /  V_A

Therefore, M_A = (4 M) (10 ml) / 20 ml = 2 M

The molarity of a base can be calculated if an acid of known concentration (standard acid) is added, drop by drop, to a specific volume of the base until the indicator changes color. At this exact point (the end point of titration) moles of H⁺ are equal to moles of OH^-

Example

What is the molarity of NaOH (aq) if 10. milliliters of 2 M HBr (aq) neutralizes exactly 5. milliliters of NaOH (aq)?

M_B = ? 

V_A = 10 ml 

M_A = 2                   

V_B = 5 ml 

From  M_A V_A = M_B V_B        M_B = M_A V_A /  V_B

Therefore, M_B = (2 M) (10 ml) / 5ml = 4 M

TEST YOUR UNDERSTANDING

1/07

Base your answers to questions 78 through 81 on the information below.

In preparing to titrate an acid with a base, a student puts on goggles and an apron. The student uses burets to dispense and measure the acid and the base in the titration. In each of two trials, a 0.500 M NaOH(aq) solution is added to a flask containing a volume of HCl(aq) solution of unknown concentration.

Phenolphthalein is the indicator used in the titration. The calculated volumes used for the two trials are recorded in the table below.

78 Write a chemical name for the acid used in the titration.

79 Using the volumes from trial 1, determine the molarity of the HCl(aq) solution.

80 Based on the information given in the table, how many significant figures should be shown in the calculated molarity of the HCl(aq) solution used in trial 2?

81 Identify one additional safety precaution the student should have taken before performing the titration.

6/07

49 Information related to a titration experiment is given in the balanced equation and table below. H₂SO₄(aq) + 2KOH(aq) ---> K₂SO₄(aq) + 2H₂O(l)

Based on the equation and the titration results, what is the concentration of the H₂SO₄(aq)? (1) 0.12 M   (2) 0.16 M (3) 0.24 M  (4) 0.96 M

8/07

Base your answers to questions 78 through 80 on the information below.

In a laboratory activity, 0.500 mole of NaOH(s) is completely dissolved in distilled water to form 400. milliliters of NaOH(aq). This solution is then used to titrate a solution of HNO₃(aq).

78 Identify the negative ion produced when the NaOH(s) is dissolved in distilled water.

79 Calculate the molarity of the NaOH(aq). Your response must include both a correct numerical setup and the calculated result.

80 Complete the equation representing this titration reaction by writing the formulas of the products.

6/02

1. If 5.0 milliliters of a 0.20 M HCl solution is required to neutralize exactly 10. milliliters of NaOH, what is the concentration of the base?

            (1) 0.10 M         (2) 0.20 M       (3) 0.30 M     (4) 0.40 M

8/02

1. When 50. milliliters of an HNO₃ solution is exactly neutralized by 150 milliliters of a 0.50 M solution of KOH, what is the concentration of HNO₃?

        (1) 1.0 M         (2) 1.5 M          (3) 3.0 M         (4) 0.5 M

1/03

 A titration setup was used to determine the unknown molar concentration of a solution of NaOH. A 1.2 M HCl solution was used as the titration standard. The following data were collected.

71. Calculate the volume of NaOH solution used to neutralize 10.0 mL of the standard HCl solution in trial 3. Show your work.

72. According to Reference Table M, what indicator would be most appropriate in determining the end point of this titration? Give one reason for choosing this indicator.

73. Calculate the average molarity of the unknown NaOH solution for all four trials. Your answer must include the correct number of significant figures and correct units.

74. Explain why it is better to use the average data from multiple trials rather than the data from a single trial to calculate the results of the titration.

6/03

66. A student recorded the following buret readings during a titration of a base with an acid:

a) In the space provided in your answer booklet, calculate the molarity of the KOH Show all work.

b) Record your answer to the correct number of significant figures.

1/04

45 A student neutralized 16.4 milliliters of HCl by adding 12.7 milliliters of 0.620 M KOH. What was the molarity of the HCl acid?
        (1) 0.168 M (2) 0.480 M (3) 0.620 M (4) 0.801 M

6/04

Base your answers to questions 65 through 67 on the information and equation below.
Antacids can be used to neutralize excess stomach acid. Brand A antacid contains the acid-neutralizing agent magnesium hydroxide, Mg(OH)₂ . It reacts with HCl( aq) in the stomach, according to the following balanced equation:

                2 HCl (aq) + Mg(OH)₂ (s) <--> MgCl₂ (aq) + 2H₂O (l)
65 Show a correct numerical setup for calculating the number of moles of Mg(OH)₂ (gram-formula mass = 58.3 grams/ mole) in an 8.40-gram sample.

66 If a person produces 0.050 mole of excess HCl in the stomach, how many moles of Mg(OH)₂ are needed to neutralize this excess hydrochloric acid?

67 Brand B antacid contains the acid-neutralizing agent sodium hydrogen carbonate. Write the chemical formula for sodium hydrogen carbonate.

8/04

29 Which process uses a volume of solution of known concentration to determine the concentration of another solution?

                (1) distillation (2) substitution (3) transmutation (4) titration

Base your answers to questions 79 through 81 on the information and data table below.

                Indigestion may be caused by excess stomach acid (hydrochloric acid). Some products used to treat indigestion contain magnesium hydroxide. The magnesium hydroxide neutralizes some of the stomach acid.

                The amount of acid that can be neutralized by three different brands of antacids is shown in the data table below.

79 Based on Reference Table F, describe the solubility of magnesium hydroxide in water.

80 Show a correct numerical setup for calculating the milliliters of HCl(aq) neutralized per gram of antacid tablet for each brand of antacid.

81 Which antacid brand neutralizes the most acid per gram of antacid tablet?

1/05

Base your answers to questions 63 through 65 on the information below.

In a titration experiment, a student uses a 1.4 M HBr(aq) solution and the indicator phenolphthalein to determine the concentration of a KOH(aq) solution. The data for trial 1 is recorded in the table below.

63 In the space provided in your answer booklet, show a correct numerical setup for calculating the molarity of the KOH(aq) solution for trial 1.

64 Why is it better to use several trials of a titration rather than one trial to determine the molarity of a solution of an unknown concentration?

65 In a second trial of this experiment, the molarity of KOH(aq) was determined to be 0.95 M. The actual molarity was 0.83 M. What is the percent error in the second trial?

1/06

8/06

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