Question 1

A 25.00 mL aliquot of 0.0900 M acetic acid is titrated with a 0.100 M KOH.Calculate the pH when 25.00 mL of base is added.Ka = 1.75 x 10−5

Accepted Answer

A) 2)30 
B) 5)27 
C) 5)71 
D) 8)73 
E) 11.70 
A) 2)30 
B) 5)27 
C) 5)71 
D) 8)73 
E) 11.70 E

Question 2

A 0.100 M oxalic acid,HO2CCO2H,solution is titrated with 0.100 M KOH.Calculate the pH when 25.00 mL of oxalic acid solution is titrated with 35.00 mL of NaOH.Ka1 = 5.4 × 10−2 and Ka2 = 5.42 × 10−5 for oxalic acid.

Accepted Answer

pH = 4.090

Question 3

Strong base solutions require standardization before use because strong bases:

Accepted Answer

A) precipitate as insoluble metal hydroxides,decreasing the hydroxide concentration. 
B) react with atmospheric carbon dioxide to form carbonates,decreasing the hydroxide concentration. 
C) react with atmospheric nitrogen to form nitrogen oxides,decreasing the hydroxide concentration. 
D) adhere to the inside surface of the solution container,reducing the hydroxide concentration. 
E) react with solution impurities to form volatile compounds,increasing the hydroxide concentration. 
A) precipitate as insoluble metal hydroxides,decreasing the hydroxide concentration. 
B) react with atmospheric carbon dioxide to form carbonates,decreasing the hydroxide concentration. 
C) react with atmospheric nitrogen to form nitrogen oxides,decreasing the hydroxide concentration. 
D) adhere to the inside surface of the solution container,reducing the hydroxide concentration. 
E) react with solution impurities to form volatile compounds,increasing the hydroxide concentration. B

Question 4

A 1.000 gram organic sample containing nitrogen is subjected to Kjeldahl analysis.The ammonia gas generated when ammonium is neutralized is distilled into 50.00 mL of 0.1000 M HCl to reform ammonium.If 17.29 mL of 0.1000 NaOH is required to titrate the unreacted HCl,calculate the percent by mass nitrogen in the original sample.

Accepted Answer

The answer of A 1.000 gram organic sample containing nitrogen...

Question 5

A 50.00 mL aliquot of a 0.1000 M propenoic acid solution,H2CCHCO2H,is titrated with 0.1250 M NaOH.Calculate the pH when 25.00 mL of NaOH is added.Ka = 5.52 × 10−5

Accepted Answer

Question 6

The end point for a titration can be determined graphically.The end point volume for a first derivative plot corresponds to:

Accepted Answer

A) the volume base added where the first derivative curve crosses the x-axis. 
B) the volume base added where the first derivative curve has maximum slope. 
C) the volume base added where the first derivative curve crosses the y-axis. 
D) the volume base added where the first derivative curve has minimum slope. 
E) the volume base added where the first derivative curve has a maximum negative slope. 
A) the volume base added where the first derivative curve crosses the x-axis. 
B) the volume base added where the first derivative curve has maximum slope. 
C) the volume base added where the first derivative curve crosses the y-axis. 
D) the volume base added where the first derivative curve has minimum slope. 
E) the volume base added where the first derivative curve has a maximum negative slope.

Question 7

For the titration of a strong acid with a strong base,which statement below is FALSE?

Accepted Answer

A) At V = 0 mL titrant added,the pH is due to initial concentration of strong acid. 
B) At the equivalence point,the pH is due to the auto-ionization of water. 
C) Between V = 0 mL and the equivalence point,the pH is due to diluted unreacted strong acid. 
D) Past the equivalence point,the pH is due to diluted excess strong base. 
E) The indicator changes color at the end point due to excess acid. 
A) At V = 0 mL titrant added,the pH is due to initial concentration of strong acid. 
B) At the equivalence point,the pH is due to the auto-ionization of water. 
C) Between V = 0 mL and the equivalence point,the pH is due to diluted unreacted strong acid. 
D) Past the equivalence point,the pH is due to diluted excess strong base. 
E) The indicator changes color at the end point due to excess acid.

Question 8

Which statement is TRUE for the leveling effect?
I Acids stronger than water react with water to form hydronium.
II Hydronium is the strongest acid in water.
III In water,strong acids have the same strength.
IV The differences in the strength of strong acids are observed in nonaqueous solvents such as acetic acid.

Accepted Answer

A) I,III,and IV 
B) II and IV 
C) II and III 
D) II,III,and IV 
E) I and III 
A) I,III,and IV 
B) II and IV 
C) II and III 
D) II,III,and IV 
E) I and III

Question 9

Which statement is false regarding the titration of a weak base with a strong acid? B + H+ → BH+

Accepted Answer

A) At the equivalence point,the pH is due to the weak acid dissociation of the conjugate acid,BH+. 
B) Past the equivalence point the pH is due to excess strong acid,H+. 
C) Before strong acid is added the pH is due to the ionization of the weak base,B. 
D) Before the equivalence point,the pH is due to a mixture of B and BH+,a buffer. 
E) The pH at the equivalence point is acidic. 
A) At the equivalence point,the pH is due to the weak acid dissociation of the conjugate acid,BH+. 
B) Past the equivalence point the pH is due to excess strong acid,H+. 
C) Before strong acid is added the pH is due to the ionization of the weak base,B. 
D) Before the equivalence point,the pH is due to a mixture of B and BH+,a buffer. 
E) The pH at the equivalence point is acidic.

Question 10

A 40.00 mL sample of 0.1000 M diprotic malonic acid is titrated with 0.0900 M KOH.What volume KOH must be added to give a pH of 5.697? Ka1 = 1.42 × 10−3 and Ka2 = 2.01 × 10−6.

Accepted Answer

A) 66.67 mL 
B) 44.44 mL 
C) 22.22 mL 
D) 88.89 mL 
E) 111.11 mL 
A) 66.67 mL 
B) 44.44 mL 
C) 22.22 mL 
D) 88.89 mL 
E) 111.11 mL

Question 11

Calculate the pH for the titration of 50.00 mL of a 0.1500 M methylamine with 35.00 mL of 0.1000 M HCl.Kb = 4.42 × 10−4

Accepted Answer

Question 12

Which statement is FALSE regarding the titration of a dibasic weak base with a strong acid? B + H+ → BH+
BH+ + H+ → BH22+

Accepted Answer

A) For V = 0,the pH is dictated by the reaction of B with H2O. 
B) Before the first equivalence point,Veq1 the buffer created by B and BH+ determines the pH. 
C) At Veq1 the pH is due to the reaction of BH+ with H2O. 
D) After the first equivalence point,Veq1 and before the second equivalence point,Veq2,the buffer created by BH+ and BH22+ determines the pH. 
E) At Veq2 the pH is due to the reaction of BH22+ with H2O. 
A) For V = 0,the pH is dictated by the reaction of B with H2O. 
B) Before the first equivalence point,Veq1 the buffer created by B and BH+ determines the pH. 
C) At Veq1 the pH is due to the reaction of BH+ with H2O. 
D) After the first equivalence point,Veq1 and before the second equivalence point,Veq2,the buffer created by BH+ and BH22+ determines the pH. 
E) At Veq2 the pH is due to the reaction of BH22+ with H2O.

Question 13

The end point for a titration can be determined graphically.The end point volume for a second derivative plot corresponds to:____________________ .

Accepted Answer

A) the volume titrant added where the second derivative curve crosses the x-axis. 
B) the volume titrant added where the second derivative curve has maximum slope. 
C) the volume titrant added where the second derivative curve crosses the y-axis. 
D) the volume titrant added where the second derivative curve has minimum slope. 
E) the volume titrant added where the second derivative curve has a maximum negative slope. 
A) the volume titrant added where the second derivative curve crosses the x-axis. 
B) the volume titrant added where the second derivative curve has maximum slope. 
C) the volume titrant added where the second derivative curve crosses the y-axis. 
D) the volume titrant added where the second derivative curve has minimum slope. 
E) the volume titrant added where the second derivative curve has a maximum negative slope.

Question 14

Which statements are TRUE for Gran plots?
I Gran plots are more accurate around end point than first and second derivative plots.
II The end point is determined using concentration/pH data prior to the end point.
III For a constant slope,where the Gran plot crosses the x-axis is the end point volume.
IV Gran plots are used with polyprotic titration curves as well as monoprotic titration curves.

Accepted Answer

A) I,II,and IV 
B) II,III,and IV 
C) I,II,and III 
D) I,II,III,and IV 
E) I,III,and IV 
A) I,II,and IV 
B) II,III,and IV 
C) I,II,and III 
D) I,II,III,and IV 
E) I,III,and IV

Question 15

Calculate the pH when 50.00 mL of 0.100 M HNO3 is titrated with 40.00 mL of 0.0750 M NaOH.

Accepted Answer

Question 16

Which statement is FALSE for indicators?

Accepted Answer

A) Indicators are acids or bases whose various protonated forms have different colors. 
B) The pH range over which an indicator changes color is the transition range 
C) Indicators are chosen to change color as close to the end point as possible. 
D) The volume difference between the equivalence point and the end point is indicator error. 
E) Indicators can change color more than once. 
A) Indicators are acids or bases whose various protonated forms have different colors. 
B) The pH range over which an indicator changes color is the transition range 
C) Indicators are chosen to change color as close to the end point as possible. 
D) The volume difference between the equivalence point and the end point is indicator error. 
E) Indicators can change color more than once.

Question 17

A 25.00 mL aliquot of 0.100 M NH3 is titrated with 21.00 mL of 0.0900 M HCl.Calculate the pH.Kb = 1.76 x 10−5

Accepted Answer

A) 5)25 
B) 8)44 
C) 8)75 
D) 5)56 
E) 7)00 
A) 5)25 
B) 8)44 
C) 8)75 
D) 5)56 
E) 7)00

Question 18

Fifty mL of 0.100 M HCl is titrated with 0.1500 M NaOH.The pH at the equivalence point is:

Accepted Answer

A) 1)00. 
B) 13.18. 
C) 1)22. 
D) 12.78. 
E) 7)00. 
A) 1)00. 
B) 13.18. 
C) 1)22. 
D) 12.78. 
E) 7)00.

Question 19

Arrange the steps below in the order required to perform a Kjeldahl Nitrogen Analysis.
I Distillation of ammonia into standard hydrochloric acid.
II Neutralize ammonium to release ammonia.
III Digest organic sample with boiling sulfuric acid to convert nitrogen to ammonium.
IV Titrate unreacted standard hydrochloric acid with sodium hydroxide.

Accepted Answer

A) III,II,I,IV 
B) II,I,IV,III 
C) I,IV,III,II 
D) IV,III,II,I 
E) III,I,IV,II 
A) III,II,I,IV 
B) II,I,IV,III 
C) I,IV,III,II 
D) IV,III,II,I 
E) III,I,IV,II

Question 20

Which statement is FALSE regarding the titration of a weak acid with a strong base? HA + OH− → A− + H2O

Accepted Answer

A) At the equivalence point,the pH is due to the hydrolysis of water by the conjugate base,A−. 
B) Past the equivalence point the pH is due to excess conjugate base,A−. 
C) Before strong base is added,the pH is due to the ionization of the weak acid,HA. 
D) Before the equivalence point,the pH is due to a mixture of HA and A−. 
E) The pH at the equivalence point is basic. 
A) At the equivalence point,the pH is due to the hydrolysis of water by the conjugate base,A−. 
B) Past the equivalence point the pH is due to excess conjugate base,A−. 
C) Before strong base is added,the pH is due to the ionization of the weak acid,HA. 
D) Before the equivalence point,the pH is due to a mixture of HA and A−. 
E) The pH at the equivalence point is basic.

Exam 12: Acid-Base Titrations

A 25.00 mL aliquot of 0.0900 M acetic acid is titrated with a 0.100 M KOH.Calculate the pH when 25.00 mL of base is added.Ka = 1.75 x 10−5

A 0.100 M oxalic acid,HO2CCO2H,solution is titrated with 0.100 M KOH.Calculate the pH when 25.00 mL of oxalic acid solution is titrated with 35.00 mL of NaOH.Ka1 = 5.4 × 10−2 and Ka2 = 5.42 × 10−5 for oxalic acid.

Strong base solutions require standardization before use because strong bases:

A 50.00 mL aliquot of a 0.1000 M propenoic acid solution,H2CCHCO2H,is titrated with 0.1250 M NaOH.Calculate the pH when 25.00 mL of NaOH is added.Ka = 5.52 × 10−5

The end point for a titration can be determined graphically.The end point volume for a first derivative plot corresponds to:

For the titration of a strong acid with a strong base,which statement below is FALSE?

Which statement is false regarding the titration of a weak base with a strong acid? B + H+ → BH+

A 40.00 mL sample of 0.1000 M diprotic malonic acid is titrated with 0.0900 M KOH.What volume KOH must be added to give a pH of 5.697? Ka1 = 1.42 × 10−3 and Ka2 = 2.01 × 10−6.

Calculate the pH for the titration of 50.00 mL of a 0.1500 M methylamine with 35.00 mL of 0.1000 M HCl.Kb = 4.42 × 10−4

Which statement is FALSE regarding the titration of a dibasic weak base with a strong acid? B + H+ → BH+ BH+ + H+ → BH22+

The end point for a titration can be determined graphically.The end point volume for a second derivative plot corresponds to:____________________ .

Calculate the pH when 50.00 mL of 0.100 M HNO3 is titrated with 40.00 mL of 0.0750 M NaOH.

Which statement is FALSE for indicators?

A 25.00 mL aliquot of 0.100 M NH3 is titrated with 21.00 mL of 0.0900 M HCl.Calculate the pH.Kb = 1.76 x 10−5

Fifty mL of 0.100 M HCl is titrated with 0.1500 M NaOH.The pH at the equivalence point is:

Which statement is FALSE regarding the titration of a weak acid with a strong base? HA + OH− → A− + H2O

The Analytical Process

Chemical Measurements

Tools of the Trade

Experimental Error

Statistics

Quality Assurance and Calibration Methods

Chemical Equilibrium

Let the Titrations Begin

Activity and the Systematic Treatment of Equilibrium

Monoprotic Acid-Base Equilibria

Polyprotic Acid-Base Equilibria

Edta Titrations

Advanced Topics in Equilibrium

Fundamentals of Electrochemistry

Electrodes and Potentiometry

Redox Titrations

Electroanalytical Techniques

Fundamentals of Spectrophotometry

Applications of Spectrophotometry

Spectrophotometers

Atomic Spectroscopy

Mass Spectrometry

Introduction to Analytical Separations

Gas Chromatography

High-Performance Liquid Chromatography

Chromatographic Methods and Capillary Electrophoresis

Gravimetric and Combustion Analysis

Sample Preparation

Filters

A 25.00 mL aliquot of 0.0900 M acetic acid is titrated with a 0.100 M KOH.Calculate the pH when 25.00 mL of base is added.K_a = 1.75 x 10⁻⁵

A 0.100 M oxalic acid,HO₂CCO₂H,solution is titrated with 0.100 M KOH.Calculate the pH when 25.00 mL of oxalic acid solution is titrated with 35.00 mL of NaOH.K_a1 = 5.4 × 10⁻² and K_a2 = 5.42 × 10⁻⁵ for oxalic acid.

A 50.00 mL aliquot of a 0.1000 M propenoic acid solution,H₂CCHCO₂H,is titrated with 0.1250 M NaOH.Calculate the pH when 25.00 mL of NaOH is added.K_a = 5.52 × 10⁻⁵

Which statement is false regarding the titration of a weak base with a strong acid? B + H⁺ → BH⁺

A 40.00 mL sample of 0.1000 M diprotic malonic acid is titrated with 0.0900 M KOH.What volume KOH must be added to give a pH of 5.697? K_a1 = 1.42 × 10⁻³ and K_a2 = 2.01 × 10⁻⁶.

Calculate the pH for the titration of 50.00 mL of a 0.1500 M methylamine with 35.00 mL of 0.1000 M HCl.K_b = 4.42 × 10⁻⁴

Which statement is FALSE regarding the titration of a dibasic weak base with a strong acid? B + H⁺ → BH⁺ BH⁺ + H⁺ → BH₂²⁺

Calculate the pH when 50.00 mL of 0.100 M HNO₃ is titrated with 40.00 mL of 0.0750 M NaOH.

A 25.00 mL aliquot of 0.100 M NH₃ is titrated with 21.00 mL of 0.0900 M HCl.Calculate the pH.K_b = 1.76 x 10⁻⁵

Which statement is FALSE regarding the titration of a weak acid with a strong base? HA + OH⁻ → A⁻ + H₂O