6 pH, a Concentration Scale for Acids and Bases

178



c h a p t er 4 Stoichiometry: Quantitative Information about Chemical Reactions



A CLOSER LOOK



Serial Dilutions



We often find in the laboratory that a solution is too

concentrated for the analytical technique we

want to use. You might want to analyze a

seawater sample for its chloride ion content,

for instance. To obtain a solution with a chloride concentration of the proper magnitude

for analysis by the Mohr method (Case Study,

page 187), for example, you might want to

dilute the sample, not once but several times.

Suppose you have 100.0 mL of a seawater sample that has an NaCl concentration of 0.550 mol/L. You transfer 10.00 mL

of that sample to a 100.0-mL volumetric

flask and fill to the mark with distilled water.

You then transfer 5.00 mL of that diluted

sample to another 100.0-mL flask and fill to

the mark with distilled water. What is the

NaCl concentration in the final 100.0-mL

sample?

The original solution contains 0.550 mol/L

of NaCl. If you remove 10.00 mL, you have

removed



Now we take 5.00 mL of the diluted solution and dilute that once again to 100.0 mL.

The final concentration is

0.00500 L × 5.50 × 10−2 mol/L

= 2.75 × 10−4 mol NaCl



This is 1/200 of the concentration of the

original solution.

A fair question at this point is why we did

not just take 1 mL of the original solution and

dilute to 200 mL. The answer is that there is



or 1/10 of the concentration of the original

solution (because we diluted the sample by a

factor of 10).



Module 9: pH covers concepts in this

section.



You have a 100.0-mL sample of a blue dye

having a concentration of 0.36 M. You

dilute a 10.0-mL sample of this to 100.0 mL

and then a 2.00-mL sample of that solution

to 100.0 mL. What is the final dye concentration?

Answer: 7.2 × 10−4 M



3

Transfer 5.00 mL



1

Transfer 10.0 mL



and the concentration in 100.0 mL of the

diluted solution is

cNaCl = 5.50 × 10−3 mol/0.1000 L

= 5.50 × 10−2 M



Question:



cNaCl = 2.75 × 10−4 mol/0.1000 L

= 2.75 × 10−3 M



0.01000 L × 0.550 mol/L

= 5.50 × 10−3 mol NaCl



less error in using larger pipets such as 5.00or 10.00-mL pipets rather than a 1.00-mL

pipet. And then there is a limitation in available glassware. A 200.00-mL volumetric

flask is not often available.



4

Fill to mark with

distilled water.



2

Fill to mark with

distilled water.



NaCl concentration

0.550 mol/L



1/10 original

concentration



100mL



100mL



Original Solution

100.0-mL seawater sample



10.0-mL sample

diluted to 100.0 mL



100mL



1/200 original

concentration



5.00-mL sample

diluted to 100.0 mL



4.6 pH,aConcentrationScaleforAcidsandBases

A sample of vinegar, which contains the weak acid acetic acid, has a hydronium ion

concentration of 1.6 × 10−3 M and “pure” rainwater has [H3O+] = 2.5 × 10−6 M.

These small values can be expressed using scientific notation, but a more convenient way to express such numbers is the logarithmic pH scale.

The pH of a solution is the negative of the base-10 logarithm of the hydronium

ion concentration.

pH ϭ Ϫlog[H3Oϩ]



• Logarithms Numbers less than 1

have negative logs. Defining pH as

−log[H3O+] produces a positive number. See Appendix A for a discussion

of logs.



(4.3)



Taking vinegar, pure water, blood, and ammonia as examples,

pH of vinegar



= −log (1.6 × 10−3 M) = −(−2.80) = 2.80



pH of pure water (at 25 °C) = −log (1.0 × 10−7 M) = −(−7.00) = 7.00

pH of blood



= −log (4.0 × 10−8 M) = −(−7.40) = 7.40



pH of household ammonia = −log (4.3 × 10−12 M) = −(−11.37) = 11.37



you see that something you recognize as acidic has a relatively low pH, whereas ammonia, a common base, has a very low hydronium ion concentration and a high pH.



kotz_48288_04_0156-0207.indd 178



11/19/10 9:09 AM



4.6 pH, a Concentration Scale for Acids and Bases







Vinegar

Soda

Orange

pH = 2.8 pH = 2.9 pH = 3.8



Blood

pH = 7.4



7



Oven cleaner

pH = 11.7



14



Photos © Cengage Learning/Charles D. Winters



0



Ammonia

pH = 11.4



179



FiguRe4.11 pH values of some common substances. Here, the “bar” is colored red at one end

and blue at the other. These are the colors of litmus paper, commonly used in the laboratory to decide

whether a solution is acidic (litmus is red) or basic (litmus is blue).



For aqueous solutions at 25 °C, acids have pH values less than 7, bases have values greater

than 7, and a pH of 7 represents a neutral solution (Figure 4.11). Blood, which your common sense tells you is likely to be neither acidic nor basic, has a pH slightly greater

than 7.

Suppose you know the pH of a solution. To find the hydronium ion concentration you take the antilog of the pH. That is,

[H3Oϩ] ϭ 10ϪpH



(4.4)



For example, the pH of a diet soda is 3.12, and the hydronium ion concentration of

the solution is



• Logs and Your Calculator All scientific calculators have a key marked

“log.” To find an antilog, use the key

marked “10x” or the inverse log. In determining [H3O+] from a pH, when you

enter the value of x for 10x, make sure

it has a negative sign.



[H3O+] = 10−3.12 = 7.6 × 10−4 M



The approximate pH of a solution may be determined using any of a variety of

dyes. Litmus paper contains a dye extracted from a type of lichen, but many other

dyes are also available (Figure 4.12a). A more accurate measurement of pH is done

with a pH meter such as that shown in Figure 4.12b. Here, a pH electrode is immersed in the solution to be tested, and the pH is read from the instrument.



© Cengage Learning/Charles D. Winters



© Cengage Learning/Charles D. Winters



FiguRe4.12 Determining pH.



(a) Some household products. Each solution

contains a few drops of a universal indicator, a

mixture of several acid–base indicators. A

color of yellow or red indicates a pH less than

7. A green to purple color indicates a pH

greater than 7.



kotz_48288_04_0156-0207.indd 179



(b) The pH of a soda is measured with a

modern pH meter. Soft drinks are often quite

acidic, owing to dissolved CO2 and other

ingredients.



11/19/10 9:09 AM



180



c h a p t er 4 Stoichiometry: Quantitative Information about Chemical Reactions



• Weak and Strong Acids and

Hydronium Ion Concentration

Because a weak acid (e.g., acetic acid)

does not ionize completely in water,

the hydronium ion concentration in an

aqueous solution of a weak acid does

not equal the concentration of the acid

(as is the case for strong acids) but

must be calculated using the principles

of chemical equilibrium (▶ Chapter 17).



eXaMPLe 4.8



pH of Solutions



Problem

(a)



Lemon juice has [H3O+] = 0.0032 M. What is its pH?



(b) Seawater has a pH of 8.30. What is the hydronium ion concentration of this solution?

(c)



A solution of nitric acid has a concentration of 0.0056 mol/L. What is the pH of this

solution?



What Do You Know? In part (a) you are given a concentration and asked to calculate the pH,

whereas the opposite is true in (b). For part (c), however, you first must recognize that HNO3 is a

strong acid and is 100% ionized in water.

Strategy Use Equation 4.3 to calculate pH from the H3O+ concentration and Equation 4.4 to

find [H3O+] from the pH.

Solution

(a)



Lemon juice: Because the hydronium ion concentration is known, the pH is found using

Equation 4.3.

pH = −log [H3O+] = −log (3.2 × 10−3) = −(−2.49) = 2.49



(b) Seawater: Here pH = 8.30. Therefore,

[H3O+] = 10−pH = 10−8.30 = 5.0 × 10−9 M

(c)



Nitric acid: Nitric acid, a strong acid (Table 3.1, page 129), is completely ionized in aqueous

solution. Because the concentration of HNO3 is 0.0056 mol/L, the ion concentrations are

[H3O+] = [NO3−] = 0.0056 M

pH = −log [H3O+] = −log (0.0056 M) = 2.25



Think about Your Answer Our answers all agree with the fact that, at 25 °C, solutions with

[H3O+] > 10−7 M have pH < 7, and those with [H3O+] < 10−7 have pH > 7. A comment on logarithms and significant figures (Appendix A) is useful. The number to the left of the decimal point

in a logarithm is called the characteristic, and the number to the right is the mantissa. The mantissa has as many significant figures as the number whose log was found. For example, the logarithm of 3.2 × 10−3 (two significant figures) is 2.49. (The significant figures are the two numbers

to the right of the decimal point.)

Check Your Understanding

(a)



What is the pH of a solution of HCl in which [HCl] = 2.6 × 10−2 M?



(b) What is the hydronium ion concentration in orange juice with a pH of 3.80?



revIeW & cHecK FOr SectIOn 4.6

1.



Which of the solutions listed below has the lowest pH?

(a)



0.10 M HCl



(b) 0.10 M NaOH

2.



2.5 × 10−5 M HNO3



(d) pure H2O



A 0.365-g sample of HCl is dissolved in enough water to give 2.00 × 102 mL of solution.

What is the pH?

(a)



2.000



(b) 0.0500



kotz_48288_04_0156-0207.indd 180



(c)



(c)



1.301



(d) 1.000



11/19/10 9:09 AM



4.7 Stoichiometry of Reactions in Aqueous Solution







181



4.7 StoichiometryofReactions

inAqueousSolution

Solution Stoichiometry

Suppose we want to know what mass of CaCO3 is required to react completely with

25 mL of 0.750 M HCl. The first step in finding the answer is to write a balanced

equation. In this case, we have a gas-forming exchange reaction involving a metal

carbonate and an aqueous acid (Figure 4.13).

+ 2 HCl(aq) n CaCl2(aq) + H2O(ℓ) +

acid 



n 



CO2(g)



+ water + carbon dioxide



salt



This problem differs from the previous stoichiometry problems in that the quantity

of one reactant (HCl) is given as a volume of a solution of known concentration

instead of as a mass in grams. Because our balanced equation is written in terms of

amounts (moles), our first step will be to determine the number of moles of HCl

present from the known information so that we can relate the amount of HCl available to the amount of CaCO3

Amount of HCl ϭ cHClVHCl ϭ



© Cengage Learning/Charles D. Winters



CaCO3(s)



metal carbonate +



FiguRe4.13 A commercial

remedy for excess stomach acid.



0.750 mol HCl

× 0.025 L HCl = 0.019 mol HCl

1 L HCl



This is then related to the amount of CaCO3 required using the stoichiometric factor from the balanced equation.

0.019 mol HCl ×



1 mol CaCO3

= 0.0094 mol CaCO3

2 mol HCl



The tablet contains calcium carbonate, which reacts with hydrochloric acid, the acid present in the

digestive system. The most obvious

product is CO2 gas, but CaCl2(aq) is

also produced.



Finally, the amount of CaCO3 is converted to a mass in grams using the molar mass

of CaCO3 as the conversion factor.

0.0094 mol CaCO3 ×



100. g CaCO3

= 0.994 g CaCO3

1 mol CaCO3



If you follow the general scheme outlined in Problem-Solving Tip 4.4 and pay

attention to the units on the numbers, you can successfully carry out any kind of

stoichiometry calculations involving concentrations.



PrOBLeM SOLvInG tIP 4.4

In Problem-Solving Tip 4.1, you learned

about a general approach to stoichiometry

problems. We can now modify that scheme

for a reaction involving solutions such as

x A(aq) + y B(aq) n products.



Stoichiometry Calculations Involving Solutions

grams reactant A

×



1 mol A

gA



grams reactant B

direct calculation

not possible



moles reactant A

× c molarity A

cA =



mol A

L soln.



Volume of soln. A



kotz_48288_04_0156-0207.indd 181



×



gB

1 mol B



moles reactant B

×



mol reactant B

mol reactant A



stoichiometric factor



×



1

cmolarity B



L soln.

1

=

cB

mol B



Volume of soln. B



11/19/10 9:09 AM



182



c h a p t er 4   Stoichiometry: Quantitative Information about Chemical Reactions



  I nteractive Example 4.9 Stoichiometry of a Reaction



Strategy Map 4.9



in Solution



PROBLEM



Calculate volume of HCl solution

required to consume given mass of

a reactant (Zn)



Problem ​Metallic zinc reacts with aqueous HCl.

Zn(s) + 2 HCl(aq) n ZnCl2(aq) + H2(g)

What volume of 2.50 M HCl, in milliliters, is required to convert 11.8 g of Zn completely to

products?



DATA/INFORMATION



• Mass of Zn

• Concentration of HCl

• Balanced equation

STEP 1.



What Do You Know? ​The balanced equation for the reaction of Zn and HCl(aq) is provided. You

know the mass of zinc and the concentration of HCl(aq).



Write balanced equation.



Strategy ​



Balanced equation is given

Amount reactant (mol)

= mass ì (1 mol/molar mass).







Calculate the amount of zinc.



•



Use a stoichiometric factor (= 2 mol HCl/1 mol Zn) to relate amount of HCl required to

amount of Zn available.



•



Calculate the volume of HCl solution from the amount of HCl and its concentration.



STEP 2.



Amount of Zn



Solution ​Begin by calculating the amount of Zn.

11.8 g Zn ×



Use stoichiometric

factor to relate moles Zn to

amount (mol) acid required.

STEP 3.



1 mol Zn

= 0.180 mol Zn

65.38 g Zn



Use the stoichiometric factor to calculate the amount of HCl required.

0.180 mol Zn ×



Amount of acid required (mol)



2 mol HCl

= 0.360 mol HCl

1 mol Zn



Use the amount of HCl and the solution concentration to calculate the volume.



Volume of acid required

= mol HCl required ×

(1 L/mol HCl).



STEP 4.



0.360 mol HCl ϫ



Volume of acid required (L)



1.00 L solution

ϭ  0.144 L HCl 

2.50 mol HCl



The answer is requested in units of milliliters, so we convert the volume to milliliters and find

that  144 mL of 2.50 M HCl  is required to convert 11.8 g of Zn completely to products.

Think about Your Answer ​You began with much less than 1 mol of zinc, and the concentration

of the HCl solution is 2.50 M. Because the reaction requires 2 mol HCl/1 mol Zn, it makes sense

that your answer should be significantly below 1 L of solution needed. Notice also that this is a

redox reaction in which zinc is oxidized (oxidation number changes from 0 to +2) and hydrogen,

in HCl(aq), is reduced (its oxidation number changes from +1 to 0).

Check Your Understanding 

I​ f you combine 75.0 mL of 0.350 M HCl and an excess of Na2CO3, what mass of CO2, in grams, is

produced?

Na2CO3(s) + 2 HCl(aq) n 2 NaCl(aq) + H2O(ℓ) + CO2(g)



Titration: A Method of Chemical Analysis

Oxalic acid, H2C2O4, is a naturally occurring acid. Suppose you are asked to determine the mass of this acid in an impure sample. Because the compound is an acid,

it reacts with a base such as sodium hydroxide.

H2C2O4(aq) + 2 NaOH(aq) n Na2C2O4(aq) + 2 H2O(ℓ)



You can use this reaction to determine the quantity of oxalic acid present in a given

mass of sample if the following conditions are met:

•

•



kotz_48288_04_0156-0207.indd 182



You can determine when the amount of sodium hydroxide added is just enough

to react with all the oxalic acid present in solution.

You know the concentration of the sodium hydroxide solution and volume that

has been added at exactly the point of complete reaction.



11/19/10 9:09 AM



4.7  Stoichiometry of Reactions in Aqueous Solution



183



Photos © Cengage Learning/Charles D. Winters







Flask containing aqueous solution of sample

being analyzed and

an indicator

(a)

Buret containing aqueous NaOH of

accurately known concentration.



(b)

A solution of NaOH is added slowly to

the sample being analyzed.



Figure 4.14   Titration of an acid in aqueous solution with a base.  (a) A buret, a volumetric

measuring device calibrated in divisions of 0.1 mL (and consequently read to the nearest 0.01 mL), is

filled with an aqueous solution of a base of known concentration. (b) Base is added slowly from the

buret to the solution containing the acid being analyzed and an indicator. (c) A change in the color of

the indicator signals the equivalence point. (The indicator used here is phenolphthalein.)



These conditions are fulfilled in a titration, a procedure illustrated in Figure 4.14.

The solution containing oxalic acid is placed in a flask along with an acid–base indicator, a dye that changes color when the pH of the reaction solution reaches a certain

value. Aqueous sodium hydroxide of accurately known concentration is placed in a

buret. The sodium hydroxide in the buret is added slowly to the acid solution in the

flask. As long as some acid is present in solution, all the base supplied from the buret

is consumed, the solution remains acidic, and the indicator color is unchanged. At

some point, however, the amount of OH− added exactly equals the amount of H3O+

that can be supplied by the acid. This is called the equivalence point. As soon as the

slightest excess of base has been added beyond the equivalence point, the solution

becomes basic, and the indicator changes color (see Figure 4.14). Example 4.10 shows

how to use the equivalence point and the other information to determine the percentage of oxalic acid in a mixture.



  I nteractive Example 4.10 Acid–Base Titration

Problem ​A 1.034-g sample of impure oxalic acid is dissolved in water and an acid–base indicator

added. The sample requires 34.47 mL of 0.485 M NaOH to reach the equivalence point. What is

the mass of oxalic acid, and what is its mass percent in the sample?

What Do You Know? ​You know the mass of oxalic acid (the formula is H2C2O4) and the volume

and concentration of NaOH solution used in the titration.



kotz_48288_04_0156-0207.indd 183



(c)

When the amount of NaOH added from

the buret equals the amount of H3O+

supplied by the acid being analyzed,

the dye (indicator) changes color.



H atom

lost as H+



H atom

lost as H+



oxalic acid H2C2O4



(−)



(−)

oxalate anion C2O42−



Oxalic acid.  Oxalic acid has two

groups that can supply an H+ ion to

solution. Hence, 1 mol of the acid

requires 2 mol of NaOH for complete

reaction.



11/19/10 9:09 AM



184



c h a p t er 4   Stoichiometry: Quantitative Information about Chemical Reactions



Strategy Map 4.10

Strategy ​



PROBLEM



Calculate the mass percent of acid

in an impure sample. Determine the

acid content using an acid-base

titration.

DATA/INFORMATION



• Mass of impure sample



containing acid.

• Volume and concentration of



base used in titration.

STEP 1.



Write balanced equation.



Balanced equation for reaction of

acid (oxalic acid) with base (NaOH)

STEP 2. Amount of base (mol)

= volume (L) × (mol/L).



Amount of base (mol)

STEP 3. Use a stoichiometric

factor to relate amount of base

(mol) to amount (mol) of acid.



Amount of acid (mol) in impure

sample

STEP 4. Mass of acid in sample

= mol acid ì (g/mol).



Mass of acid in impure sample







Write a balanced chemical equation for this acid–base reaction.



•



Calculate the amount of NaOH used in the titration from its volume and concentration.



•



Use the stoichiometric factor defined by the equation to determine the amount of H2C2O4.



•



Calculate the mass of H2C2O4 from the amount and its molar mass.



•



Determine the percent by mass of H2C2O4 in the sample.



Solution ​The balanced equation for the reaction of NaOH and H2C2O4 is

H2C2O4(aq) + 2 NaOH(aq) n Na2C2O4(aq) + 2 H2O(ℓ)

and the amount of NaOH is given by

Amount of NaOH ϭ cNaOH × VNaOH ϭ



0.485 mol NaOH

× 0.03447 L ϭ 0.0167 mol NaOH

L



The balanced equation for the reaction shows that 1 mol of oxalic acid requires 2 mol of

sodium hydroxide. This is the required stoichiometric factor to obtain the amount of oxalic

acid present.

0.0167 mol NaOH ×



1 mol H2C2O4

= 0.00836 mol H2C2O4

2 mol NaOH



The mass of oxalic acid is found from the amount of the acid and its molar mass.

0.00836 mol H2C2O4 ϫ



90.04 g H2C2O4

ϭ  0.753 g H2C2O4 

1 mol H2C2O4



This mass of oxalic acid represents 72.8% of the total sample mass.

0.753 g H2C2O4

ϫ 100% ϭ  72.8% H2C2O4 

1.034 g sample



STEP 5.



Mass % of acid in sample

= (g acid/g sample)100%.

Mass % acid in impure sample



Think about Your Answer ​Problem Solving Tip 4.4 outlines the procedure used to solve this

problem.

Check Your Understanding ​

A 25.0-mL sample of vinegar (which contains the weak acid acetic acid, CH3CO2H) requires

28.33 mL of a 0.953 M solution of NaOH for titration to the equivalence point. What is the mass

of acetic acid, in grams, in the vinegar sample, and what is the concentration of acetic acid in the

vinegar?

CH3CO2H(aq) + NaOH(aq) n NaCH3CO2(aq) + H2O(ℓ)



Standardizing an Acid or Base

In Example 4.10 the concentration of the base used in the titration was given. In

actual practice this usually has to be found by a prior measurement. The procedure

by which the concentration of an analytical reagent is determined accurately is

called standardization, and there are two general approaches.

One approach is to weigh accurately a sample of a pure, solid acid or base

(known as a primary standard) and then titrate this sample with a solution of the base

or acid to be standardized (Example 4.11). An alternative approach to standardizing a solution is to titrate it with another solution that is already standardized (see

“Check Your Understanding” in Example 4.11). This is often done using standard

solutions purchased from chemical supply companies.



kotz_48288_04_0156-0207.indd 184



11/19/10 9:09 AM



4.7  Stoichiometry of Reactions in Aqueous Solution







Example 4.11



185



Standardizing an Acid by Titration



Problem ​Sodium carbonate, Na2CO3, is a base, and an accurately weighed sample can be used

to standardize an acid. A sample of sodium carbonate (0.263 g) requires 28.35 mL of aqueous HCl

for titration to the equivalence point. What is the concentration of the HCl?

What Do You Know? ​The concentration of the HCl(aq) solution is the unknown in this problem.

You know the mass of Na2CO3 and the volume of HCl(aq) solution needed to react completely

with the Na2CO3. You need the molar mass of Na2CO3 and a balanced equation for the reaction.

Strategy ​

•



Write a balanced equation for this acid–base reaction.



•



Calculate the amount of Na2CO3 from the mass and molar mass.



•



Use the stoichiometric factor (from the equation) to find the amount of HCl(aq).



•



The amount of HCl divided by the volume of solution (in liters) gives its concentration

(mol/L).



Solution ​The balanced equation for the reaction is written first.

Na2CO3(aq) + 2 HCl(aq) n 2 NaCl(aq) + H2O(ℓ) + CO2(g)

Calculate the amount of the base, Na2CO3, from its mass.

0.263 g Na2CO3 ×



1 mol Na2CO3

= 0.00248 mol Na2CO3

106.0 g Na2CO3



Next, use the stoichiometric factor to calculate the amount of HCl in 28.35 mL.

0.00248 mol Na2CO3 ϫ



2 mol HCl required

ϭ 0.00496 mol HCl

1 mol Na2CO3 available



Finally, the 28.35-mL (0.02835-L) sample of aqueous HCl contains 0.00496 mol of HCl, so the

concentration of the HCl solution is 0.175 M.

[HCl ] ϭ



0.00496 mol HCl

ϭ  0.175 M HCl 

0.02835 L



Think about Your Answer ​Sodium carbonate is commonly used as a primary standard. It can be

obtained in pure form, can be weighed accurately, and it reacts completely with strong acids.

Check Your Understanding 

​ ydrochloric acid, HCl, can be purchased from chemical supply houses with a concentration of

H

0.100 M, and this solution can be used to standardize the solution of a base. If titrating 25.00 mL

of a sodium hydroxide solution to the equivalence point requires 29.67 mL of 0.100 M HCl, what

is the concentration of the base?



Determining Molar Mass by Titration

In Chapter 2 and in this chapter we used analytical data to determine the empirical

formula of a compound. The molecular formula could then be derived if the molar

mass were known. If the unknown substance is an acid or a base, it is possible to

determine the molar mass by titration.



  I nteractive Example 4.12 Determining the Molar Mass

of an Acid by Titration

Problem ​To determine the molar mass of an organic acid, HA, we titrate 1.056 g of HA with

standardized NaOH. Calculate the molar mass of HA assuming the acid reacts with 33.78 mL of

0.256 M NaOH according to the equation

HA(aq) + NaOH(aq) n NaA(aq) + H2O(ℓ)



kotz_48288_04_0156-0207.indd 185



11/19/10 9:09 AM



186



c h a p t er 4   Stoichiometry: Quantitative Information about Chemical Reactions



Strategy Map 4.12

What Do You Know?  ​You know the mass of the sample of unknown acid, and the volume and

concentration of NaOH(aq). From the equation given, you know that the acid and base react in

a 1∶1 fashion.



PROBLEM



Calculate the molar mass of an acid,

HA, using an acid-base titration.



Strategy  ​The key to this problem is to recognize that the molar mass of a substance is the ratio

of the mass of the sample (g) to its amount (mol). You know the mass, but you need to determine

the amount equivalent to that mass. The equation informs you that 1 mol of HA reacts with 1 mol

of NaOH, so the amount of HA equals the amount of NaOH used in the titration. And, the

amount of NaOH can be calculated from its concentration and volume.



DATA/INFORMATION



• Mass of acid sample.

• Volume and concentration of



base used in titration

STEP 1.



Solution  ​Let us first calculate the amount of NaOH used in the titration.



Write balanced equation.



Amount of NaOH ϭ cNaOHVNaOH ϭ



Balanced equation for reaction of

acid (HA) with base (NaOH)



0.256 mol

ϫ 0.03378 L ϭ 8 .65 × 10Ϫ3 mol NaOH

L



Next, recognize from the balanced equation that the amount of NaOH used in the titration is the

same as the amount of acid titrated. That is,



Amount of base (mol)

= volume (L) ϫ (mol/L).

STEP 2.



8 .65 ϫ 10Ϫ3 mol NaOH ϫ



Amount of base (mol)



1 mol HA

ϭ 8 .65 ϫ 1 0Ϫ3 mol HA

1 mol NaOH



Finally, calculate the molar mass of HA.



Use a stoichiometric

factor to relate amount of base

(mol) to amount (mol) of acid.

STEP 3.



Molar mass of acid ϭ



Amount of acid HA (mol)



1 .056 g HA

ϭ  122 g/mol 

8 .65 × 10Ϫ3 mol HA



Think about Your Answer  ​Note that we assumed the acid had the composition HA.

That is, it could lose one H+ ion per molecule. If the acid had the composition H2A (such as

oxalic acid, page 183) or H3A (for example, citric acid, page 129), the calculated molar mass

would not be correct.



STEP 4. Molar mass = mass of

acid in sample/amount of acid

in sample.



Check Your Understanding ​



Molar mass of acid HA



An acid reacts with NaOH according to the net ionic equation

HA(aq) + OH−(aq) n A−(aq) + H2O(ℓ)

Calculate the molar mass of HA if 0.856 g of the acid requires 30.08 mL of 0.323 M NaOH.



Titrations Using Oxidation–Reduction Reactions

Analysis by titration is not limited to acid–base chemistry. Many oxidation–reduction

reactions go rapidly to completion in aqueous solution, and methods exist to determine their equivalence points.



Example 4.13 Using an Oxidation–Reduction Reaction in a Titration

Problem  ​The iron in a sample of an iron ore can be converted quantitatively to the iron(II) ion,

Fe2+, in aqueous solution, and this solution can then be titrated with aqueous potassium permanganate, KMnO4. The balanced, net ionic equation for the reaction occurring in the course of

this titration is





MnO4−(aq) + 5 Fe2+(aq) + 8 H3O+(aq)  n  Mn2+(aq) + 5 Fe3+(aq) + 12 H2O(ℓ)

purple



colorless



colorless



pale yellow



A 1.026-g sample of iron-containing ore requires 24.35 mL of 0.0195 M KMnO4 to reach the

equivalence point. What is the mass percent of iron in the ore?

What Do You Know?  ​You know the concentration and volume of the KMnO4 solution used to

titrate Fe2+(aq) to the equivalence point. The stoichiometric factor relating amounts of KMnO4

and Fe2+(aq) is derived from the balanced equation.



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187



4.7 Stoichiometry of Reactions in Aqueous Solution







How Much Salt Is There in Seawater?



Saltiness is one of the basic

taste sensations, and a taste

of seawater quickly reveals it is salty. How did

the oceans become salty?

A result of the interaction of atmospheric CO2 and water is hydronium ions

and bicarbonate ions.

CO2(g) + H2O(ℓ) n H2CO3(aq)

H2CO3(aq) + H2O(ℓ)

uv H3O+(aq) + HCO3−(aq)



Indeed, this is the reason rain is normally

acidic, and this slightly acidic rainwater can

then cause substances such as limestone or

corals to dissolve, producing calcium ions

and more bicarbonate ions.

CaCO3(s) + H3O+(aq) n

Ca2+(aq) + HCO3−(aq) + H2O(ℓ)



Sodium ions arrive in the oceans by a

similar reaction with sodium-bearing minerals such as albite, NaAlSi3O6. Acidic rain

falling on the land extracts sodium ions that

are then carried by rivers to the ocean.

The average chloride content of rocks in

the earth’s crust is only 0.01%, so only a

minute proportion of the chloride ion in the

oceans can come from the weathering of

rocks and minerals. What then is the origin

of the chloride ions in seawater? The answer

is volcanoes. Hydrogen chloride gas, HCl, is

a constituent of volcanic gases. Early in

Earth’s history, the planet was much hotter,

and volcanoes were much more widespread.



The HCl gas emitted from these volcanoes is

very soluble in water and quickly dissolves

to give a dilute solution of hydrochloric acid.

The chloride ions from dissolved HCl gas

and sodium ions from weathered rocks are

the source of the salt in the sea.

Suppose you are an oceanographer, and

you want to determine the concentration of

chloride ions in a sample of seawater. How

can you do this? And what results might you

find?

There are several ways to analyze a solution for its chloride ion content; among

them is the classic “Mohr method.” Here, a

solution containing chloride ions is titrated

with standardized silver nitrate. You know

that the following reaction should occur:



John C. Kotz



CASE STUDY



Ag+(aq) + Cl−(aq) n AgCl(s)



and will continue until the chloride ions have

been precipitated completely. To detect the

equivalence point of the titration of Cl− with

Ag+, the Mohr method involves the addition

of a few drops of a solution of potassium

chromate. This “indicator” works because

silver chromate is slightly more soluble than

AgCl, so the red Ag2CrO4 precipitates only

after all of the AgCl is precipitated.

2 Ag+(aq) + CrO42–(aq) n Ag2CrO4(s)



The appearance of the red color of Ag2CrO4

(◀ Figure 3.12d) signals the equivalence point.



Question:

Using the following information, calculate

the chloride ion concentration in a sample of

seawater.

a. Volume of original seawater sample

= 100.0 mL.

b. A 10.00-mL sample of the seawater

was diluted to 100.0 mL with distilled

water.

c. 10.00 mL of the diluted sample was

again diluted to 100.0 mL.

d. A Mohr titration was performed on

50.00 mL of the diluted sample (from

step c) and required 26.25 mL of

0.100 M AgNO3. What was the chloride ion concentration in the original

seawater sample?

The answer to this question is available in

Appendix N.



Strategy

•



Use the volume and concentration of the KMnO4 solution to calculate the amount of KMnO4

used in the titration.



•



Use the stoichiometric factor to determine the amount of Fe2+ from the amount of KMnO4.



•



Convert the amount of Fe2+ to mass of iron using the molar mass of iron.



•



Calculate the mass percent of iron in the sample.



Solution First, calculate the amount of KMnO4.

Amount of KMnO4 ϭ cKMnO4 ϫ VKMnO4 ϭ



0.0195 mol KMnO4

ϫ 0.02435 L ϭ 0.000475 mol

L



Use the stoichiometric factor to calculate the amount of iron(II) ion.

0.000475 mol KMnO4 ×



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5 mol Fe2 +

ϭ 0.000237 mol Fe2+

1 mol KMnO4



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188



c h a p t er 4 Stoichiometry: Quantitative Information about Chemical Reactions



Next, calculate the mass of iron.

0.00237 mol Fe2 + ×



55.85 g Fe2 +

ϭ 0.133 g Fe2+

1 mol Fe2+



Finally, the mass percent can be determined.

0.133 g Fe2+

× 100% ϭ 12.9% iron

1.026 g sample



© Cengage Learning/Charles D. Winters



Think about Your Answer The mass of the sample is 1.026 g, so the mass of iron in the sample

must be less than this, as confirmed by the answer. The titration of iron(II) ions with KMnO4 is a

useful analytical reaction because it is easy to detect when all the iron(II) ion has reacted. The

MnO4− ion is a deep purple color, but when it reacts with Fe2+ the color disappears because the

reaction product, Mn2+, is colorless. Therefore, KMnO4 solution is added from a buret until the

initially colorless, Fe2+-containing solution just turns a faint purple color (due to unreacted

KMnO4), the signal that the equivalence point has been reached.

Check Your Understanding



Using an oxidation–reduction

reaction for analysis by titration.

Purple, aqueous KMnO4 is added to a

solution containing Fe2+. As KMnO4

drops into the solution, colorless

Mn2+ and pale yellow Fe3+ form.

Here, an area of the solution containing unreacted KMnO4 is seen. As

the solution is mixed, this disappears

until the equivalence point is

reached.



Vitamin C, ascorbic acid (C6H8O6), is a reducing agent. One way to determine the ascorbic acid

content of a sample is to mix the acid with an excess of iodine,

C6H8O6(aq) + I2(aq) + 2 H2O(ℓ) n C6H6O6(aq) + 2 H3O+(aq) + 2 I−(aq)

and then titrate the iodine that did not react with the ascorbic acid with sodium thiosulfate. The

balanced, net ionic equation for the reaction occurring in this titration is

I2(aq) + 2 S2O32−(aq) n 2 I−(aq) + S4O62−(aq)

Suppose 50.00 mL of 0.0520 M I2 was added to the sample containing ascorbic acid. After the

ascorbic acid/I2 reaction was complete, the I2 not used in this reaction required 20.30 mL of

0.196 M Na2S2O3 for titration to the equivalence point. Calculate the mass of ascorbic acid in the

unknown sample.



revIeW & cHecK FOr SectIOn 4.7

1.



A 25-mL sample of 0.40 M HCl is added to 32 mL of 0.30 M NaOH. Is the resulting solution

acidic, basic, or neutral?

(a)



2.



acidic



(b) basic



(c)



neutral



What volume of 0.250 M NaOH is required to react completely with 0.0100 moles of H2SO4?

(a)



80.0 mL



(b) 60.0 mL



(c)



40.0 mL



(d) 125 mL



4.8 Spectrophotometry

Solutions of many compounds are colored, a consequence of the absorption of light

(Figure 4.15). It is possible to measure, quantitatively, the extent of light absorption

and to relate this to the concentration of the dissolved solute. This kind of experiment, called spectrophotometry, is an important analytical method and you may

have used it in your laboratory course.

Every substance absorbs or transmits certain wavelengths of radiant energy but

not others (Figures 4.15 and 4.16). For example, nickel(II) ions (and chlorophyll)

absorb red and blue/violet light while they transmit green light. Your eyes “see” the

transmitted or reflected wavelengths, those not absorbed, as the color green. Furthermore, the specific wavelengths absorbed and transmitted are characteristic for

a substance, so a spectrum serves as a “fingerprint” of the substance that can help

identify an unknown.



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