Chapter 8 Acids and Bases

Chapter 8 Acids and Bases
8.1 Acids and Bases

Acids Arrhenius acids produce H+ ions in water H2O(l)
HCl(g)  H+(aq) + Cl(aq) are electrolytes have a sour taste turn litmus red neutralize bases

Naming Acids Acids with H and a nonmetal are named with the prefix hydro and end with ic acid. HCl hydrochloric acid Acids with H and a polyatomic ion are named by changing the end of the name of the polyatomic ion from ate to ic acid or ite to ous acid ClO3 chlorate HClO3 chloric acid ClO2 chlorite HClO2 chlorous acid

Names of Common Acids

Learning Check Select the correct name for each of the following acids. 1. HBr A. bromic acid B. bromous acid C. hydrobromic acid 2. H2CO3 A. carbonic acid B. hydrocarbonic acid C. carbonous acid 3. HBrO2 A. bromic acid B. hydrobromous acid C. bromous acid

Solution 1. HBr Br, bromide C. hydrobromic acid
The name of an acid with H and a nonmetal uses the prefix hydro and ends with ic acid. 2. H2CO3 CO32, carbonate A. carbonic acid An acid with H and a polyatomic ion ending in ate is called an ic acid. 3. HBrO2 BrO2, bromite C. bromous acid An acid with H and a polyatomic ion ending in ite is called an ous acid.

Bases Arrhenius bases produce OH ions in water taste bitter or chalky
are electrolytes feel soapy and slippery neutralize acids

Naming Bases Bases with OH ions are named as the hydroxide of the metal in the formula. NaOH sodium hydroxide KOH potassium hydroxide Ba(OH)2 barium hydroxide Al(OH)3 aluminum hydroxide Fe(OH)3 iron(III) hydroxide

Learning Check Match the formulas with the names.
1. ___HNO2 A. iodic acid 2. ___Ca(OH)2 B. sulfuric acid 3. ___H2SO4 C. sodium hydroxide 4. ___HIO3 D. nitrous acid 5. ___NaOH E. calcium hydroxide

Solution Match the formulas with the names.
1. _D__HNO2 D. nitrous acid 2. _E__Ca(OH)2 E. calcium hydroxide 3. _B__H2SO4 B. sulfuric acid 4. _A__HIO3 A. iodic acid 5. _C__NaOH C. sodium hydroxide

Brønsted–Lowry Acids and Bases
According to the Brønsted–Lowry theory, acids donate a proton (H+) bases accept a proton (H+)

NH3, a Brønsted–Lowry Base
In the reaction of ammonia and water, NH3 is the base that accepts H+ H2O is the acid that donates H+

Characteristics of Acids and Bases

Learning Check In each of the following identify the Brønsted–Lowry acid and base: A. HNO3(aq) + H2O(l)  H3O+(aq) + NO3(aq) B. HF(aq) + H2O(l) H3O+ (aq) + F(aq)

Solution In each of the following identify the Brønsted–Lowry acid and base: A. HNO3(aq) + H2O(l)  H3O+(aq) + NO3(aq) acid base B. HF(aq) + H2O(l) H3O+ (aq) + F(aq) acid base

Learning Check Identify each as a characteristic of an
A. acid or B. base. ____1. has a sour taste ____2. produces OH in aqueous solutions ____3. has a chalky taste ____4. is an electrolyte ____5. produces H+ in aqueous solutions

Solution Identify each as a characteristic of an A. acid or B. base.
A 1. has a sour taste B 2. produces OH in aqueous solutions B 3. has a chalky taste A, B 4. is an electrolyte A 5. produces H+ in aqueous solutions

Conjugate Acid–Base Pairs
In any acid–base reaction, there are two conjugate acid–base pairs. Each pair is related by the loss and gain of H+. One pair occurs in the forward direction. One pair occurs in the reverse direction. conjugate acid–base pair 1 HA + B A + BH+ conjugate acid–base pair 2

In this acid–base reaction, an acid, HF, donates H+ to form its conjugate base, F a base, H2O, accepts H+ to form its conjugate acid, H3O+ there are two conjugate acid–base pairs

In this acid–base reaction, the first conjugate acid–base pair is HF, which donates H+ to form its conjugate base, F the other conjugate acid–base pair is H2O, which accepts H+ to form its conjugate acid, H3O+ each pair is related by a loss and gain of H+

In the reaction of NH3 and H2O, one conjugate acid–base pair is NH3/NH4+ the other conjugate acid–base is H2O/H3O+

Learning Check 1. Write the conjugate base of each of the following:
A. HBr B. H2S C. H2CO3 2. Write the conjugate acid of each of the following: A. NO2 B. NH3 C. OH

Solution 1. Write the conjugate base of each of the following:
Subtract an H+ from each acid to get the conjugate base: A. HBr – H+  Br B. H2S – H+  HS C. H2CO3 – H+  HCO3

Solution 2. Write the conjugate acid of each of the following:
Add an H+ to each base to get the conjugate acid. A. NO2 + H+  HNO2 B. NH3 + H  NH4+ C. OH + H+  H2O

Learning Check Identify the sets that contain acid–base conjugate pairs. 1. HNO2, NO2 2. H2CO3, CO32 3. HCl, ClO4 4. HS, H2S 5. NH3, NH4+

Solution Identify the sets that contain acid–base conjugate pairs.
1. HNO2, NO2 acid, conjugate base 2. H2CO3, CO32 not acid–base conjugate pair 3. HCl, ClO4 not acid–base conjugate pair 4. HS, H2S base, conjugate acid 5. NH3, NH4+ base, conjugate acid

Learning Check 1. The conjugate base of HCO3 is
A. CO32 B. HCO3 C. H2CO3 2. The conjugate acid of HCO3 is A. CO32 B. HCO3 C. H2CO3 3. The conjugate base of H2O is A. OH B. H2O C. H3O+ 4. The conjugate acid of H2O is

Solutions 1. The conjugate base of HCO3 is A. CO32 B. HCO3 C. H2CO3
2. The conjugate acid of HCO3 is A. CO32 B. HCO3 C. H2CO3 3. The conjugate base of H2O is A. OH B. H2O C. H3O+ 4. The conjugate acid of H2O is

8.2 Strengths of Acids and Bases

Acids A strong acid completely ionizes (100%) in aqueous solutions.
HCl(g) + H2O(l)  H3O+(aq) + Cl(aq) A weak acid dissociates only slightly in water to form a few ions in aqueous solutions. H2CO3(aq) + H2O(l) H3O+(aq) + HCO3(aq)

Strong Acids In water, the dissolved molecules of HA, a strong acid,
dissociate into ions 100% give large concentrations of H3O+ and the anion (A) HCl(g) + H2O(l)  H3O+(aq) + Cl(aq)

Strong Acids Strong acids make up six of all the acids
have weak conjugate bases

Weak Acids In weak acids, only a few molecules dissociate
most of the weak acid remains as the undissociated (molecular) form of the acid the concentrations of the H3O+ and the anion (A) are small H2CO3(aq) + H2O(l) H3O+(aq) + HCO3(aq)

Weak Acids Weak acids make up most of the acids
have strong conjugate bases

Strong and Weak Acids In an HCl solution, the strong acid HCl dissociates 100%. A solution of the weak acid HC2H3O2 contains mostly molecules and a few ions.

Comparing Strong and Weak Acids

Strong Bases Strong bases
are formed from metals of Groups 1A (1) and 2A (2) include LiOH, NaOH, KOH, and Ca(OH)2 dissociate completely in water KOH(s)  K+(aq) + OH(aq)

Learning Check Identify each of the following as a strong or weak acid or base. A. HBr B. HNO2 C. NaOH D. H2SO4 E. Cu(OH)2

Solution Identify each of the following as a strong or weak acid or base. A. HBr strong acid B. HNO2 weak acid C. NaOH strong base D. H2SO4 strong acid E. Cu(OH)2 weak base

Learning Check Identify the stronger acid in each pair. A. HNO2 or H2S
B. HCO3 or HBr C. H3PO4 or H3O+

Solution Identify the stronger acid in each pair.
A. HNO2 or H2S HNO2 is the stronger acid. B. HCO3 or HBr HBr is the stronger acid. C. H3PO4 or H3O+ H3O+ is the stronger acid.

8.3 Ionization of Water In a neutral solution, [H3O+] and [OH] are equal. In acidic solutions, the [H3O+] is greater than the [OH]. In basic solutions, the [OH] is greater than the [H3O+].

Ionization of Water In water,
H+ is transferred from one H2O molecule to another one water molecule acts as an acid, while another acts as a base H2O H2O H3O OH water water hydronium hydroxide ion (+) ion ()

Pure Water Is Neutral In pure water,
the ionization of water molecules produces small but equal quantities of H3O+ and OH ions molar concentrations are indicated in brackets, as [H3O+] and [OH] [H3O+] = 1.0  107 M [OH−] = 1.0  107 M

Acidic Solutions Adding an acid to pure water increases the [H3O+]
causes the [H3O+] to exceed 1.0  107 M decreases the [OH−]

Basic Solutions Adding a base to pure water increases the [OH−]
causes the [OH−] to exceed 1.0  107M decreases the [H3O+]

Comparison of [H3O+] and [OH−]

Ion Product of Water, Kw The ion product constant, Kw, for water
is the product of the concentrations of the hydronium and hydroxide ions Kw = [H3O+] [OH] can be obtained from the concentrations in pure water Kw = [1.0  107 M]  [1.0  107 M] =  1014

[H3O+] and [OH−] in Solutions
In neutral, acidic, or basic solutions, the Kw is always 1.0 x 1014.

Guide to Calculating [H3O+]

Calculating [H3O+] What is the [H3O+] of a solution if [OH] is 5.0  108 M? Step 1 Write the Kw for water. Kw = [H3O+][OH] = 1.0  1014 Step 2 Solve the Kw expression for unknown [H3O+]. [H3O+] =  1014 [OH]

Calculating [H3O+] What is the [H3O+] of a solution if [OH] is 5.0  108 M? Step 3 Substitute in the known [H3O+] or [OH] and calculate. [H3O+] = 1.0  1014 = 2.0  107 M 5.0  10−8

Learning Check If lemon juice has [H3O+] of 2  103 M, what is the [OH] of the solution? A. 2  1011 M B. 5  1011 M C. 5  1012 M

Solution If lemon juice has [H3O+] of 2  103 M, what is the [OH] of the solution? Step 1 Write the Kw for water. Kw = [H3O+ ][OH] = 1.0  1014 Step 2 Solve the Kw expression for unknown [OH]. [OH] =  1014 [H3O+]

Solution If lemon juice has [H3O+] of 2  103 M, what is the [OH] of the solution? Step 3 Substitute in the known [H3O+] or [OH] and calculate. [OH] = 1.0  10 = 5  1012 M 2  103 The answer is C, 5  1012 M.

Learning Check The [OH] of an ammonia solution is 4.0  102 M. What is the [H3O+] of the solution? A. 2.5  1011 M B. 2.5  1012 M C. 2.5  1013 M

Solution The [OH] of an ammonia solution is 4.0  102 M. What is the [H3O+] of the solution? Step 1 Write the Kw for water. Kw = [H3O+][OH] = 1.0  1014 Step 2 Solve the Kw expression for unknown [H3O+]. [H3O+] =  1014 [OH]

Solution The [OH] of an ammonia solution is 4.0  102 M. What is the [H3O+] of the solution? Step 3 Substitute in the known [H3O+] or [OH] and calculate. [ H3O+] =  1014 = 2.5  1013 M 4.0  102 The answer is C, 2.5  1013 M.

8.4 The pH Scale

Rain Is Naturally Acidic
Natural rain is slightly acidic with a pH of 5.6. Two reactions that take place in the atmosphere account for this: CO2(g) + H2O(l) H2CO3(aq) H2CO3(aq) + H2O(l) H3O+(aq) + HCO3(aq)

The pH Scale The pH of a solution
is used to indicate the acidity of a solution has values that usually range from 0 to 14 is acidic when the values are less than 7 is neutral with a pH of 7 is basic when the values are greater than 7

pH of Everyday Substances
On the pH scale, values below 7.0 are acidic, a value of 7.0 is neutral, and values above 7.0 are basic.

Learning Check Identify each solution as acidic, basic, or neutral.
A. ___ HCl with a pH = 1.5 B. ___ pancreatic fluid, [H3O+] = 1  108 M C. ___ Sprite soft drink, pH = 3.0 D. ___ pH = 7.0 E. ___ [OH] = 3  1010 M F. ___ [H3O+ ] = 5  1012

Solution Identify each solution as acidic, basic , or neutral.
A. ___ HCl with a pH = acidic B. ___ pancreatic fluid, [H3O+] = 1  108 M basic C. ___ Sprite soft drink, pH = acidic D. ___ pH = neutral E. ___ [OH] = 3  1010 M basic F. ___ [H3O+ ] = 5  1012 basic

Acid Rain Forms from Fossil Fuels
In some parts of the U.S. and Europe, the burning of fossil fuels and coal releases sulfur impurities that form SO2 and SO3 gases. These gases react with moisture in the air to form acid rain, lowering the pH of soil, lakes, and streams. 2 SO2(g) + O2(g) SO3(g) SO3(g) + H2O(l) H2SO4(aq)

Acid Rain Forms from Fossil Fuels
When the pH of lakes and streams falls below 4.5–5, most fish and plant life die. Trees and forests are damaged as the acid rain breaks down the waxy coating on leaves, interfering with photosynthesis. Acid rain has severely damaged forests in Eastern Europe.

Calculating pH of Solutions
pH is the negative log of the hydronium ion concentration. pH = log [H3O+] Example: For a solution with [H3O+] = 1  10 4 pH = log [1  10 4 ] pH = [4.0] pH = 4.0 Note: The number of decimal places in the pH equals the significant figures in the coefficient of [H3O+]. SF: 1  104

Significant Figures in pH Calculations
When expressing log values, the number of decimal places in the pH is equal to the number of significant figures in the coefficient of [H3O+]. [H3O+] = 1  10 4 pH = 4.0 [H3O+] = 8.0  106 pH = 5.10 [H3O+] = 2.4  108 pH = 7.62

Guide to Calculating pH of Solutions

Calculating pH Find the pH of a solution with a [H3O+] of 1.0  103:
Step 1 Enter the [H3O+]. Enter 1  103 by pressing 1 (EE) 3. The EE key gives an exponent of 10. Enter change sign (+/ key or – key). Step 2 Press the log key and change the sign. log (1  103) = [3]

Calculating pH Find the pH of a solution with a [H3O+] of 1.0  103:
Step 3 Adjust the number of significant figures on the right of the decimal point to equal the SF in the coefficient. The answer is 3 with 2 decimal places: 3.00. (Two significant figures in 1.0  103)

Learning Check What is the pH of coffee if the [H3O+] is 1  105 M?
A. pH = 9.0 B. pH = 7.0 C. pH = 5.0

Solution Step 1 Enter the [H3O+]. Enter 1  105 by pressing 1 (EE) 5.
What is the pH of coffee if the [H3O+] is 1  105 M? Step 1 Enter the [H3O+]. Enter 1  105 by pressing 1 (EE) 5. The EE key gives an exponent of 10. Enter change sign (+/ key or – key). Step 2 Press the log key and change the sign. log (1  105) = [5]

Solution What is the pH of coffee if the [H3O+] is 1  105 M? Step 3 Adjust the number of significant figures on the right of the decimal point to equal the SF in the coefficient. The pH is 5 with 1 decimal place: 5.0, answer C. (One significant figure in 1  105)

Learning Check The [H3O+] of tomato juice is 2  104 M.
What is the pH of the solution? A B C

Solution 1. The [H3O+] of tomato juice is 2  104 M.
Step 1 Enter the [H3O+]. Enter 2  104 by pressing 2 (EE) 4. The EE key gives an exponent of 10. Enter change sign (+/ key or – key). Step 2 Press the log key and change the sign. log (2  104) = [3.7] Step 3 Adjust the number of significant figures on the right of the decimal point to equal the SF in the coefficient. pH = log [ 2  104] = 3.7 The answer is B.

[H3O+], [OH], and pH Values

Calculating [H3O+] from pH
The [H3O+] can be expressed by using the pH as the negative power of 10. [H3O+] = 1  10pH For pH = 3.0, the [H3O+] = 1  103 On a calculator: 1. Enter the pH value 2. Change sign 3.0 3. Use the inverse log key (or 10x) to obtain the [H3O+]. = 1  103 M

Learning Check 1. What is the [H3O+] of a solution with a pH of 10.0?
A. 1  104 M B. 1  1010 M C. 1  1010 M 2. What is the [OH] of a solution with a pH of 2.00? A. 1.0  102 M B. 1.0  1012 M C. 2.0 M

Solution 1. What is the [H3O+] of a solution with a pH of 10.0?
Using equation: 1  10pH = 1  1010 M The answer is C, 1  1010 M. 2. What is the [OH] of a solution with a pH of 2.00? [H3O+] = 1.0  102 M 1  10pH [OH] = 1.0  1014 = 1.0  1012 M 1.0  102 The answer is B, 1.0  1012 M.

8.5 Reactions of Acids and Bases

Acids and Metals Acids react with metals
such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn to produce hydrogen gas and the salt of the metal Molecular equations: 2K(s) HCl(aq)  2KCl(aq) + H2(g) metal acid salt hydrogen gas Zn(s) HCl(aq)  ZnCl2(aq) + H2(g) metal acid salt hydrogen gas

Learning Check Write a balanced equation for the reaction of magnesium metal with HCl(aq). Label the metal, acid, and salt.

Learning Check Write a balanced equation for the reaction of magnesium metal with HCl(aq). Mg(s) HCl(aq)  MgCl2(aq) + H2(g) metal acid salt

Acids and Carbonates Acids react
with carbonates and hydrogen carbonates to produce carbon dioxide gas, a salt, and water 2HCl(aq) + CaCO3(s)  CO2(g) + CaCl2(aq) + H2O(l) acid carbonate carbon salt water dioxide HCl(aq) + NaHCO3(s)  CO2(g) + NaCl (aq) + H2O(l) acid bicarbonate carbon salt water

Learning Check Write a balanced equation for the following reactions:
A. MgCO3(s) + HBr(aq)  B. HCl(aq) + NaHCO3(aq) 

Solution Write a balanced equation for the following reactions:
A. MgCO3(s) + 2HBr(aq)  MgBr2(aq) + CO2(g) + H2O(l) carbonate acid salt carbon water dioxide B. HCl(aq) + NaHCO3(aq)  NaCl(aq) + CO2(g) + H2O(l) acid bicarbonate salt carbon water

Neutralization Reactions
In a neutralization reaction, an acid reacts with a base to produce salt and water the acid HCl reacts with NaOH to produce salt and water the salt formed is the anion from the acid and cation of the base HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) acid base salt water

Neutralization Reactions
In neutralization reactions, if we write the strong acid and strong base as ions, we see that H+ reacts with OH− to form water, leaving the ions Na+ and Cl in solution: H+(aq) + Cl(aq) + Na+(aq) + OH(aq)  Na+(aq) + Cl(aq) + H2O(l) the overall reaction is H3O+ from the acid and OH from the base form water: H+(aq) OH(aq)  H2O(l)

Guide for Balancing Neutralization Reactions

Balancing Neutralization Reactions
Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid. Step 1 Write the reactants and products. Mg(OH)2 + HNO3 Step 2 Balance the H+ in the acid with the OH in the base. Mg(OH) HNO3 Step 3 Balance the H2O with H+ and the OH. Mg(OH) HNO3  salt + 2H2O Step 4 Write the salt from the remaining ions. Mg(OH) HNO3  Mg(NO3) H2O

Learning Check Select the correct group of coefficients for each of the following neutralization equations. 1. HCl(aq) + Al(OH)3(aq)  AlCl3(aq) + H2O(l) A. 1, 3, 3, B. 3, 1, 1, C. 3, 1, 1, 3 2. Ba(OH)2(aq) + H3PO4(aq)  Ba3(PO4)2(s) + H2O(l) A. 3, 2, 2, B. 3, 2, 1, C. 2, 3, 1, 6

Solution 1. HCl(aq) + Al(OH)3(aq)  AlCl3(aq) + H2O(l)
Step 1 Write the reactants and products. HCl + Al(OH)3 Step 2 Balance the H+ in the acid with the OH in the base. 3HCl + Al(OH)3 Step 3 Balance the H2O with H+ and the OH. 3HCl + Al(OH)3  salt + 3H2O Step 4 Write the salt from the remaining ions. 3HCl(aq) + Al(OH)3(aq)  AlCl3(aq) + 3H2O(l) The answer is C. 3, 1, 1, 3.

Solution 2. Ba(OH)2(aq) + H3PO4(aq)  Ba3(PO4)2(s) + H2O(l)
Step 1 Write the reactants and products. Ba(OH)2 + H3PO4 Step 2 Balance the H+ in the acid with the OH in the base. 3Ba(OH)2 + 2H3PO4 Step 3 Balance the H2O with H+ and the OH. 3Ba(OH)2 + 2H3PO4  salt + 6H2O Step 4 Write the salt from the remaining ions. 3Ba(OH)2(aq) + 2H3PO4(aq)  Ba3(PO4)2(s) + 6H2O(l) The answer is B. 3, 2, 1, 6.

Basic Compounds in Antacids

Learning Check Write the neutralization reactions for stomach acid, HCl, and the ingredients in Mylanta. Mylanta: Al(OH)3 and Mg(OH)2

Solution Write the neutralization reactions for stomach acid, HCl, and
the ingredients in Mylanta. Mylanta: For Al(OH)3: Step 1 Write the reactants and products. Al(OH)3 + HCl Step 2 Balance the H+ in the acid with the OH in the base. Al(OH)3 + 3HCl Step 3 Balance the H2O with H+ and the OH. Al(OH)3 + 3HCl  salt + 3H2O Step 4 Write the salt from the remaining ions. Al(OH)3(aq) + 3HCl(aq)  AlCl3(aq) + 3H2O(l)

Solution Write the neutralization reactions for stomach acid, HCl, and
the ingredients in Mylanta. Mylanta: For Mg(OH)2: Step 1 Write the reactants and products. Mg(OH)2 + HCl Step 2 Balance the H+ in the acid with the OH in the base. Mg(OH)2 + 2HCl Step 3 Balance the H2O with H+ and the OH. Mg(OH)2 + 2HCl  salt + 2H2O Step 4 Write the salt from the remaining ions. 2HCl(aq) + Mg(OH)2(aq)  MgCl2(aq) + 2H2O(l)

Acid–Base Titration Titration
is a laboratory procedure used to determine the molarity of an acid uses a base such as NaOH to neutralize a measured volume of an acid Base  NaOH Acid  Solution

Indicator An indicator is added to the acid in the flask
causes the solution to change color when the acid is neutralized

Endpoint of Titration At the endpoint,
the indicator gives the solution a permanent pink color the volume of the base used to reach the endpoint is measured the molarity of the acid is calculated using the neutralization equation for the reaction

Guide to Calculations for Acid–Base Titrations

Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Step 1 State given and needed quantities. Given: 18.5 mL of M NaOH 10.0 mL of HCl Needed: Molarity of HCl

What is the molarity of an HCl solution if 18.5 mL of M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Step 2 Write a plan to calculate molarity or volume. molarity equation molarity NaOH coefficients HCl 18.5 mL  L  moles NaOH  moles HCl  L HCl

What is the molarity of an HCl solution if 18.5 mL of M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Step 3 State equalities and conversion factors including concentration. 1 L = 1000 mL 0.225 mole NaOH and mole HCl x 1 L NaOH mole NaOH

What is the molarity of an HCl solution if 18.5 mL of M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Step 4 Set up the problem to calculate the needed quantity mL NaOH  1 L NaOH  mole NaOH 1000 mL NaOH 1 L NaOH  1 mole HCl = mole of HCl 1 mole NaOH MHCl = mole HCl = M HCl L HCl

Learning Check Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) A mL B mL C mL

Solution Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Step 1 State given and needed quantities. Given: M H2SO4 50.0 mL of 1.00 M KOH Needed: mL of H2SO4

Solution Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Step 2 Write a plan to calculate molarity or volume. molarity equation molarity KOH coefficients H2SO4 50.0 mL  L  moles KOH  moles H2SO4  L H2SO4

Solution Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Step 3 State equalities and conversion factors including concentration. 1 L = 1000 mL 1 L H2SO and mole H2SO4 2.00 mole H2SO mole KOH

Solution Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) Step 4 Set up the problem to calculate the needed quantity. 50.0 mL  1 L = L 103 mL L KOH  1.00 mole KOH  1 mole H2SO4 1 L KOH mole KOH  1 L H2SO  mL = 12.5 mL. 2.00 mole H2SO L H2SO4 The answer is A.

8.6 Buffers

Buffers When an acid or base is added to water, the pH changes drastically. In a buffer solution, the pH is maintained; pH does not change when acid or base is added.

How Buffers Work Buffers work because they
resist changes in pH from the addition of acid or base in the body, absorb H3O+ or OH from foods and cellular processes to maintain pH are important in the proper functioning of cells and blood maintain a pH close to 7.4 in blood. A change in the pH of the blood affects the uptake of oxygen and cellular processes

Components of a Buffer A buffer solution
contains a combination of acid–base conjugate pairs For example: HA and NaA may contain a weak acid and a salt of its conjugate base typically has equal concentrations of a weak acid and its salt may also contain a weak base and a salt of the conjugate acid

How Buffers Work In the acetic acid/acetate buffer with acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2), the salt produces acetate ions and sodium ions NaC2H3O2(aq)  C2H3O2(aq) + Na+(aq) the salt is added to provide a higher concentration of the conjugate base C2H3O2 than the weak acid alone HC2H3O2(aq) + H2O(l) C2H3O2(aq) + H3O+(aq) Large amount Large amount

Function of a Weak Acid in a Buffer
The function of the weak acid in a buffer is to neutralize a base. The acetate ion produced in the neutralization reaction adds to the concentration of acetate already in solution from the salt. HC2H3O2 + OH  C2H3O2 + H2O acetic acid base acetate ion water

Function of Conjugate Base in a Buffer
The function of the acetate ion, C2H3O2, is to neutralize H3O+ from acids. The acetic acid produced contributes to the available weak acid. C2H3O2 + H3O+  HC2H3O2 + H2O acetate ion acid acetic acid water

Working Buffers Buffers work because
the weak acid in a buffer neutralizes bases and the conjugate base in the buffer neutralizes acids the pH of the solution is maintained The buffer described here consists of about equal concentrations of acetic acid (HC2H3O2) and its conjugate base, acetate ion (C2H3O2). Adding H3O+ to the buffer reacts with C2H3O2 whereas adding OH neutralizes HC2H3O2. The pH of the solution is maintained as long as the added amounts of acid or base are small compared to the concentrations of the buffer components.

Learning Check Which of the following combinations make a buffer solution? A. HCl and KCl B. H2CO3 and NaHCO3 C. H3PO4 and NaCl D. HC2H3O2 and KC2H3O2

Solution Which of the following combinations make a buffer solution?
A buffer consists of a weak acid and salt of its conjugate base. A. HCl and KCl Not a buffer. B. H2CO3 and NaHCO3 Buffer, a weak acid and its salt. C. H3PO4 and NaCl Not a buffer. D. HC2H3O2 and KC2H3O2 Buffer, a weak acid and its salt.

Chapter 8 Acids and Bases

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Chapter 8 Acids and Bases

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