Chapter 10 Acids and Bases

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

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

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+

Conjugate Acid–Base Pairs In the reaction of HF and H2O, one conjugate acid–base pair is HF/F− the other conjugate acid–base pair is H2O/H3O+ each pair is related by a loss and gain of H+

Conjugate Acid–Base Pairs (continued) In the reaction of NH3 and H2O, one conjugate acid–base pair is NH3/NH4+ the other conjugate acid–base is H2O/OH– each pair is related by a loss and gain of H+

Chapter 10 Acids and Bases 10.2 Strengths of Acids and Bases

Strengths of Acids Strong acids completely ionize (100%) in aqueous solutions. HCl(g) + H2O(l) H3O+(aq) + Cl−(aq) dissociates into ions gives H3O+ and the anion (A–)

Weak Acids dissociate only slightly in water to form a solution of mostly molecules and a few ions. remain mostly as the undissociated (molecular) form have low concentrations of H3O+ and anion (A–) HA(aq) + H2O(l) H3O+(aq) + A−(aq) H2CO3(aq) + H2O(l) H3O+(aq) + HCO3−(aq)

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

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)

Weak Bases Weak bases dissociate only slightly in water form only a few ions in water NH3(g) + H2O(l) NH4+(aq) + OH−(aq)

Acid Dissociation Constant, Ka In a weak acid, the rate of the dissociation of the acid is equal to the rate of the association. HA + H2O H3O+ + A– The equilibrium expression is: Keq = [H3O+][A–] [HA][H2O] The concentration of H2O is constant, the Ka expression for a weak acid is: Ka = [H3O+][A–] [HA]

Writing Ka for a Weak Acid Write the Ka for H2S. 1. Write the equation for the dissociation of H2S: H2S(aq) + H2O(l) H3O+(aq) + HS−(aq) 2. Set up the Ka expression: Ka = [H3O+][HS−] [H2S]

Chapter 10 Acids and Bases 10.3 Ionization of Water

Ionization of Water In the ionization of 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− .. .. .. .. H:O: + H:O: H:O:H+ + :O:H− H H H water water hydronium hydroxide ion(+) ion(-)

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−] is obtained from the concentrations in pure water Kw = [1.0 x 10−7 M][1.0 x 10−7 M] = 1.0 x 10−14

Guide to Calculating [H3O+]

Example of Calculating [H3O+] What is the [H3O+] of a solution if [OH−] is 5.0 x 10−8 M? STEP 1 Write the Kw for water: Kw = [H3O+ ][OH− ] = 1.0 x 10−14 STEP 2 Rearrange the Kw expression for [H3O+]: [H3O+] = 1.0 x 10−14 [OH−] STEP 3 Substitute the known [OH−] and calculate: [H3O+] = 1.0 x 10−14 = 2.0 x 10−7 M [5.0 x 10−8]

Chapter 10 Acids and Bases 10.4 The pH Scale

Calculating pH Mathematically, pH is the negative log of the hydronium ion concentration pH = −log [H3O+] For a solution with [H3O+] = 1 x 10−4, pH = −log [1 x 10−4 ] pH = [4.0] pH = 4.0

Significant Figures in pH 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+]. coefficient decimal places [H3O+] = 1 x 10−4 pH = 4.0 [H3O+] = 8.0 x 10−6 pH = 5.10 [H3O+] = 2.4 x 10−8 pH = 7.62

Guide to Calculating pH

Example 1: Calculating [H3O+] from pH Calculate the [H3O+] for a pH value of 8.0. [H3O+] = 1 x 10−pH For pH = 8.0, the [H3O+] = 1 x 10−8 STEP 1 Enter the pH value, change sign: –8.0 STEP 2 Convert pH to concentration: Use 2nd function key and then10x key or inverse key and then log key 1 −08 STEP 3 Adjust the significant figures in the coefficient (1 digit following decimal point = 1 digit in the coefficient): [H3O+] = 1 x 10−8 M

Example 2: Calculating [H3O+] from pH Calculate the [H3O+] for a pH of 3.80. STEP 1 Enter the pH value, change sign: –3.80 STEP 2 Convert pH to concentration: Use 2nd function key and then10x key or inverse key and then log key 1.584893 −04 STEP 3 Adjust the significant figures in the coefficient (2 digit following decimal point = 2 digit in the coefficient): [H3O+] = 1.6 x 10−4 M

Chapter 10 Acids and Bases 10.5 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 Equations: 2K(s) + 2HCl(aq) 2KCl(aq) + H2(g) Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

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) HCl(aq) + NaHCO3(s) CO2(g) + NaCl (aq) + H2O(l)

Neutralization Reactions Neutralization is the reaction of an acid such as HCl and a base such as NaOH HCl(aq) + H2O(l) H3O+ (aq) + Cl−(aq) NaOH(aq) Na+ (aq) + OH−(aq) the H3O+ from the acid and the OH− from the base to form water H3O+(aq) + OH−(aq) 2H2O(l)

Neutralization Equations In the equations for neutralization, an acid and a base produce a salt and water. acid base salt water HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 2HCl(aq) + Ca(OH)2(aq) CaCl2(aq) + 2H2O(l)

Guide to Balancing an Equation for Neutralization

Balancing Neutralization Reactions Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid. STEP 1 Write the base and acid formulas: Mg(OH)2(aq) + HNO3(aq) STEP 2 Balance OH– and H+: Mg(OH)2(aq) + 2HNO3(aq) STEP 3 Balance with H2O: Mg(OH)2(aq) + 2HNO3(aq) salt + 2H2O(l) STEP 4 Write the salt from remaining ions: Mg(OH)2(aq) + 2HNO3(aq) Mg(NO3)2(aq) + 2H2O(l)

Guide to Calculating Molarity

Calculating Molarity from a Titration with a Base What is the molarity of an HCl solution if 18.5 mL of a 0.225 M NaOH are required to neutralize 10.0 mL HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 1 Given: 18.5 mL (0.0185 L) of 0.225 M NaOH 10.0 mL of HCl Need: M of HCl STEP 2 Plan: L of NaOH moles of NaOH moles of HCl M of HCl

Calculating Molarity from a Titration with a Base (continued) STEP 3 State equalities and conversion factors: 1 L of NaOH = 0.225 mole of NaOH 1 L of NaOH and 0.225 mole NaOH 0.225 mole NaOH 1 L of NaOH 1 mole of NaOH = 1 mole of HCl 1 mole of NaOH and 1 mole HCl 1 mole HCl 1 mole of NaOH STEP 4 Set up the problem to calculate moles of HCl: 0.0185 L NaOH x 0.225 mole NaOH x 1 mole HCl 1 L NaOH 1 mole NaOH = 0.00416 mole of HCl

Calculating Molarity from a Titration with a Base (continued) STEP 4 (continued) Calculate the volume in liters of HCl: 10.0 mL HCl = 0.0100 L HCl Calculate the molarity of HCl: 0.00416 mole HCl = 0.416 M HCl 0.0100 L HCl

Chapter 10 Acids and Bases 10.7 Buffers

Buffers Buffers 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 in blood maintain a pH close to 7.4; a change in the pH of the blood affects the uptake of oxygen and cellular processes

Components of a Buffer The components of a buffer solution are acid–base conjugate pairs can be a weak acid and a salt of its conjugate base typically have equal concentrations of the weak acid and its salt can also be a weak base and a salt of its conjugate acid

Summary of Buffer Action Buffer action occurs because the weak acid in a buffer neutralizes base the conjugate base in the buffer neutralizes acid the pH of the solution is maintained

pH of a Buffer The [H3O+] in the Ka expression is used to determine the pH of a buffer. Weak acid + H2O H3O+ + conjugate base Ka = [H3O+][conjugate base] [weak acid] [H3O+] = Ka x [weak acid] [conjugate base] pH = log [H3O+]

Guide to Calculating pH of a Buffer

Example of Calculating Buffer pH The weak acid H2PO4 in a blood buffer H2PO4/HPO42 has a Ka = 6.2 x 108. What is the pH of the buffer if [H2PO4] = 0.20 M and [HPO42] = 0.20 M? STEP 1 Write the Ka expression for: H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq) Ka = [HPO42][H3O+] [H2PO4] STEP 2 Rearrange the Ka for [H3O+]: [H3O+] = Ka x [H2PO4] [HPO42] Dihydrogen phosphate ion and hydrogen phosphate ion

Example of Calculating Buffer pH (continued) STEP 3 Substitute [HA] and [A]: [H3O+] = 6.2 x 108 x [0.20 M] = 6.2 x 108 [0.20 M] STEP 4 Use [H3O+] to calculate pH: pH =  log [6.2 x 10 8] = 7.21