Chapter 19: Acids and Bases

I. Acids and Bases - properties Arrhenius theory Acids, bases, and salts break down in H2O Theory: acids produce H+ ions (protons) bases produce OH- ions (hydroxides)

I. Acids and Bases - properties BrØnstad-Lowry theory Any donor of H+= acid; any acceptor = base Ex: HCl + H2O H3O+ + Cl- acid base hydronium Acid + Base Conjugate Acid + Conj. Base Ex: NH3 + H2O NH4+ + OH- base acid conj. acid conj. base

I. Acids and Bases - properties Amphoterics – compounds that sometimes act as acids, sometimes as bases Ex: HCl + H2O  H3O+ + Cl- NH3 + H2O  NH4+ + OH-

I. Acids and Bases - properties Acidic & basic anhydrides (w/o water) Acidic anhydride = oxide producing acid w/ water SO2 + H2O  H2SO3 Basic anhydride = oxide producing base w/ water Na2O + H2O  2 NaOH

I. Acids and Bases - properties Acid/base strength Acids that ionize completely into + and – ions = strong acid (pH of 0 – 2) Bases that ionize completely into + and – ions = strong base (pH of 12 – 14) Weak acid/base = pH 2.1 – 6.9; 7.1 – 11.9 Neutral = pH of 7

A quick look at water… H2O - amphoteric Sometimes loses H+  OH- Sometimes gains H+  H3O+ (hydronium ion)

Acid/Base problems Chapter 19 (3, 4, 6, 8, 53, 56, 57, 84, 85, 88, 89)

Neutralization reactions H3PO4 + NaOH  H2O + Na3PO4 H2SO4 + Ba(OH)2  H2O + BaSO4 HCl + KOH  H2O + KCl HNO3 + Ca(OH)2  H2O + Ca(NO3)2 #H+ ions X Macid X Vol acid = #OH- ions X Mbase X Volbase (Your book will have you balance the equations and run mole ratios instead of the above method, so beginning problems will look different)

Neutralization calculation example How many mL of 8 M H2SO4 is needed to neutralize 750 mL of 12 M NaOH? Known: 2 H+, 1 OH- (2) (8) (X) = 1 (750 mL) (12) X = 562.5 mL .25 L of 5 M H3PO4 is used to neutralize 1.5L of Ba(OH)2. What is the concentration of the base? Known: 3 H+, 2 OH- (3) (5) (.25L) = (2) (X) (1.5L) X = 1.25 M

Homework: P. 673 (35-38, 41, 73) This will take you about 15 minutes!

Acidic vs. Basic solutions Acid = more [H+] than [OH-]; acidic [H+] > 1.0 x 10-7 Base = fewer [H+] than [OH-]; alkaline [H+] < 1.0 x 10-7 [H+] X [OH-] = 1.0 X 10-14 Ex: If [H+] = 1.0 x 10-5M, is it an acid, base, or neutral? What is the [OH-] of solution?

Ex: If [H+] = 1. 0 x 10-5M, is it an acid, base, or neutral Ex: If [H+] = 1.0 x 10-5M, is it an acid, base, or neutral? What is the [OH-] of solution? 1 -1 -2 -7 -10 -14 0 Acidic, since 10-5 > 10-7

Ex: If [H+] = 1. 0 x 10-5M, is it an acid, base, or neutral Ex: If [H+] = 1.0 x 10-5M, is it an acid, base, or neutral? What is the [OH-] of solution? Known: [H+] x [OH-] = 1.0 x 10-14 [OH-] = 1.0 x 10-14 / 1.0 x 10-5 = 1.0 x 10-9

[ ] Homework: P. 655 (10 & 11)

pH Concept pH = - log [H+] Example: H2O is neutral [H+] = 1 x 10-7M; ph = 7.0 pH = - log (1 x 10-7) = - (log 1 + log 10-7)  log 10x 7- = - (0.0 + (-7.0)) = 7.0 These two are opposites of each other

pH Concept Question: What is the pH of a solution with a [H+] of 1 x 10-4 M? pH = - log (1 x 10-4) = - (log 1 + log 10-4) = - (0.0 + (-4)) = 4.0

pH Concept Ex: What is the pH of a solution with a [H+] of 1.0 x 10-10? pH = - log [H+] = - log (1.0 x 10-10) = - (log 1 + log 10-10) = - (0.0 + (-10.00)) = 10

pH Concept Ex: What is the pH of a solution with a [H+] of 2.7 x 10-8? pH = - log (2.7 X 10-8) = - ( log 2.7 + log 10x 8-) = - (.43136 + (-8)) = 7.56 pH

pH Concept Ex: What is the pH of a solution that is .0050 M [H+]? Convert to scientific notation 1st! = 5.0 x 10-3 pH = - (log 5 + log 10-3) = - ( log 5 + log 10x 3-) = - (.69897 + (-3)) = 2.30 pH

pH Concept The pH of a solution is 6. What is the [H+]? - log [H+] = pH 6 (switch signs!) log [H+] = - 6.00 antilog = 10x (– 6.00) = .000001 = 1 x 10-6 [H+]

pH Concept What is the [H+] if the pH is 3.70? - log [H+] = pH 3.70 antilog = 10x (– 3.70) = .000199526 = 2.00 x 10-4

pH Problems P. 657 (12-17, 22, 24, 63-65, 86)

pOH Concept pOH = - log [OH-] pH + pOH = 14 pH = 14 – pOH Also recall: [H+] X [OH-] = 1.0 X 10-14

pH 14 - _ pOH [H+] [OH-] pH = -log [H+] pOH = -log [OH-] [H+] X [OH-] = 1 X 10-14 [OH-]

pOH Concept Ex: What is the pH if [OH-] = 4.0 x 10-11M? pH = 1 x 10-14/ 4 x 10-11 = .00025 = 2.5 x 10-4M

pOH Concept Ex: What is the [OH-] if the pOH is 11.22? log [OH-] = 11.22 log [OH-] = -11.22 [OH-] = antilog -11.22 -11.22 10x = 6.03 X 10-12

Last one! What is the [H+] if the pOH is 2.38?

A Quick Review… Arrhenius acids/bases – H+ or OH- BrØnstad-Lowry – H+ donated or accepted Anhydrides – add water! If OH, base; if H, acid MgO SO2 ZnO K2O CO2 Neutralizations: #H+ ions X Macid X Vol acid = #OH- ions X Mbase X Volbase pH, pOH, [H+], [OH-]pH Calculations