Acids and Bases

Some Properties of Acids Produce H+ (as H3O+) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) Taste sour Corrode metals Electrolytes React with bases to form a salt and water pH is less than 7 Turns blue litmus paper to red “Blue to Red A-CID”

Some Properties of Bases Produce OH- ions in water Taste bitter, chalky Are electrolytes Feel soapy, slippery React with acids to form salts and water pH greater than 7 Turns red litmus paper to blue “Basic Blue”

Acid/Base definitions Definition 1: Arrhenius Arrhenius acid is a substance that produces H+ (H3O+) in water Arrhenius base is a substance that produces OH- in water 4.3

Acid/Base Definitions Definition #2: Brønsted – Lowry Acids – proton donor Bases – proton acceptor A “proton” is really just a hydrogen atom that has lost it’s electron!

+ Cl H O acid base conjugate acid conjugate base A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor + Cl H O acid base conjugate acid conjugate base conjugate acid-base pairs

A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor conjugate acid conjugate base base acid

ACID-BASE THEORIES The Brønsted definition means NH3 is a BASE in water — and water is itself an ACID

Conjugate Pairs

Learning Check! Label the acid, base, conjugate acid, and conjugate base in each reaction: HCl + OH-    Cl- + H2O H2O + H2SO4    HSO4- + H3O+

The ph scale is a way of expressing the strength of acids and bases The ph scale is a way of expressing the strength of acids and bases. instead of using very small numbers, we just use the negative power of 10 on the molarity of the H+ (or OH-) ion. Under 7 = acid 7 = neutral Over 7 = base

(Remember that the [ ] mean Molarity) Calculating the pH pH = - log [H+] (Remember that the [ ] mean Molarity) Example: If [H+] = 1 X 10-10 pH = - log 1 X 10-10 pH = - (- 10) pH = 10 Example: If [H+] = 1.8 X 10-5 pH = - log 1.8 X 10-5 pH = - (- 4.74) pH = 4.74

Try These! pH = - log [H+] pH = - log 0.15 pH = - (- 0.82) pH = 0.82 pH = - log 3 X 10-7 pH = - (- 6.52) pH = 6.52 Find the pH of these: A 0.15 M solution of Hydrochloric acid 2) A 3.00 X 10-7 M solution of Nitric acid

pH calculations – Solving for H+ If the pH of Coke is 3.12, [H+] = ??? Because pH = - log [H+] then - pH = log [H+] Take antilog (10x) of both sides and get 10-pH = [H+] [H+] = 10-3.12 = 7.6 x 10-4 M *** to find antilog on your calculator, look for “Shift” or “2nd function” and then the log button

More About Water Equilibrium constant for water = Kw H2O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION Equilibrium constant for water = Kw Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC

More About Water and so [H3O+] = [OH-] = 1.00 x 10-7 M Autoionization Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC In a neutral solution [H3O+] = [OH-] and so [H3O+] = [OH-] = 1.00 x 10-7 M

pOH Since acids and bases are opposites, pH and pOH are opposites pOH does not really exist, but it is useful for changing bases to pH. pOH looks at the perspective of a base pOH = - log [OH-] Since pH and pOH are on opposite ends, pH + pOH = 14

[H3O+], [OH-] and pH What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10-3 M) pOH = - log 0.0010 pOH = 3 ; pH + pOH = 14 pH + 3 = 14 pH = 11

Strong and Weak Acids/Bases Generally divide acids and bases into STRONG or WEAK ones. STRONG ACID: HNO3 (aq) + H2O (l)  H3O+ (aq) + NO3- (aq) HNO3 is about 100% dissociated in water. Weak acids are much less than 100% ionized in water. *One of the best known is acetic acid = CH3CO2H

Strong and Weak Acids/Bases Strong Base: 100% dissociated in water. NaOH (aq)  Na+ (aq) + OH- (aq) Strong bases are the group I hydroxides Calcium, strontium, and barium hydroxides are strong, but only soluble in water to 0.01 M Weak base: less than 100% ionized in water One of the best known weak bases is ammonia NH3 (aq) + H2O (l) ↔ NH4+ (aq) + OH- (aq)

Strong Acid Weak Acid 15.4

What is the pH of a 2 x 10-3 M HNO3 solution? HNO3 is a strong acid – 100% dissociation. Start 0.002 M 0.0 M 0.0 M HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq) End 0.0 M 0.002 M 0.002 M pH = -log [H+] = -log(0.002) = 2.7 What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution? Ba(OH)2 is a strong base – 100% dissociation. Start 0.018 M 0.0 M 0.0 M Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq) End 0.0 M 0.018 M 0.036 M pOH = -log (0.036) =1.44 ; pH + pOH = 14 pH = 14-1.44 = 12.56 15.4

Equilibria Involving Weak Acids and Bases Consider acetic acid, HC2H3O2 (HOAc) HC2H3O2 + H2O ↔ H3O+ + C2H3O2 - Acid Conj. base (K is designated Ka for ACID)

Ionization Constants for Acids/Bases

Equilibrium Constants for Weak Acids Weak acid has Ka < 1

Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calculate the equilibrium concentration of HOAc, H3O+, OAc-, and the pH. Ka= 1.8 x 105 Step 1. Define equilibrium concentration in ICE table. [HOAc] [H3O+] [OAc-] initial change equilib 1.00 0 0 -x +x +x 1.00-x x x

Equilibria Involving A Weak Acid Step 2. Write Ka expression This is a quadratic. Solve using quadratic formula.

Equilibria Involving A Weak Acid Step 3. Solve Ka expression First assume x is very small because Ka is so small. Now we can more easily solve this approximate expression.

Equilibria Involving A Weak Acid Step 3. Solve Ka approximate expression x2 = 1.8 x 10-5 x = [H3O+] = [OAc-] = 4.2 x 10-3 M pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37

Equilibrium Constants for Weak Bases Weak base has Kb < 1

Equilibria Involving A Weak Base You have 0.010 M NH3. Calc. the pH. NH3 + H2O ↔ NH4+ + OH- Kb = 1.8 x 10-5 Step 1. Define equilibrium concs. in ICE table [NH3] [NH4+] [OH-] initial change equilib 0.010 0 0 -x +x +x 0.010 - x x x

Equilibria Involving A Weak Base Step 2. Solve the equilibrium expression Assume x is small, so x2 = 1.8 x 10-5 x 0.010 = 1.8x10-7 x= 4.2 x 10-4

Equilibria Involving A Weak Base Step 3. Calculate pH [OH-] = 4.2 x 10-4 M so pOH = - log [OH-] = 3.37 Because pH + pOH = 14 pH = 14 – pOH pH = 10.63

Types of Acid/Base Reactions: Summary

15.4

For a monoprotic acid HA Ionized acid concentration at equilibrium Initial concentration of acid x 100% percent ionization = For a monoprotic acid HA Percent ionization = [H+] [HA]0 x 100% [HA]0 = initial concentration 15.5

Ionization Constants of Conjugate Acid-Base Pairs HA (aq) H+ (aq) + A- (aq) Ka A- (aq) + H2O (l) OH- (aq) + HA (aq) Kb H2O (l) H+ (aq) + OH- (aq) Kw KaKb = Kw Weak Acid and Its Conjugate Base Ka = Kw Kb Kb = Kw Ka 15.7

Exercise Calculate the pH of 1 moldm-3 ethanoic acid with Ka is 1.7x10-5 CH3CO2H (aq)  CH3CO2- (aq) + H+ (aq)

\ Calculate the pH of 1 moldm-3 ethanoic acid with Ka is 1.7x10-5 CH3CO2H (aq)  CH3CO2- (aq) + H+ (aq) Before 1 0 0 Equi 1-x x x 1.7x10-5 = x.x 1-x ~ 1 So, x = 4.12 x 10-3 pH = -log (4.12 x 10-3 ) ; pH = 2.4 \

Exercise Calculate the pH of 0.1 mol dm-3 ethanoic acid. CH3CO2H (aq)  CH3CO2- (aq) + H+ (aq)

Calculate the pH of 0.1 mol dm-3 ethanoic acid. Using the same method, we get: Ka = x.x 0.1-x ~ 0.1 So, 1.7 x 10-5 = x.x 0.1 x= 1.30 x 10-3 pH = -log 1.30x10-3 ; pH = 2.9

Exercise What are the concentrations of H3O+ and OH- ions in 0.05 mol dm-2 HCl? HCl (aq) + H2O  H3O+ (aq) + Cl - (aq)

Example What are the concentrations of H3O+ and OH- ions in 0.05 mol dm-2 HCl? HCl (aq) + H2O  H3O+ (aq) + Cl - (aq) [H3O+] = [HCl] = 0.05 mol dm-2 (complete dissociation) Kw = [H3O+] [OH- ] 1.0 x 10-14 = [0.05 mol dm-2 ] [OH- ] [OH- ] = 2.0 x 10-13mol dm-3

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