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Acid/Base Chemistry Part 3 (5.4- 5.5) Science 10 CT05D05 Resource: Brown, Ford, Ryan, IB Chem

Acid/Base Chemistry Part 3 (5.4-5.5)

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Acid/Base Chemistry Part 3 (5.4-5.5). Science 10 CT05D05 Resource: Brown, Ford, Ryan, IB Chem. Topic 05 – Acids/Bases. 5.1 Solutions 5.2 Definitions of Acids and Bases 5.3 Properties of Acids and Bases 5.4 Calculating pH, pOH , H+, OH- 5.5 Neutralization equations - PowerPoint PPT Presentation

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Page 1: Acid/Base Chemistry Part 3 (5.4-5.5)

Acid/Base Chemistry

Part 3 (5.4-5.5)Science 10CT05D05Resource: Brown, Ford, Ryan, IB Chem

Page 2: Acid/Base Chemistry Part 3 (5.4-5.5)

Topic 05 – Acids/Bases 5.1 Solutions 5.2 Definitions of Acids and Bases 5.3 Properties of Acids and Bases 5.4 Calculating pH, pOH, H+, OH- 5.5 Neutralization equations 5.6 Titrations

Page 3: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 Calculating pH, pOH, H+, OH-

5.4.1 Calculate the concentration of ions (H+ and OH-) and acidity (pH and pOH) of strong acids and bases

5.4.2 Calculate the above of a mixture of strong acids and bases

Page 4: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 – pH Scale Proposed

Søren Sørenson Denmark Biochemist Early 20th century Proposed the pH

scale

Page 5: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 - Concentrations of [H+]

The more acidic a solution is, the more H+ ions are donated into solutionActual concentrations of the hydronium ion, [H+], are often very smallTherefore, Sørenson proposed a manipulation of the concentration of H+ in a way that made the data much more simple to relateThe pH scale is based on the logarithm of the concentration of H+

Page 6: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 - Concentration We represent concentration by molarity,

therefore the concentration of an acid and a base will give us the following information: 0.45 M HCl = 0.45 mol [H+]

From an acid we find the [H+] We can calculate pH directly

0.65 M NaOH = 0.65 mol [OH-] From a base we find the [OH-] We can calculate pOH directly

If we know the concentration of one, we can find it of the other: [H+] [OH-] = Kw = 1 x 10-14

Page 7: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 - Calculating pH

The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration (in mol/L). pH = -log[H3O+] pH = -log[H+]This calculation results in a pH scale, 0-14Therefore, the pH range of solutions are as follows: Acidic Solutions, pH < 7.0 (0.0-

6.9) Neutral Solution, pH = 7.0 (7.0) Basic Solutions, pH > 7.0 (7.1-

14)

Page 8: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 – Problems with scale

Find the pH of a 12M solution of HCl pH = -log (12M) = -1.07 Usually the pH of 12 Molar HCl represented

as 0.00 and not -1.07? Any solution with a pH or pOH calculation

that results in a negative number you are welcome to round to a pH or pOH of zero (0).

For our purposes in this class, the scale will be from 0-14!

Page 9: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 - The pH scale is logarithmic

What does the logarithm scale mean? The logarithm is base 10, so a change in one

value has 10x the effect Since pH 7 is neutral

pH 5 is 10 x more acidic than pH 6 pH 4 is 100 x more acidic than pH 6 pH 3 is 1000 x more acidic than pH 6 pH 9 is 10 x more alkaline than pH 8 pH 10 is 100 x more alkaline than pH 8 pH 11 is 1000 x more alkaline than pH 8

Page 10: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 - What is pOH? pOH is the opposite of pH, and a measure of alkalinity (how basic)If pH goes down, pOH goes up. The pOH scale is (opposite pH): Basic Solutions: pOH < 7.0 (0.0-

6.9) Neutral Solutions: pOH = 7.0 (7.0) Acidic Solutions: pOH > 7.0 (7.1-

14)pH + pOH = 1414 – pH = pOHCalculated: pOH = -log [OH-]

Page 11: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 – Common Solutions

Page 12: Acid/Base Chemistry Part 3 (5.4-5.5)

Practice #1

What is the pOH of a 0.005 M Mg(OH)2 ?What is the pH ?

-log [OH-] = pOH-log [0.005] = 2.3 = pOH

14 – pOH = pH14 – 2.3 = 11.7 = pH

Page 13: Acid/Base Chemistry Part 3 (5.4-5.5)

Practice #2If 20g of NH3 are dissolved in 2.5 L of distilled water, what would the pH of the solution be?

20g NH3 17g NH3

1 mol NH3 = 1.17 mol NH3

1.17 mol NH3 2.5 L solution

= 0.47 M NH3

-log [OH-] = pOH-log [0.47] = pOH =

0.3214 – pOH = pH

14 – 0.32 = 13.68

Page 14: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 - Strong Acids and Bases

Strong AcidsHClO4 (perchloric acid)HI (hydroiodic acid)HBr (hydrobromic acid)HCl (hydrochloric acid)H2SO4 (sulfuric acid)HNO3 (nitric acid)

Strong Bases• LiOH (lithium hydrox.)• NaOH (sodium hydrox.)• KOH (potassium

hydrox.)• RbOH (rubidium

hydrox.)• CsOH (cesium hydrox.)• Ca(OH)2 (calcium

hydrox.)• Sr(OH)2 (strontium

hydrox.)• Ba(OH)2 (barium

hydrox.)

Mainly the acids of halides! Mainly the bases of

alkali metals and some alkaline earths!

We will calculate primarily with strong monoprotic acids and bases (BOLD)

Page 15: Acid/Base Chemistry Part 3 (5.4-5.5)

Applicable Solution Def’s

Strong acids and bases will completely ionize in water and we can therefore use a “” yields symbol in the equation. (for weak acids and bases, an equilibrium arrow would be used ⇌)Ionization: Process by which a neutral compound is split into charged particles by action when dissolved in liquid waterEquilibrium: When reactants and products are in a constant ratio. The forward and reverse reactions occur at the same rate when a system is in equilibrium.

Page 16: Acid/Base Chemistry Part 3 (5.4-5.5)

Ionization ReactionsCompletion: For those that include a strong acid or strong base, the reaction will run to completion and can be shown as such with a generic ‘yields’ symbol ()

Equilibrium: For those that include a weak acid or base or do not go to completion, the reaction can be represented by an equilibrium symbol ()

Page 17: Acid/Base Chemistry Part 3 (5.4-5.5)

5.5 Neutralization equations

5.5.1 Balance simple acid base equations 5.5.2 Conjugate Acid/Base pairs

Page 18: Acid/Base Chemistry Part 3 (5.4-5.5)

5.5 - Neutralization When equal concentrations of H+ and OH- are

added to one another, a neutral solution results NaOH + HCl NaCl + H2O Base + Acid Salt + Water

In equal amounts this is always the case, whether the acid is strong or weak, as long as concentrations are taken into account

Try the following, assume complete neutralization: Mg(OH)2 + HCl KOH + H2SO4

Page 19: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 - Aqueous Solutions of A&B

• What happens when you put an acid or a base into water?

• Each have the property of being electrolytes so will therefore dissociate

• Water itself can act as an acid or a base H+ + H2O H3O+ (or H+ + OH- H2O) H3O+ H+ + H2O (or H2O H+ + OH- )

Page 20: Acid/Base Chemistry Part 3 (5.4-5.5)

5.4 – Conjugate Acid/Base When acids and bases go through the process

of donating or accepting protons, they then switch roles as they can easily reverse the reaction

Acids become conjugate bases Bases become conjugate acids Strong become weak Weak become strong

Page 21: Acid/Base Chemistry Part 3 (5.4-5.5)

Arrhenius acid is a substance that produces H+ (H3O+) in water

Arrhenius base is a substance that produces OH- in water

Page 22: Acid/Base Chemistry Part 3 (5.4-5.5)

A Brønsted acid is a proton donorA Brønsted base is a proton acceptor

acidbase conj. acid conj. base

A Brønsted acid must contain at least one ionizable proton!

Page 23: Acid/Base Chemistry Part 3 (5.4-5.5)

5.5 - Conjugate Acids and Bases

• Conjugate pairs are two substances that differ by one H+ (they gain or lose one PROTON)

HCl (acid) Cl- (conjugate base) NH3 (base) NH4

+ (conjugate acid)

• When an acid loses a proton it becomes its conjugate base

H2O OH- or HCl H+ + Cl-

• When a base gains a proton it becomes its conjugate acid

H2O H3O+ or NH3 + H+ NH4+

Page 24: Acid/Base Chemistry Part 3 (5.4-5.5)

5.5 - Conjugate pairsConjugate base of the acid H2SO4

HSO4-Conjugate acid of the base HS-

H2SConjugate acid of H2O

H3O+

Conjugate base of H2OOH-

Page 25: Acid/Base Chemistry Part 3 (5.4-5.5)

5.5 - Bronsted-Lowry Acids and Bases

NH4+ + OH- NH3 + H2O

Acid Base C. Base C. Acid

C2H3O2- + H3O+ HC2H3O2 + H2O

Base Acid C. Acid C. Base