acid bronsted

Properties of acids and basesProperties of acids and bases• Get 8 test tubes. Rinse all tubes well with water.

Add acid to four tubes, base to the other four.• Touch a drop of base to your finger. Record the

feel in the chart (on the next slide). Wash your hands with water. Repeat for acid.

• Use a stirring rod, add base to the litmus and pH papers (for pH paper use a colour key to find a number). Record results. Repeat for acid.

• Into the four base tubes add: a) two drops of phenolphthalein, b) 2 drops of bromothymol, c) a piece of Mg, d) a small scoop of baking soda. Record results. Repeat for acid.

• Clean up (wash tubes, pH/litmus paper in trash).

BubblesNRBaking soda

BubblesNRMagnesium

*Yellow*BlueBromothymol

*Cloudy/ white

*PinkPhenolphthalein

RedBlueLitmus (blue or red)

114pH (# from the key)

Not slipperySlipperyFeel (choose slippery

or not slippery)

SourBitterTaste

HCl(aq)NaOH(aq)ObservationsObservations

*Usually, but not always

pHpH• There are many ways to consider acids and

bases. One of these is pH. Read pg. 368-70.• [H+] is critical in many chemical reactions.• A quick method of denoting [H+] is via pH.• By definition pH = – log [H+], [H+] = 10-pH

• The pH scale, similar to the Richter scale, describes a wide range of values

• An earthquake of “6” is 10 as violent as a “5”• Thus, the pH scale condenses possible

values of [H+] to a 14 point scale (fig. 2, p370)• Also, it is easier to say pH = 7 vs. [H+] = 1 x 10–7

Calculations with pHCalculations with pH

Ans: 4.2

3.98 x 10–8 M

Try questions 2 and 6 (a-b) on page 375

Q: What is the pH if [H+]= 6.3 x 10–5? pH = – log [H+]

(‘6.3’, ‘exp’ or ‘EE’, ‘5’, ‘+/-’, ‘log’, ‘+/-’)(‘-’, ‘log’, ‘6.3’, ‘exp’ or ‘EE’, ‘-’, ‘5’)

Q: What is the [H+] if pH = 7.4?[H+] = 10–pH mol/L

(’10’, ‘xy’, ‘7.4’, ‘+/-’, ‘=‘)(’10’, ‘^’, ‘-’, ‘7.4’, ‘=‘)

Pg. 375Pg. 3752 a) pH = – log [H+] = – log [1x10–8] = 8.0 b) pH = – log [H+] = – log [1x10–7] = 7.0 c) pH = – log [H+] = – log [2.5x10–6] = 5.60 d) pH = – log [H+] = – log [1.3x10–4] = 3.89

6 a) [H+] = 10–pH = 10–5.4 = 4 x 10–6 mol/L b) [H+] = 10–pH = 10–5.72 = 1.9 x 10–6 mol/L

Historical views on acidsHistorical views on acids

• The more recent Bronsted-Lowry concept is that acids are H+ (proton) donors and bases are proton acceptors

Ionization

+Cl HH

HO

+H

HH O Cl+

• O (e.g. H2SO4) was originally thought to cause acidic properties. Later, H was implicated, but it was still not clear why CH4 was neutral.

• Arrhenius made the revolutionary suggestion that some solutions contain ions & that acids produce H3O+ (hydronium) ions in solution.

The Bronsted-Lowry conceptThe Bronsted-Lowry concept• In this idea, the ionization of an acid by water

is just one example of an acid-base reaction.

• Acids and bases are identified based on whether they donate or accept H+.

• “Conjugate” acids and bases are found on the products side of the equation. A conjugate base is the same as the starting acid minus H+.

+Cl HH

HO

+H

HH O Cl+

acid base conjugate acid conjugate base

conjugate acid-base pairs

Practice problemsPractice problemsIdentify the acid, base, conjugate acid, conjugate base, and conjugate acid-base pairs:

• Reference: pg. 386 – 387• Try Q18 (p389), Q 8 & 11 (p392): do as above

acid base conjugate acidconjugate baseHC2H3O2(aq) + H2O(l) C2H3O2

–(aq) + H3O+(aq)


acidbase conjugate acidconjugate baseOH

–(aq) + HCO3–(aq) CO3

2–(aq) + H2O(l)


acid base conjugate acidconjugate baseHF(aq) + SO3

2–(aq) F–(aq) + HSO3–(aq)


acidbase conjugate acidconjugate baseCO3

2–(aq) + HC2H3O2(aq) C2H3O2–(aq) + HCO3

–(aq)


acid base conjugate acidconjugate baseH3PO4(aq) + OCl

–(aq) H2PO4–(aq) + HOCl(aq)


Answers: question 18Answers: question 18(a)

(b)

(c)

acid base conjugate baseconjugate acidHCO3

–(aq) + S2–(aq) HS–(aq) + CO32–(aq)


baseacid conjugate acidconjugate baseH2CO3(aq) + OH

–(aq) HCO3–(aq) + H2O(l)


acid base conjugate acidconjugate baseH3O+(aq) + HSO3

–(aq) H2O(l) + H2SO3(aq)


8a)

8b)

11a)

base acid conjugate baseconjugate acidOH

–(aq) + HSO3–(aq) H2O(l) + SO3

2–(aq)


11b)

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