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Strengths of acids and bases, Lecture notes of Chemistry

Usually Ka very small, ex : Ka for acetic acid = 1.8x10-5. For NH4. + it is 5.6x10-10. Which is the stronger weak acid? Weak base equilbrium.

Typology: Lecture notes

2021/2022

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๎˜
๎˜‚๎˜๎˜Œ๎˜“๎˜#"%๎˜๎˜’๎˜Œ๎˜Š๎˜š๎˜๎˜˜๎˜Œ๎˜$๎˜‰๎˜
๎˜†๎˜š๎˜˜๎˜Œ๎˜”๎˜Ž๎˜š๎˜๎˜™๎˜๎˜•๎˜๎˜๎˜ˆ๎˜Š๎˜๎˜‹๎˜™๎˜๎˜ˆ๎˜”๎˜‹๎˜๎˜‰๎˜ˆ๎˜™๎˜Œ๎˜™๎˜
๎˜…
๎˜ˆ
๎˜๎˜ˆ๎˜”๎˜‹๎˜๎˜…
๎˜‰
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๎˜–๎˜„๎˜๎˜ˆ๎˜”๎˜‹๎˜๎˜–๎˜„๎˜๎˜Š๎˜ˆ๎˜’๎˜Š๎˜๎˜’๎˜ˆ๎˜›๎˜•๎˜”๎˜™
๎˜
๎˜†๎˜š๎˜˜๎˜Œ๎˜”๎˜Ž๎˜š๎˜๎˜™๎˜๎˜•๎˜๎˜๎˜ˆ๎˜Š๎˜๎˜‹๎˜™๎˜๎˜ˆ๎˜”๎˜‹๎˜๎˜‰๎˜ˆ๎˜™๎˜Œ๎˜™๎˜
Acids and Bases are categorized into two general types:๎˜
Strong and weak. Refers to the tendency to donate or accept H+๎˜
Strong acids and bases dissociate 100%: HCl, HNO3; NaOH, KOH๎˜
Weak acids and bases do not dissociate 100%: CH3COOH, HF, NH3๎˜
Among weak acids and bases, the rule of thumb is:๎˜
โ€œthe stronger the conjugate, the weaker the acid or baseโ€๎˜
๎˜†๎˜š๎˜˜๎˜Œ๎˜”๎˜Ž๎˜š๎˜๎˜™๎˜๎˜•๎˜๎˜๎˜ˆ๎˜Š๎˜๎˜‹๎˜™๎˜!๎˜๎˜‰๎˜ˆ๎˜™๎˜Œ๎˜™ ๎˜
Conjugate acids ๎˜๎˜Conjugate bases๎˜
HCl ๎˜๎˜๎˜๎˜Cl-๎˜
HAc ๎˜๎˜๎˜๎˜Ac-๎˜
H2O๎˜๎˜๎˜๎˜OH-๎˜
Increasing strength๎˜
Increasing strength๎˜
๎˜ƒ๎˜๎˜™๎˜™๎˜•๎˜Š๎˜๎˜ˆ๎˜›๎˜•๎˜”๎˜๎˜•๎˜๎˜๎˜™๎˜š๎˜˜๎˜•๎˜”๎˜Ž๎˜๎˜ˆ๎˜Š๎˜๎˜‹๎˜™๎˜๎˜๎˜”๎˜๎˜ž๎˜ˆ๎˜š๎˜Œ๎˜˜๎˜
Strong acids dissociate 100% in water. (Classic example:HCl)๎˜
๎˜๎˜ 100%๎˜
๎˜HCl + H2O --->H3O+ + Cl- 100%๎˜
Other examples: ๎˜ HNO3+ H2O --->H3O+ + NO3-๎˜
HBr, HI, H2SO4 (remember all these strong acids)๎˜
Often we just write: HCl--> H+ + Cl- ; HNO3-->H++NO3-๎˜
๎˜ƒ๎˜๎˜™๎˜™๎˜•๎˜Š๎˜๎˜ˆ๎˜›๎˜•๎˜”๎˜๎˜•๎˜๎˜๎˜ž๎˜Œ๎˜ˆ๎˜‘๎˜๎˜ˆ๎˜Š๎˜๎˜‹๎˜™๎˜๎˜๎˜”๎˜๎˜ž๎˜ˆ๎˜š๎˜Œ๎˜˜๎˜
Weak acids do NOT ionize 100% (classic example: acetic acid;
contains methyl -CH3- attached to carboxyl group -COOH):๎˜
๎˜CH3-COOH + H2O < = = = > H3O+ + CH3-COO-๎˜
Generic eqโ€™n for weak acid: ๎˜HA + H2O < = = = > H3O+ + A- ๎˜
Called โ€œ Ka equilibriumโ€. Ka is โ€œacid dissociation constantโ€.
What is the expression for Ka ? Ka = [H3O+][A-]
/
[HA]๎˜
The larger the Ka, the stronger the weak acid.๎˜
Usually Ka very small, ex : Ka for acetic acid = 1.8x10-5
๎˜
For NH4+ it is 5.6x10-10. Which is the stronger weak acid?
๎˜
๎˜‡๎˜Œ๎˜ˆ๎˜‘๎˜๎˜‰๎˜ˆ๎˜™๎˜Œ๎˜๎˜Œ๎˜—๎˜๎˜๎˜’๎˜‰๎˜˜๎˜๎˜๎˜“๎˜
For weak bases: โ€œKb equilibriumโ€. Kb =โ€œbase ionization
constant: Generic equation๎˜
๎˜B + H2O < = = = > BH+ + OH- ๎˜
Kb = [BH+][OH-]
/
[B]๎˜
The greater Kb is, the stronger the base.๎˜
Usually Kb very small, example: NO2- has Kb = 2.2x10-11๎˜
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Acids and Bases are categorized into two general types: Strong and weak. Refers to the tendency to donate or accept H+ Strong acids and bases dissociate 100%: HCl, HNO 3 ; NaOH, KOH Weak acids and bases do not dissociate 100%: CH 3 COOH, HF, NH 3  Among weak acids and bases, the rule of thumb is: โ€œthe stronger the conjugate, the weaker the acid or baseโ€

Conjugate acids  Conjugate bases HCl    Cl-

HAc    Ac - 

H 2 O  Increasing strength  OH - 

 Increasing strength

Strong acids dissociate 100% in water. (Classic example:HCl)   (^) 100% HCl + H 2 O --->H 3 O +^ + Cl-^ 100% Other examples:  HNO 3 + H 2 O --->H 3 O +^ + NO 3 -  HBr, HI, H 2 SO 4 (remember all these strong acids) Often we just write: HCl--> H+^ + Cl-^ ; HNO 3 -->H ++NO 3 - 

Weak acids do NOT ionize 100% (classic example: acetic acid; contains methyl - CH (^) 3- attached to carboxyl group - COOH ): CH 3 -COOH + H 2 O < = = = > H 3 O +^ + CH 3 -COO -  Generic eqโ€™n for weak acid: HA + H 2 O < = = = > H 3 O +^ + A-^ 

Called โ€œ Ka equilibriumโ€. Ka is โ€œ acid dissociation constantโ€.

What is the expression for Ka? Ka = [H 3 O +][A -^ ] / [HA]

The larger the Ka, the stronger the weak acid. Usually K (^) a very small, ex : Ka for acetic acid = 1.8x10-5 For NH 4 +^ it is 5.6x10-10^. Which is the stronger weak acid?

For weak bases: โ€œKb equilibriumโ€. Kb =โ€œbase ionization constant: Generic equation B + H 2 O < = = = > BH +^ + OH -^ 

K b = [BH +][OH -^ ] / [B]

The greater Kb is, the stronger the base. Usually K (^) b very small, example: NO 2 -^ has K (^) b = 2.2x10-11

 ! ! ^ 

Note: Ka = [H 3 O +][A -^ ]/[HA] for acid HA.

For itโ€™s conjugate, A-^ , it is: Kb = [HA][OH -^ ] / [A-^ ] 

So if multiply Ka of weak acid by Kb of its conjugate, then :

K (^) aK (^) b = ( [H 3 O +][A -^ ]/[HA] ) ( [HA][OH -^ ] / [A-^ ] ) 

KaKb = [H 3 O+][ OH-^ ] = Kw = 1.0 x 10-14 

Because of wide range of [H 3 O +], itโ€™s convenient to express concentration levels of [H+] by its exponent, using logarithmic scale. This is the โ€œpH scaleโ€. Definition of pH: pH = -log[H+] = -log[H 3 O+] For example: whatโ€™s the pH of pure water?  Note: pure water: [H 3 O +] = 1.0x10-7^ M pH = -log {[H 3 O +] } = -log(1.0x10-7^ M) = 7.00 This is the neutral pH

If you have an acidic solution with [H 3 O +] = 2.0 x 10-4M^. Whatโ€™s the pH? 

pH = -log(2.0 x 10 -4^ M) = 3.70

In general, if pH < 7.00 we say solution is acidicโ€ฆ

If pH > 7.00 the solution is basic.

What is [H+] if you are given the pH? 

[H+] = 10-pH 

Definition: pOH = -log[OH-^ ]  Before: [H 3 O +][OH -^ ] = 1.0x10-14 (so if you know [H 3 O +], you can know [OH -^ ]) take -log of both sides: -log{[H 3 O +][OH -^ ] } = -log{1.0x10-14^ } -log[H 3 O +] - log[OH -^ ] = 14.00 Or, pH + pOH = 14.00  (so if you know pH, you can know pOH)

[H 3 O +] [ OH -^ ] = 1.0 x 10-14

pH = -log[H 3 O +] and pOH = -log[OH -^ ] 

pH + pOH = 14.00

K (^) a = [H +][A -^ ]/[HA] K (^) b = [HA][OH -^ ]/[A-^ ]

K (^) aK (^) b = 1.0 x 10-14^ 

[H +] = 10-pH^  [OH -^ ] = 10-pOH

There are at least 5 scenarios weโ€™ll encounter involving pH calculations. You are expected to MASTER these calculations.

  1. Strong acid solutions
  2. Strong base solutions
  3. Pure Weak acid solutions
  4. Pure weak base solutions
  5. Buffer solutions

What is the pH of 0.10 M ammonia? (Kb = 1.8x10-5M)

Answer: Use the K (^) b equilibrium. NH 3 + H 2 O<==>NH 4 ++OH - 

    0.10-x^  x^  x

x 2 /(0.10-x) = 1.8x10-5^. Using 5% rule:

x 2  (1.8x10-5^ )(0.10) => x =1.3x10-3^ =[OH -^ ]

pOH = 2.87 => pH = 14.00-2.87 = 11.13

Poly protic acids can donate more than 1 proton. Examples: H 2 CO 3 (carbonic acid); H 3 PO 4 (phosphoric acid). If you place H 2 CO 3 in water you will have the following Ka equilibrium:  H 2 CO 3 + H 2 O < = = > H 3 O +^ + HCO 3 -^ K (^) a1 = 4.2x10-7 If you place NaHCO 3 in water you have: HCO 3 -^ + H 2 O < = = > H 3 O +^ + CO 3 2-^ K (^) a2 = 4.8x10-11 There are 2 Kaโ€™s