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Strengths of acids and bases, Lecture notes of Chemistry

University of Texas - Dallas 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

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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, HNO3; NaOH, KOH

Weak acids and bases do not dissociate 100%: CH3COOH, HF, NH3

Among weak acids and bases, the rule of thumb is:

“the stronger the conjugate, the weaker the acid or base”

! 

Conjugate acids Conjugate bases

HCl Cl-

HAc Ac-

H2OOH-

Increasing strength



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-



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 eq’n for weak acid: HA + H2O < = = = > H3O+ + A- 

Called “ Ka equilibrium”. Ka is “acid dissociation constant”.

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?

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

For weak bases: “Kb equilibrium”. Kb =“base ionization

constant: Generic equation

B + H2O < = = = > BH+ + OH- 

Kb = [BH+][OH-]

/

[B]

The greater Kb is, the stronger the base.

Usually Kb very small, example: NO2- has Kb = 2.2x10-11

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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: “the stronger the conjugate, the weaker the acid or base”

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 eq’n for weak acid: HA + H 2 O < = = = > H 3 O +^ + A-^

Called “ Ka equilibrium”. Ka is “ acid dissociation constant”.

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: “Kb equilibrium”. Kb =“base 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 it’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 +], it’s convenient to express concentration levels of [H+] by its exponent, using logarithmic scale. This is the “pH scale”. Definition of pH: pH = -log[H+] = -log[H 3 O+] For example: what’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^. What’s the pH?

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

In general, if pH < 7.00 we say solution is acidic…

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 we’ll encounter involving pH calculations. You are expected to MASTER these calculations.

Strong acid solutions
Strong base solutions
Pure Weak acid solutions
Pure weak base solutions
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 Ka’s