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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, 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: โthe 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 eqโn for weak acid: HA + H 2 O < = = = > H 3 O +^ + A-^
Called โ Ka equilibriumโ. Ka is โ acid dissociation constantโ.
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: โKb equilibriumโ. Kb =โbase ionization constant: Generic equation B + H 2 O < = = = > BH +^ + OH -^
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 itโs 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 +], itโs convenient to express concentration levels of [H+] by its exponent, using logarithmic scale. This is the โpH scaleโ. Definition of pH: pH = -log[H+] = -log[H 3 O+] For example: whatโs 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^. Whatโs the pH?
pH = -log(2.0 x 10 -4^ M) = 3.70
In general, if pH < 7.00 we say solution is acidicโฆ
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 weโll encounter involving pH calculations. You are expected to MASTER these calculations.
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 Kaโs