Unit #3: Acids and BasesAndrea HougEmily Lichko

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•Arrhenius Acid = releases an H+•Arrehenius Base = releases an OH-•Brönsted-Lowry Acid = proton donor•Brönsted-Lowry Base = proton acceptor•Lewis Acid = electron acceptor•Lewis Base = electron donor

Brönsted - Lowry

Arrhenius

Lewis

Definitions of Acids and Bases

•The strength of an acid or a base depends on how much it dissociates in water•Kw = Ka•Kb= 1E-14= [H+][OH-]

•EX: Acetic Acid’s Ka = 1.8 E-5. Kb = Kw/Ka = (1E-14)/(1.8E-5) = 5.6 E -10

•Strong Acids dissociate 100%•Weak Acids typically dissociate less than 5%• % dissociation = •The stronger the acid or base, the weaker its conjugate

%100

0

HA

H eq

Strong Acids Strong Bases

HCl Alkaline metals with OH-

HBr Ca(OH)2

HI Sr(OH)2

HNO3 Ba(OH)2

H2SO4

HClO4

Dissociation

http://myphlip.pearsoncmg.com/phproducts/student/ab2page.cfm?vbcid=9029&vid=10007#oa231470

Neutralization•Combining an acid with a base produces water and a salt•For strong acids and bases:

•NH+CaVa = NOH-CbVb

•EX: How much 0.3000M NaOH is needed to neutralize 23.98mL of 0.8000M HNO3?

Answer:

(1)(0.8000M)(23.98mL) = (1)(0.3000M)(x)X= (0.8000M)(23.98mL)/(0.3000M) =63.95mL HCl

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Calculating pH

pH pOH

[H+] [OH-]

-lo

g [H

+ ] =p

H

14-pH = pOH

14-pOH= pH

-log [OH

-] =pOH

1 E-14 / [H+] = [OH-]

1 E-14 / [OH-] = [H+]

Henderson-Hasselbalch equation

acid

basepKpH a log

base

acidpKpOH b log

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TitrationsStrong acid / strong base Weak acid / strong base

Polyprotic acid / strong base

Back Titration: when base is not very soluble,

dissolve in excess acid and titrate back with base until excess acid is used up (endpoint)

http://www.sparknotes.com/chemistry/acidsbases/titrations/section1.html

ICE chartEX: pH of original 0.100 weak acid solution, Ka given

HA + H2O H3O+ + A-

I 0.100M 0 0

C -x +x +x

E 0.100M x x

Ka = x2/0.100 x= [H3O+] -log[H3O+] =pH

HA

AOHK

3

Use to find new pH, at any point in titration, by finding initial amount of A- from base added.

EX: (5mL)(0.250M NaOH) = 1.25mmol

AmphiproticAmphiprotic: Can act as either an acid or a base

If Ka > Kb then it acts as an acid at the equivalence pointIf Kb > Ka then it acts as a base at the equivalence point (for diprotic acids, always compare Ka2 to Kb1 since the will follow one of these equilibria at the equivalence point

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Buffers•Buffer

-weak acid base and its conjugate-minimizes change in pH when acid or base is added

•Buffer capacity -the quantity of acid or base that a buffer can neutralize before pH begins to change significantly•Common ion effect -a slightly soluble salt’s solubility decreases with the addition of a solute with a common ion

EX1:Make a buffer with pH 3.50

Choose an acid with a Ka close to 1E -3.5, like HOCNHOCN + H2O ↔ H3O+ + OCN-

Ka = [H3O+][OCN-] / [HOCN]With pH 3.50, [H3O+] = 3.16 E -4

3.5 E -4 = [3.16 E-4][OCN-] / [HOCN][HOCN] = 0.9035[OCN-]

This is the ratio. The amounts depend on desired buffer capacity….

Buffers

EX 2: Make it so the pH of the buffer changes by 0.50 pH unit when 0.50 mol acid is added.

Answer: pH is allowed to change to 3.00, so [H3O+] will change to 1 E-3. [OCN-] will decrease and [HOCN] will increase.

3.5 E -4 = ([1 E-3]•([OCN-]-0.50)) /(0.9035[OCN-]+0.5)1 E-3[OCN-] - 5 E-4 = 3.16 E-4[OCN-] + 1.75 E-46.837722 E-4[OCN-] =6.75 E-4[OCN-] = 0.987 M

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