Acid-base equilibrium

pH scale, weak acids bases, salt of weak acid and bases

pH scale

• pH is the measure of the acidity or alkalinity of a solution

• pH is a measurement of the concentration of hydrogen ions in a solution. So, low pH values are associated with solutions with high concentrations of hydrogen ions, while high pH values occur for solutions with low concentrations of hydrogen ions.

pH scale

• The pH scale is an inverse logarithmic representation of hydrogen proton (H+) concentration.

• The pH of solution was defined by Danish chemist S. P. L. Sørensen as

pH = − log10[H + ]

pH scale

Example 1

Calculate the pH of a 2.0 x 10-3 M solution

of HCl.

pH scale

Solution[H+] = 2.0 x 10-3 M

pH = − log10[H + ]

= − log10[2.0 x 10-3 M ]

= 3 - log10 2.0

= 3 – 0.30

= 2.7

pH scale

• There is also pOH, in a sense the opposite of pH, which measures the concentration of OH− ions, or the alkalinity.

• A similar definition is made for the hydroxyl ion concentration

pOH = − log10[OH- ]

pH scale

• -log Kw = - log [H+][OH-]

= -log [H+] – log [OH-]

pKw = pH + pOH

pH scale

• PLEASE NOTE THAT AT 25˚C

[H+] x [OH-] = 10-14

pH + pOH = 14

pH scale

• Example 2

Calculate the pOH and the pH of a 5.0 x

10-2 M solution of NaOH

pH scale

• Solution Method 1

[OH-] = 5.0 x10-2 M

pOH = - log(5.0 x10-2 )

= 2 – log 5.0

= 1.3

pH scale

• Solution continued

Since

pH + 1.3 = 14.00

pH = 12.7

pH + pOH = 14

pH scaleSolution Method 2

Since

[H+] = 1.0 x 10-14 = 2.0 x 10-13

5.0 x 10-12

pH = - log (2.0 x 10-13) = 13 – log 2.0

= 13 – 0.30 = 12.7

[H+] x [OH-] = 10-14

pH scale

• When [H+] = [OH-] the solution is NEUTRAL. – pH@pOH of 7

• When [H+] < [OH-] the solution is ALKALINE– pH@pOH greater than 7

• When [H+] > [OH-] the solution is ACIDIC– pH@pOH less then 7

pH scale

Weak Acid

• A weak acid is an acid that does not completely donate all of its hydrogens when dissolved in water.

• These acids have higher pH compared to strong acids, which release all of their hydrogens when dissolved in water.

Weak Acid

• The acidity constant for acetic acid at 25oC is 1.75 x 10-5

• [H+][OAc-] = 1.75 x 10-5

[HOAc]• When acetic acid ionizes, it dissociates to equal portion

of H+ and OAc- by such an amount will always be equal 1.75 x 10-5

• The equilibrium concentrations of reactants and products are related by the Acidity constant expression (Ka)

• [H+][OAc-] = Ka[HOAc]

Weak Acid

• example 3

Calculate the pH of a 1.00 x 10-3 M solution

Of Acetic Acid

Weak Bases• A weak base is a chemical base that only

partially ionize in water.

Weak Acid and Weak Base

Example 4

Calculate the pH and pOH for a 1.00 x 10-3

M solution of ammonnia.

Kb ,basicity constant is 1.75 x 10-5 at 25oC

Salts of Weak Acids and Bases

• The salt of a weak acid for example NaOAc is strong electrolyte, like all salt and completely ionizes.

• In addition, the anion of the salt of a weak acid is a Brønsted base which will accept protons.

• It partially hydrolyzed in water to form hydroxide ion and the corresponding undissociated acid.

Salts of Weak Acids and Bases

• If the salt hydrolyzes that salt is consider as a weak base.

• The weaker the conjugate acid, the stronger the conjugate base, that is, the more strongly the salt will combine with a proton, as from the water , to shift the ionization to the right.

Salts of Weak Acids and Bases

• Example 5

Calculate the pH of a 0.25 M solution of ammonium chloride.

Salts of Weak Acids and Bases

Solution.

Write the equilibria

NH4Cl NH4+ + Cl- (ionization)

NH4+ + H20 ↔ NH4OH + H+ (hydrolysis)

(NH4+ + H20 ↔ NH3 + H30 + )

Salts of Weak Acids and Bases

[NH4OH][H+ ] = Ka = Kw = 1.0 x 10 -14

NH4 Kb 1.7 x 10-5

= 5.7 x 10-10

Let x represent the concentration of [NH4OH] and

[H+ ] at equilibrium. Then at equilibrium,

[NH4OH] = [H+ ] = x

[NH4+] = CNH4+ - x = 0.25 - x

Salts of Weak Acids and Bases

Since CNH4+ » Ka , neglect x compared to CNH4+ Then,

(x)(x) = 5.7 x 10^-10

0.25

X = √ 5.7 x 10^-10 x 0.25 = 1.2 x 10-5 M

The NH4OH formed is undissociated and does no

contribute to the pH

[H+] = 1.2 x 10-5 M

pH = - log (1.2 x 10-5 M) = 5 – 0.08 = 4.92