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Acid-base equilibrium

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Presentation on theme: "Acid-base equilibrium"— Presentation transcript:

1 Acid-base equilibrium
pH scale, weak acids bases, salt of weak acid and bases

2 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.

3 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 + ]

4 pH scale Example 1 Calculate the pH of a 2.0 x 10-3 M solution of HCl.

5 pH scale Solution [H+] = 2.0 x 10-3 M pH = − log10[H + ]
= − log10[2.0 x 10-3 M ] = 3 - log = 3 – 0.30 = 2.7

6 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- ]

7 pH scale -log Kw = - log [H+][OH-] = -log [H+] – log [OH-]
pKw = pH + pOH edit

8 pH scale PLEASE NOTE THAT AT 25˚C [H+] x [OH-] = 10-14 pH + pOH = 14

9 pH scale Example 2 Calculate the pOH and the pH of a 5.0 x
10-2 M solution of NaOH

10 pH scale Solution Method 1 [OH-] = 5.0 x10-2 M pOH = - log(5.0 x10-2 )
= 1.3

11 pH scale pH + pOH = 14 Solution continued Since pH + 1.3 = 14.00

12 pH scale [H+] x [OH-] = 10-14 Solution Method 2 Since
[H+] = 1.0 x = 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

13 pH scale When [H+] = [OH-] the solution is NEUTRAL.
of 7 When [H+] < [OH-] the solution is ALKALINE greater than 7 When [H+] > [OH-] the solution is ACIDIC less then 7

14 pH scale

15 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.

16 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

17 Weak Acid example 3 Calculate the pH of a 1.00 x 10-3 M solution
Of Acetic Acid

18 Weak Bases A weak base is a chemical base that only partially ionize in water.

19 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

20 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.

21 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.

22 Salts of Weak Acids and Bases
Example 5 Calculate the pH of a 0.25 M solution of ammonium chloride.

23 Salts of Weak Acids and Bases
Solution. Write the equilibria NH4Cl NH Cl- (ionization) NH4+ + H20 ↔ NH4OH + H+ (hydrolysis) (NH4+ + H20 ↔ NH3 + H30 + )

24 Salts of Weak Acids and Bases
[NH4OH][H+ ] = Ka = Kw = 1.0 x NH Kb 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 = x

25 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 – = 4.92

26 Thats All Folks! Any Question(s)?!


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