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Acids and Bases.

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Presentation on theme: "Acids and Bases."— Presentation transcript:

1 Acids and Bases

2 An Acid is a substance that can release a proton or hydrogen ion (H+ )when dissolved in water”
HCl H+ + Cl- . A base is a substance that accepts a proton .

3 A base can be defined as a substance that can release OH- ions when dissolved in water . NaOH OH- + Na+ . “ A Base is a substance that can release a Hydroxyl ion when dissolved in water”

4 Example of conjugate acid – base pairs CH3COOH / CH3COO- H2CO3 / HCO3-

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9 Weak acids and aeak bases

10 Weak acids . Weak acids and bases dissociate weakly in solutions thus they are predominantly present in the molecular form. Dissociation of weak acids Ka; In aqueous solutions weak acids ionize to a very limited extent as follows; HA + H2O H3O+ + A- Ki = [H3O+ ] [A- ] [H2O][HA] Since [H2O] is constant a new constant ka can be introduced , ka = [H3O+ ] [A- ] [HA]

11 Dissociation of water Water molecules can dissociate in water as follows, H2O H + + OH- The tendency of water to ionize is expressed as Ki Ki = [ H + ] [OH- ] [ H2O] Calculation of Ki ; Since 1 mole of H2O weighs 18g , thus the number of molecules in 1l of water (1000g) No of moles of water in 1 l = wt / MWt = 1000/ 18 = 55.56mole So the molarity of pure water is 55.5M. Since the probability that a hydrogen is present as an ion is 1.8 x10-9 thus the Molar concentration of both [ H + ] and [OH- ] can be calculated as 1.8 x x = 1.0 x M . Thus Ki = [ H + ] [OH- ] = x = x M . [ H2O]

12 Ionization of water To calculate for kw ; Ki = [ H + ] [OH- ] [ H2O]
Ki [ H2O] = [ H + ] [OH- ] 1.8 x x = 1x = [ H + ] [OH- ] So kw = [ H + ]+ [OH- ] = 1x10-14 Taking -log of both sides -log kw = -log [ H + ] -log[OH- ] p kw = p [ H + ] + p[OH- ] = = 14


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