# ACIDS AND BASES Ionization of Water H O H H + O H -

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ACIDS AND BASES Ionization of Water H O H H + O H -

Describe the equilibrium between hydronium and hydroxide ions in water. Define the ion product of water (K w ). Describe the concentrations of hydronium and hydoxide ions. Calculate the concentration of hydronium or hydroxide ions.

Water is amphoteric HA + H 2 O (l) H 3 O + (aq) + A¯ (aq) or B + H 2 O (l) BH + (aq) + OH¯ (aq)

Water dissociates into ions - self-ionization. When water particles collide - ions form. H 2 O (l) H + (aq) + OH¯ (aq) H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH¯ (aq) or

ion product for water, K w K W = [H 3 O + ][OH¯] K W = [H + ][OH¯] H 2 O (l) H + (aq) + OH¯ (aq)

In room temperature water, the concentration of hydrogen ions and hydroxide ions, are each 1.0 x 10 -7 mol/L Water is neutral - these two must be a 1:1 ratio Any solution in which the [H + ] and [OH - ] are 1.0 x 10 -7 M is described as a neutral solution. Like all eqbm constants, the value of K w varies with temperature. K w = [H + ][OH - ] = 1.00 x 10 -14

Like any reversible reaction, Le Chatelier's applies: Add a SA – increase [H + ] decrease [OH - ] Add a SB – increase [OH - ] decrease [H + ] This means that H + and OH¯ are BOTH present in any solution - whether they are acidic or basic. OH - H2OH2O H+H+ +

If 2.5 moles of hydrochloric acid is dissolved in 5.0 L of water, what is the concentration of the hydroxide ions? Since HCl is a strong acid (100% dissociation): [H 3 O + ] = 0.50 mol/L. 5.0 L = 0.50 mol/L 2.5 mol n V M= HCl (s) H + (aq) + Cl¯ (aq)

[OH - ] = 2.0 x 10 -14 M K w = [H 3 O + ] [OH - ] 1.0 x 10 -14 = [0.50 [OH - ] + 1.0 x 10 -7 ]]

0.40 g of NaOH is dissolved in water to make a solution with a volume of 1.0 L. What is the hydronium ion concentration in this solution? NaOH = 40.0 g/mol 0.40 g 1 L 40.0 g 1 mol = 0.01 mol/L NaOH (s) Na + (aq) + OH¯ (aq)

[NaOH] = [OH – ] = 0.010 mol/L Substitute into the equilibrium law and solve for hydronium ion concentration: [H + ] = 1.0 x 10 -12 M K w = [OH - ] [H 3 O + ] 1.0 x 10 -14 = [0.010 [H 3 O + ] + 1.0 x 10 -7 ]]

H 2 O (l) H + (aq) + OH¯ (aq) K W = [H + ][OH¯] = 1.00 x 10 -14 Add a SA – increase [H + ] decrease [OH - ] Add a SB – increase [OH - ] decrease [H + ]

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