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ACIDS AND BASES. Ionization of Water. Describe the relationship between the hydronium and hydroxide ion concentrations in water Include: the ion product of water, K w. Additional KEY Terms. Water is amphoteric. H +. HA + H 2 O ( l ) H 3 O + ( aq ) + A¯ ( aq ) - PowerPoint PPT Presentation
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ACIDS AND BASES
Ionization of Water
• Describe the relationship between the hydronium and hydroxide ion concentrations in water
Include: the ion product of water, Kw
Additional KEY Terms
• Water is amphoteric
HA + H2O(l) H3O+(aq) + A¯(aq)
or
B + H2O(l) BH+(aq) + OH¯(aq)
H+
H+
• Water also dissociates into ions - self-ionization• Water particles collide - ions form
H2O(l) H+(aq) + OH¯(aq)
H2O(l) + H2O(l) H3O+(aq) + OH¯(aq)
ion product for water, Kw
KW = [H3O+][OH¯]
KW = [H+][OH¯]
H2O(l) H+(aq) + OH¯(aq)
• Water reaches equilibrium with its ions
Like all constants, the value of Kw varies with temperature.
Water is neutral - these ions must be a 1:1 ratio
Kw = [H+][OH-] = 1.00 x 10-14
[H+] = [OH-] = 1.00 x 10-7 M
At room temperature:
Like any reversible reaction, Le Chatelier's applies:
Add a SB – increase [OH-] decrease [H+]
OH- H2O H+ +
Kw = [H+][OH-] = 1.00 x 10-14
Add a SA – increase [H+] decrease [OH-]
This means that H+ and OH¯ are BOTH present in any solution -
whether they are acidic or basic.
If 2.5 moles of hydrochloric acid is dissolved in 5.0 L of water, what is the [hydroxide ions]?
Since HCl is a strong acid (100% dissociation):
[H+] = 0.50 M
5.0 L= 0.50 mol/L
2.5 molnV
M=
HCl (s) H+(aq) + Cl¯(aq)
[OH-] = 2.0 x 10-14 M
Kw =
[H3O+]
[OH-]
1.0 x 10-14 =
[0.50
[OH-]+ 1.0 x 10-7]]
H2O(l) H+(aq) + OH¯(aq)
How does the addition of HCl affect WATER’s equilibrium?
Amount contributed by self-ionization of
water
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 g1 L 40.0 g
1 mol= 0.010 M
NaOH (s) Na+(aq) + OH¯(aq)
Since NaOH is a strong base(100% dissociation):[OH-] = 0.010 M
[H+] = 1.0 x 10-12 M
Kw =
[OH-]
[H3O+]
1.0 x 10-14 =
[0.010
[H3O+]+ 1.0 x 10-7]]
How does addition of NaOH affect WATER’s equilibrium?
H2O(l) H+(aq) + OH¯(aq)
KW = [H+][OH¯]
= 1.00 x 10-14
Add a SA – increase [H+] decrease [OH-]
Add a SB – increase [OH-] decrease [H+]
CAN YOU / HAVE YOU?
• Describe the relationship between the hydronium and hydroxide ion concentrations in water
Include: the ion product of water, Kw
Additional KEY Terms
