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pH And pOH
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Water Water autoionizes H 2 O + H 2 O H 3 O+ + OH- Hydrogen ion concentration determines the acidity of the solution Kc = [H3O+][OH-] ionization constant
![Water Water autoionizes H 2 O + H 2 O H 3 O+ + OH- Hydrogen ion concentration determines the acidity of the solution Kc = [H3O+][OH-] ionization constant](http://images.slideplayer.com/32/10002656/slides/slide_2.jpg "Water Water autoionizes H 2 O + H 2 O H 3 O+ + OH- Hydrogen ion concentration determines the acidity of the solution Kc = [H3O+][OH-] ionization constant")
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Ion-product Constant Kc = [H3O+][OH-] could be written as Kc = [H+][OH-] To indicate this is for water, we write it as Kw Kw is the ion product constant, the product of the concentration of H+ and OH- at a particular temperature In pure water at 25 o C, Kw = 1.0 x 10 -14
![Ion-product Constant Kc = [H3O+][OH-] could be written as Kc = [H+][OH-] To indicate this is for water, we write it as Kw Kw is the ion product constant, the product of the concentration of H+ and OH- at a particular temperature In pure water at 25 o C, Kw = 1.0 x](http://images.slideplayer.com/32/10002656/slides/slide_3.jpg "Ion-product Constant Kc = [H3O+][OH-] could be written as Kc = [H+][OH-] To indicate this is for water, we write it as Kw Kw is the ion product constant, the product of the concentration of H+ and OH- at a particular temperature In pure water at 25 o C, Kw = 1.0 x")
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Example 1 Concentrations of OH- in household ammonia is 0.0025 M. Calculate the concentration of H+ ions. Kw = [H+][OH-] so [H+] = Kw/[OH-] [H+] = 1.0 x 10 -14 /0.0025 = 4.0 x 10 -12 M Since [H+]<1.0 x 10 -7 M, this is basic
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pH and pOH pH measures [H+]: pH = -log[H+] pOH measures [OH-]: pOH = -log[OH-] pH + pOH = 14.00
![pH and pOH pH measures [H+]: pH = -log[H+] pOH measures [OH-]: pOH = -log[OH-] pH + pOH = 14.00](http://images.slideplayer.com/32/10002656/slides/slide_5.jpg "pH and pOH pH measures [H+]: pH = -log[H+] pOH measures [OH-]: pOH = -log[OH-] pH + pOH = 14.00")
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Example 2 What is the pH of a 1.0 x 10 -3 M NaOH solution? NaOH → Na+ + OH- so 1.0 x 10 -3 M NaOH produces 1.0 x 10 -3 M [OH-] Kw = [H+][OH-] [H+] = Kw/[OH-] = (1.0 x 10 -14 )/(1.0 x 10 -3 ) [H+]=1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11
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Example 3 Determine the hydronium ion concentration of an aqueous solution with pH 4.0 pH=-log [H+] [H+] = antilog -pH [H+] = antilog -4.0 [H+]=1.0 x 10 -4
![Example 3 Determine the hydronium ion concentration of an aqueous solution with pH 4.0 pH=-log [H+] [H+] = antilog -pH [H+] = antilog -4.0 [H+]=1.0 x 10 -4](http://images.slideplayer.com/32/10002656/slides/slide_7.jpg "Example 3 Determine the hydronium ion concentration of an aqueous solution with pH 4.0 pH=-log [H+] [H+] = antilog -pH [H+] = antilog -4.0 [H+]=1.0 x 10 -4")
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Example 4 Determine the hydronium ion concentration of an aqueous solution with pH 4.0 pH=-log [H+] [H+] = antilog -pH [H+] = antilog -4.0 [H+]=1.0 x 10 -4
![Example 4 Determine the hydronium ion concentration of an aqueous solution with pH 4.0 pH=-log [H+] [H+] = antilog -pH [H+] = antilog -4.0 [H+]=1.0 x 10 -4](http://images.slideplayer.com/32/10002656/slides/slide_8.jpg "Example 4 Determine the hydronium ion concentration of an aqueous solution with pH 4.0 pH=-log [H+] [H+] = antilog -pH [H+] = antilog -4.0 [H+]=1.0 x 10 -4")
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