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Ch17_pH_2

# Ch17_pH_2 - MORE ABOUT WATER 1 Autoionization H2O can...

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Unformatted text preview: MORE ABOUT WATER 1 Autoionization H2O can function as both an ACID and a BASE. 2 MORE ABOUT WATER OH-OH Kw = [OH -] = 1.00 x 10 -14 at 25 Le Chatelier predicts equilibrium shifts to the ____________. [H3O+] < 10-7 at equilibrium. Set up a concentration table. oC In a neutral solution [H 3O+] = [OH-] In neutral solution so Kw = [H3O+]2 = [OH-]2 In pure water there can be AUTOIONIZATION In AUTOIONIZATION + Calculating [H3O+] & [OH--] 3 You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3O+] and [OH -]. Solution 2 H2O(liq) ¸ H3O+(aq) + OH-(aq) 0 0.0010 initial initial change change equilib equilib +x x and so [H 3O+] = [OH -] = 1.00 x 10 -7 M 4 + Calculating [H3O+] & [OH--] 3 You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3O+] and [OH -]. Solution 2 H2O(liq) ¸ H3O+(aq) + OH-(aq) [H3O+] = Kw / 0.0010 = 1.0 x 10 -11 M +x 0.0010 + x Kw = (x) (0.0010 + x) Because x << 0.0010 M, assume [OH-] = 0.0010 M [H3O+] = Kw / 0.0010 = 1.0 x 10 -11 M 3 You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3O+] and [OH -]. Solution 2 H2O(liq) ¸ H3O+(aq) + OH-(aq) + H33O+ HO [H 3O+] + Calculating [H3O+] & [OH--] 3 BASIC This solution is _____________ because + [H3O +] < [OH --] 3 Page 1 5 + [H3O+], [OH--] and pH 3 A common way to express acidity and basicity is with pH pH = log (1/ [H3O+])) = - log [H3O+] log In a neutral solution, In [H3O+] = [OH -] = [H 1.00 x 10 -7 at 25 oC 1.00 pH = -log (1.00 x 10 -7) = - (-7) = 7 (-7) 6 + [H3O+], [OH--] and pH 3 What is the pH of the What 0.0010 M NaOH solution? 0.0010 [H3O+] = 1.0 x 10 -11 M 7 + [H3O+], [OH--] and pH 3 8 Other pX Scales pX Scales If the pH of Coke is 3.12, it is ____________. In general pX = -log X Because pH = - log [H 3O+] then and so pOH = - log [OH -] pOH log log [H3O+] = - pH Kw = [H3 O+] [OH-] = 1.00 x 10-14 at 25 oC pH = - log (1.0 x 10 -11) = 11.00 Take antilog and get Take the log of both sides General conclusion — [H3O+] = 10-pH -log (10-14) = - log [H3O+] + (-log [OH -]) Basic solution Neutral Acidic solution Acidic pH > 7 pH pH = 7 pH pH < 7 [H3O +] = 10 -3.12 = 7.6 x 10 -4 M 7.6 Page 2 14 = pH + pOH 14 9 ...
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