lecture 10asf

Lecture 10asf - pH of Weak Acids Calculate the pH and percent dissociation of 0.10 M CH3COOH CH3COOH H2O H3O CH3COOStart 0.10 0 0 Change-x x x

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pH of Weak Acids Calculate the pH and percent dissociation of 0.10 M CH 3 COOH CH 3 COOH + H 2 O H 3 O + + CH 3 COO - K a = 1.8 x 10 -5 Start 0.10 0 0 Change -x +x +x equil 0.10 -x x x x 10 . 0 x 2 - = 1.8 x 10 -5 Using the approximation that x is small compared with 0.10, we should be able to solve quickly: x = [H 3 O + ] = 1.34 x 10 -3 M pH = - log[H 3 O + ] = -log (1.34 x 10 -3 ) = 2.87 Percent ionization = 10 . 0 10 x 34 . 1 3 - x 100 = 1.34% Since the percent ionization is less than 5%, our approximation is considered valid. If we solved exactly, we would get x = 1.33 x 10 -3 M, very close to our approximate solution. Calculate the pH and percent dissociation of 0.0010 M CH 3 COOH CH 3 COOH + H 2 O H 3 O + + CH 3 COO - K a = 1.8 x 10 -5 Start 0.0010 0 0 Change -x +x +x equil 0.0010 -x x x x 0010 . 0 x 2 - = 1.8 x 10 -5 Using the approximation that x is small compared with 0.0010, we should be able to solve quickly: x = [H 3 O + ] = 1.34 x 10 -4 M Percent ionization = 0010 . 0 10 x 34 . 1 4 - x 100 = 13.4%
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Since the percent ionization is more than 5%, our approximation is suspect. Using successive approximations, or solving the quadratic equation exactly, we get x = [H 3 O + ] = 1.26 x 10 -4 pH = - log[H 3 O + ] = -log (1.26 x 10 -4 ) = 3.90 Percent ionization = 0010 . 0 10 x 26 . 1 4 - x 100 = 12.6% Comparing 0.10 M CH 3 COOH and 0.0010 M CH 3 COOH [H 3 O + ] pH % ionization 0.10 M CH 3 COOH 1.34 x 10 -3 2.87 1.34% 0.0010 M CH 3 COOH 1.26 x 10 -4 3.90 12.6% The more dilute acid has lower [H 3 O + ] and consequently a higher pH, but it also has a higher percent ionization. Calculate the pH and percent dissociation of 0.10 M HCN HCN + H 2 O H 3 O + + CN - K a = 6.2 x 10 -10 0.10 - x x x We have set up our equilibrium calculation in one line. We start with 0.10M HCN and x reacts. This forms x in both H 3 O + and CN - and leaves over 0.10-x in HCN. To save time and space, further equilibrium calculations shown here will be done in one line. x 10 . 0 x 2 - = 6.2 x 10 -10 Solving, using the usual approximation: x = [H 3 O + ] = 7.9 x 10 -6 M pH = - log[H 3 O + ] = -log(7.9 x 10 -6 ) = 5.10 Percent ionization = 10 . 0 10 x 9 . 7 6 - x 100 = 0.0079 % Compare the answers for 0.10 M CH 3 COOH and 0.10 M HCN.
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These are both weak acids, both HCN is a weaker acid (smaller K a ) than CH 3 COOH. The HCN solution has a lower [H 3 O + ], a higher pH and a smaller percent ionization—all consistent with a smaller value of K a . A 0.020 M solution of acid HX is measured to have pH = 2.90. Calculate the K a and percent ionization of HX. In the previous examples, we were given the concentration of a weak acid and its K a value and were asked to calculate the pH and percent ionization. In this example, we are given the concentration and pH of a weak acid, and asked to calculate its K a and percent ionization. Let’s first set up a general approach to this type of problem.
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This note was uploaded on 04/04/2008 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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Lecture 10asf - pH of Weak Acids Calculate the pH and percent dissociation of 0.10 M CH3COOH CH3COOH H2O H3O CH3COOStart 0.10 0 0 Change-x x x

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