AP Chapter 15 - The Common Ion Effect When the salt with...

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1 Chapter 15 Applying equilibrium The Common Ion Effect c When the salt with the anion of a weak acid is added to that acid, c It reverses the dissociation of the acid. c Lowers the percent dissociation of the acid. c The same principle applies to salts with the cation of a weak base. c The calculations are the same as last chapter. Buffered solutions c A solution that resists a change in pH. c Either a weak acid and its salt or a weak base and its salt. c We can make a buffer of any pH by varying the concentrations of these solutions. c Same calculations as before. c Calculate the pH of a solution that is .50 M HAc and .25 M NaAc (Ka = 1.8 x 10 -5 ) Na + is a spectator and the reaction we are worried about is HAc H + + Ac - c Choose x to be small x c We can fill in the table x x -x 0.50-x 0.25+x Initial 0.50 M 0 0.25 M Final c Do the math c Ka = 1.8 x 10 -5 1.8 x 10 -5 = x (0.25+x) (0.50-x) c Assume x is small = x (0.25) (0.50) x = 3.6 x 10 -5 c Assumption is valid c pH = -log (3.6 x 10 -5 ) = 4.44 HAc H + + Ac - x x x -x 0.50-x 0.25+x Initial 0.50 M 0 0.25 M Final Adding a strong acid or base c Do the stoichiometry first. –Use moles not molar c A strong base will grab protons from the weak acid reducing [HA] 0 c A strong acid will add its proton to the anion of the salt reducing [A - ] 0 c Then do the equilibrium problem. c What is the pH of 1.0 L of the previous solution when 0.010 mol of solid NaOH is added?
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2 HAc H + + Ac - c In the initial mixture M x L = mol c 0.50 M HAc x 1.0 L = 0.50 mol HAc c Adding 0.010 mol OH - will reduce the HAc and increase the Ac - by 0.010 mole c Because it is in 1.0 L, we can convert it to molarity 0.50 mol 0.25 mol c 0.25 M Ac - x 1.0 L = 0.25 mol Ac - 0.49 mol 0.26 mol HAc H + + Ac - c In the initial mixture M x L = mol c 0.50 M HAc x 1.0 L = 0.50 mol HAc c 0.25 M Ac - x 1.0 L = 0.25 mol Ac - c Adding 0.010 mol OH - will reduce the HAc and increase the Ac - by 0.010 mole c Because it is in 1.0 L, we can convert it to molarity 0.50 mol 0.25 mol 0.49 M 0.26 M HAc H + + Ac - c Fill in the table 0.50 mol 0.25 mol 0.49 M 0.26 M HAc H + + Ac - x x x -x 0.49-x 0.26+x Initial 0.49 M 0 0.26 M Final c Do the math c Ka = 1.8 x 10 -5 1.8 x 10 -5 = x (0.26+x) (0.49-x) c Assume x is small = x (0.26) (0.49) x = 3.4 x 10 -5 c Assumption is valid c pH = -log (3.4 x 10 -5 ) = 4.47 HAc H + + Ac - x x x -x 0.49-x 0.26+x Initial 0.49 M 0 0.26 M Final Notice c If we had added 0.010 mol of NaOH to 1 L of water, the pH would have been. c 0.010 M OH - c pOH = 2 c pH = 12 c But with a mixture of an acid and its conjugate base the pH doesn’t change much c Called a buffer. General equation c Ka = [H + ] [A - ] [HA] c so [H + ] = Ka [HA] [A - ] c The [H + ] depends on the ratio [HA]/[A - ] c taking the negative log of both sides c pH = -log(Ka [HA]/[A - ]) c pH = -log(Ka)-log([HA]/[A - ]) c pH = pKa + log([A - ]/[HA])
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3 This is called the Henderson- Hasselbach equation c pH = pKa + log([A - ]/[HA]) c pH = pKa + log(base/acid) c Works for an acid and its salt
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AP Chapter 15 - The Common Ion Effect When the salt with...

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