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

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1 Chapter 15 Applying equilibrium The Common Ion Effect circle6 When the salt with the anion of a weak acid is added to that acid, circle6 It reverses the dissociation of the acid. circle6 Lowers the percent dissociation of the acid. circle6 The same principle applies to salts with the cation of a weak base. circle6 The calculations are the same as last chapter. Buffered solutions circle6 A solution that resists a change in pH. circle6 Either a weak acid and its salt or a weak base and its salt. circle6 We can make a buffer of any pH by varying the concentrations of these solutions. circle6 Same calculations as before. circle6 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 - circle6 Choose x to be small x circle6 We can fill in the table x x -x 0.50-x 0.25+x Initial 0.50 M 0 0.25 M Final circle6 Do the math circle6 Ka = 1.8 x 10 -5 1.8 x 10 -5 = x (0.25+x) (0.50-x) circle6 Assume x is small = x (0.25) (0.50) x = 3.6 x 10 -5 circle6 Assumption is valid circle6 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 circle6 Do the stoichiometry first. –Use moles not molar circle6 A strong base will grab protons from the weak acid reducing [HA] 0 circle6 A strong acid will add its proton to the anion of the salt reducing [A - ] 0 circle6 Then do the equilibrium problem. circle6 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 - circle6 In the initial mixture M x L = mol circle6 0.50 M HAc x 1.0 L = 0.50 mol HAc circle6 Adding 0.010 mol OH - will reduce the HAc and increase the Ac - by 0.010 mole circle6 Because it is in 1.0 L, we can convert it to molarity 0.50 mol 0.25 mol circle6 0.25 M Ac - x 1.0 L = 0.25 mol Ac - 0.49 mol 0.26 mol HAc H + + Ac - circle6 In the initial mixture M x L = mol circle6 0.50 M HAc x 1.0 L = 0.50 mol HAc circle6 0.25 M Ac - x 1.0 L = 0.25 mol Ac - circle6 Adding 0.010 mol OH - will reduce the HAc and increase the Ac - by 0.010 mole circle6 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 - circle6 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 circle6 Do the math circle6 Ka = 1.8 x 10 -5 1.8 x 10 -5 = x (0.26+x) (0.49-x) circle6 Assume x is small = x (0.26) (0.49) x = 3.4 x 10 -5 circle6 Assumption is valid circle6 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 circle6 If we had added 0.010 mol of NaOH to 1 L of water, the pH would have been. circle6 0.010 M OH - circle6 pOH = 2 circle6 pH = 12 circle6 But with a mixture of an acid and its conjugate base the pH doesn’t change much circle6 Called a buffer. General equation circle6 Ka = [H + ] [A - ] [HA] circle6 so [H + ] = Ka [HA] [A - ] circle6 The [H + ] depends on the ratio [HA]/[A - ] circle6 taking the negative log of both sides circle6 pH = -log(Ka [HA]/[A - ]) circle6 pH = -log(Ka)-log([HA]/[A - ]) circle6 pH = pKa + log([A - ]/[HA])
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3 This is called the Henderson- Hasselbach equation circle6 pH = pKa + log([A - ]/[HA]) circle6 pH = pKa + log(base/acid) circle6 Works for an acid and its salt circle6 Like HNO 2 and NaNO 2 circle6 Or a base and its salt circle6 Like NH 3 and NH 4 Cl circle6
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