Chem+162-2016+Chapter+18+Acid-base+_+solub+equilibria+lecture+notes - CHEMISTRY 162-2016 CHAPTER 18 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA PROF

Chem+162-2016+Chapter+18+Acid-base+_+solub+equilibria+lecture+notes

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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 1 CHEMISTRY 162-2016 CHAPTER 18 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA PROF. E. TAVSS
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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 2 CHEMICAL, ACID AND BASE EQUILIBRIA transp pH [H + ] 14 Basic 1x10 -14 H 2 O + H 2 O H 3 O + + OH - K w = 1x10 -14 7 Neutral K w = [H + ][OH - ] = 1 x 10 -14 pH + pOH = 14 K w = K a x K cb = K ca x K b = 1 x 10 -14 * *a & b = acid and base; ca & cb = conjugate acid and conjugate base pK a + pK b = 14 0 Acidic 1x10 o A pH scale of 1 to 13 is common, but pH’s of -1 to +15 are possible. pH = -log[H + ] pH [H + ] [H + ] = 10 -pH pOH = -log[OH - ] [OH - ] = 10 -pOH pOH [OH - ] Henderson-Hasselbalch equation: pH = pK a + log([B]/[A]) where acid = acid or conjugate acid, and base = base or conjugate base. Also, pOH = pK base + log([Acid]/[Base]) pH + pOH = 14 pH + pOH = 14 [H + ][OH - ] = 1 x 10 -14 [H + ][OH - ] = 1 x 10 -14 pOH = -log[OH - ] [OH - ] = 10 -pOH pH = -log[H + ] [H + ] = 10 -pH
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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 3 MATHEMATICAL TRICKS FOR AVOIDING QUADRATIC EQUATIONS RESULTING FROM SOLVING ICE TABLE PROBLEMS (1) Small K rule. page 33. 2CO 2 (g) 2CO(g) + O 2 (g) If the equilibrium constant is small, such as 10 -3 K, then this means that there is very little substance on the right side of the equation (which is usually the product side) and virtually all of the substance is on the left side (which is usually the reactant side). Therefore, drop the X in the ICE table when the X is in a format such as Y-X on the reactant side, or Y+X on the product side, because no significant amount of new product will be forming. The requirement for using the small K rule is that the error caused by this approximation must be < 5% of the value of Y. (2) Large K rule. page 34. 2NO(g) + Cl 2 (g) 2NOCl(g) If the equilibrium constant is large, such as 10 3 , this means that almost all of the substances are on the product side. Handle this ICE table by first finding the limiting reactant and then bringing the reaction to completion. This will result in the concentration of the limiting reactant becoming zero and the concentrations of the product becoming large. (There will be some times when a zero concentration is obtained for a substance, that we are required to have a finite value for that substance. In that case we would have to bring the reaction slightly back to the left, i.e., back to a finite equilibrium value, using the “Right-to-Left K Rule.) (3) Right-to-left K Rule. After using the large K rule and bringing the reaction to completion, sometimes it’s necessary to go from right to left (hence, “Right-to-Left K Rule) in order to provide a finite value of a substance having “0” concentration on the left side of the equilibrium equation. When going from right to left, realize that K is large, so there is a large quantity on the right side and a small quantity on the left side of the equation. Therefore, to avoid a quadratic equation, one may drop the “-X” from a larger number on the right side (X would be very small in going from right to left when K is large) and one may correspondingly drop the “+X” from a larger number on the left side.
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