Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut.

Acids and Bases Weak Bases Arrhenius says that Bases react with water to produce hydroxide ion. What are the other definitions of Bases?

Acids and Bases Weak Bases Since bases dissociate into different ions, the equilibrium expression is different. where K b is the base equilibrium constant.

Acids and Bases Weak Bases There are two types of Weak bases: 1)Amines: Remember these? -Amines are molecules that have at least one nitrogen – hydrogen bond. The Nitrogen atom will donate a lone pair (Lewis Base) and accept a proton (Bronsted-Lowry Base) 2)Anions of Weak Acids: - In solution, the metal cation will be a spectator. The anion however will be the conjugate base of a weak acid, thus the solution becomes basic.

Acids and Bases Examples of Weak Bases: NaClO + H 2 O  What are the products? Na + and ClO - The Sodium ion will be spectator ion, but the chlorite ion is the conjugate base of HClO. The Hydrogen isn’t there to be donated, but the presence of the ClO - ion is! Thus it behaves like a base! NaC 2 H 3 O 2 + H 2 O  What are the products? Na+ = spectator C 2 H 3 O 2 - = conjugate base = BASIC Sol.

Acids and Bases pH of Basic Solutions What is the pH of a 0.15 M solution of NH 3 ? To solve this, you must realize that this is a Weak BASE, and we want the pH (H + ions) Working with a weak base, we must use the K B – the base equilibrium constant. Where can we find the K B ? Look on our tables:

Acids and Bases K A x K B = Magic!!! We can use the charts that have K A values to determine the K B values. We again can go back to the autoionization of water and prove this, but this is what you need to know: K A x K B = 1.0 x

Acids and Bases K a and K b are linked: Combined reaction = ?

Acids and Bases K a and K b are linked: Combined reaction = ?

Acids and Bases K a and K b K a and K b are related in this way: K a  K b = K w Therefore, if you know one of them, you can calculate the other.

Acids and Bases K A + K B = Magic!!! Looking at the chart, We see the K A for ammonia is 5.8 x , so what is the Kb? K A x K B = 1.0 x x = 1.7 x10 -5 = K B 5.8 x

Acids and Bases Back to the Original Problem! What is the pH of a 0.15 M solution of NH 3 ? [NH 4 + ] [OH − ] [NH 3 ] K b = = 1.7  [NH 3 ][NH 4 + ][OH − ] Initial Equilibrium xxx

Acids and Bases Trying Something New For many (NOT ALL) of these weak acid/base problems, subtracting “X” from the initial concentration is SOOOO small, that it doesn’t change the calculation very much. So small in fact that we can ELIMINATE it from the problem. We will test a shortcut here, then come back and make sure it is ok to do so…. (x) 2 (0.15-x) 1.7  =

Acids and Bases Notice the denominator! X is Gone! (x)(x) (0.15) 1.7  = (1.7  ) (0.15) = x x10 -6 = x 2  2.55 x10 -6 = x 1.6 x10 -3 M = x = [OH - ]

Acids and Bases Check to see if it’s ok to eliminate X! We need to make sure that eliminating X from the initial concentration does not affect the answer very much. (0.15-x)  x10 -3 =.1484 Notice that the value didn’t change very much, but how much is acceptable?

Acids and Bases Rule for Eliminating X We need to make sure that eliminating X from the initial concentration did not change the answer very much. If X / Initial Concentration is less than 5 %, then OK to eliminate. If X / Initial Concentration is greater than 5 %, must go back and use the quadratic equation.

Acids and Bases Eliminate X Double Check! X = [OH-] = 1.6 x Initial Concentration = 0.15 (1.6 x10 -3 / 0.15) x 100 = 1.06 % 1.06 % < 5 % so OK to eliminate!

Acids and Bases Oh Yeah, the Original Question… What is the pH of a 0.15 M solution of NH 3 ? X = [OH – ] = 1.6  M pOH = –log (1.6  ) pOH = 2.80 pH = – 2.80 pH = 11.20

Acids and Bases A M solution of niacin has a pH of (a) What percentage of the acid is ionized in this solution? (b) What is the acid-dissociation constant, K a, for niacin? PRACTICE EXERCISES 1. Niacin, one of the B vitamins, has the following molecular structure: 2. What is the pH of (a) a M solution of NaOH, (b) a M solution of Ca(OH) 2 ? What percentage of the bases are ionized? 3. Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a M HF solution. Ka for HF is 6.8 x10 -4.