Review 1: Written the conjugate base and acid for the following acids and bases.

Review 2: Name the acid, base, conjugate acid and base for each reaction. CH 3 NH 2 + H 2 O  CH 3 NH OH - H 3 PO 4 + H 2 O  H 3 O + + H 2 PO 4 - HC 2 H 3 O 2 + H 2 O  H 3 O + + C 2 H 3 O 2 - CH 3 NH H 2 PO 4 -  H 3 PO 4 + CH 3 NH 2 base acid conjugate acid conjugate base

Some common Acids Most acids are Oxyacids These are acids in which the acidic hydrogen is attached to an oxygen. Organic acids are ones that have a carbon atom backbone. They commonly contain the carboxyl group COOH. Acetic acid is an example

Water as an Acid and a Base H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) Acid base Some substances can act as both acids and bases, they are called amphoteric. Water is an example. Water can react with itself to form ions The equilibrium expression for this reaction is: K w = [H 3 O + ] [OH - ] this is called the dissociation constant K w = (1.0  )(1.0  ) K w = (1.0  ) Only two H 2 O molecules out of every billion ionize. The concentration of is [H 3 O + ] and [OH - ] is 1.0  mol L -1 K w : K = constant w = water

But at 25  C the K w of any aqueous solution (no matter what it contains) will equal 1.0  [H 3 O + ]  [OH - ] must = 1  10 –14. In pure water, [H 3 O + ] = [OH - ] = 1  10 –7 Kw = (1  10 –7 ) (1  10 –7 ) = 1  10 –14 Kw = 1  10 –14 So if [H 3 O + ] goes up, [OH - ] must go down so that the product of the two is still 1  10 – In a neutral solution [H 3 O + ] = [OH - ] 2. In an acidic solution [H 3 O + ]  [OH - ] 3. In a basic solution [OH - ]  [H 3 O + ] There are three possibilities in an aqueous solution.

Calculating Ion Concentrations in water solutions: Kw = [H 3 O + ] [OH - ] = 1  10 –14 What is the [H 3 O + ] or [OH - ] in each of the following solutions, also state if the solution is neutral, acidic, or basic. a. 1  10 –5 M OH - b. 1  10 –7 M OH - c M H + a. b. c.

The pH scale is used to compare the strength of different acids and bases. The strength of an acid is based on the [H 3 O + ]. But these values are often very small numbers E.g. 1.0  mol L -1 = To express small numbers conveniently, chemists use the “p scale” which is based on logarithim (base 10) 9.00 on the p scale pH = -log[H + ]

[H + ] = 1  = mol L -1 [H + ] = 1  = 0.1 mol L -1 pH = -log [H + ] = 14 pH = -log [H + ] = 1

So you can convert the concentrations of [H + ] and [OH - ] in solutions in terms of pH and pOH pH = - log [H + ]pOH = - log [OH - ] pH + pOH = 14 So if you know the [H + ] then you can calculate the pH of a solution.

Calculating the pH and pOH using a calculator: 1. What is the pH of a solution where [H + ] =1.0  Step 1: Enter the number 1.0  Step 2: Push the log key-9.00 Step 3: Push the +/- key 9.00 pH = What is the pOH of the above solution? pH + pOH = 14 pOH = 14 - pH = 14 – 9.00 pOH = 5.00

3. What is the pH of a solution with [OH - ] =1.0  Kw = [H 3 O + ] [OH - ] = 1  10 –14 [H + ] = 1  10 –14 / 1.0  = 1.0  Step 1: Enter the number 1.0  Step 2: Push the log key-8.00 Step 3: Push the +/- key Calculating [H + ] from the pH pH = 8.00 Step 1: Enter the pH 7.0 Step 2: Push the +/- key-7.00 Step 3: Push the inv log keys1.0  [H + ] = 1.0  10 -7