 Acids are substances that: ◦ taste sour. ◦ react w/ bases to form salts and water. ◦ are electrolytes. ◦ turn blue litmus red. ◦ produce H 3 O +1 ions in aqueous solution. ◦ donate H +1 ions (protons).

 Bases are substances that: ◦ taste bitter. ◦ react w/ acids to form salts and water. ◦ are electrolytes. ◦ turn red litmus blue. ◦ produce OH -1 ions in aqueous solutions. ◦ accept H +1 ions (protons).

 H 2 O + H 2 O  H 3 O +1 + OH -1 ◦ Extremely small K eq. ◦ K eq = K w = 1.0 x ◦ Only % of H 2 O molecules autoionize.  In pure water, [H 3 O +1 ] = 1.0 x Molar. ◦ [OH -1 ] also equals 1.0 x M.  Less than 2 H 3 O +1 and OH -1 ions per billion water molecules. ◦ Since [H 3 O +1 ] = [OH -1 ], water is pH neutral.  Adding an acid or a base upsets this balance.

H O H H O H - +

Adding an Acid to Water H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O H2OH2OH2OH2OH 3 O +1 H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OOH -1 H2OH2OH2OH2OH2OH2OH2OH2OH2OH2O H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O HA A -1 H 3 O +1 A -1 H 3 O +1 A -1 H2OH2O H 3 O +1 A -1 H2OH2O

H 3 O +1 H2OH2O H2OH2O H2OH2O H2OH2OH2OH2O H2OH2O OH -1 Adding a Base to Water H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O H2OH2OOH -1 H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O A -1 H2OH2O OH -1 HA

As [H 3 O +1 ] Increases, [OH -1 ] Decreases 1x M 1x M 1x M 1x M 1x10 -9 M 1x10 -8 M 1x10 -7 M 1x10 -6 M 1x10 -5 M 1x10 -4 M 1x10 -3 M 1x10 -2 M 1x10 -1 M 1x M 1x M 1x M 1x M 1x10 -9 M 1x10 -8 M 1x10 -7 M 1x10 -6 M 1x10 -5 M 1x10 -4 M 1x10 -3 M 1x10 -2 M 1x10 -1 M [H 3 O +1 ] [OH -1 ] Neutral Solution Acid added to neutral solution Base added to neutral solution

 In any aqueous solution: ◦ [H 3 O +1 ] [OH -1 ] = 1x ◦ As [H 3 O +1 ] goes up, [OH -1 ] must decrease. ◦ As [OH -1 ] goes up, [H 3 O +1 ] must decrease.  In other words, adding an acid to water causes the solution to become more acidic and less basic.  Adding a base to water causes the solution to become less acidic and more basic.

 If [H 3 O +1 ] = 1x10 -3 M, what is [OH -1 ]? ◦ [H 3 O +1 ][OH -1 ] = 1x ◦ (1x10 -3 M)[OH -1 ] = 1x ◦ [OH -1 ] = (1x ) / (1x10 -3 ) ◦ [OH -1 ] = 1x M  If [OH -1 ] = 1x10 -8 M, what is [H 3 O +1 ]? ◦ [H 3 O +1 ][OH -1 ] = 1x ◦ [H 3 O +1 ](1x10 -8 M) = 1x ◦ [H 3 O +1 ] = (1x ) / (1x10 -8 M) ◦ [H 3 O +1 ] = 1x10 -6 M

 Acidity ◦ How much H 3 O +1 is dissolved in a sol’n.  Remember, acids increase [H 3 O +1 ] in solutions. ◦ Acidity = pH.

 pH = power of Hydrogen ◦ negative logarithmic (powers of ten) scale.  pH = -log 10 [H 3 O +1 ] ◦ If [H 3 O +1 ] = 1x10 -1 M,  pH = -log(1x10 -1 M) = 1 ◦ If [H 3 O +1 ] = 1x10 -2 M,  pH = -log(1x10 -2 M) = 2 ◦ If [H 3 O +1 ] = 1x10 -3 M,  pH = -log(1x10 -3 M) = 3

 The logarithm of a number is the power to which you would have to raise a base to equal that number. ◦ Unless otherwise indicated, assume the base is 10.  log(100) = 2 ◦ because 10 2 = 100  log(1000) = 3 ◦ because 10 3 = 1000  log(0.001) = -3 ◦ because =  log( ) = -6 ◦ because =

 The [H 3 O +1 ] and [OH -1 ] of an aqueous solution can vary by a very large degree. ◦ [H 3 O +1 ] = 1 M for a very acidic soln ◦ [H 3 O +1 ] = 1x10 -7 M for a neutral soln ◦ [H 3 O +1 ] = 1x M for a very basic soln  1 M is ten million times greater than 1x10 -7 M. ◦ If you tried to plot both concentrations on the same graph, 1x10 -7 M would barely register above zero. ◦ If 1x10 -7 M was 1 mm above the 0 mark, the axis would have to be ten kilometers (six miles) tall to show 1 M.  Logarithms allow us to compare numbers that are widely different by thinking of them as powers of ten.

This graph shows pH as a function of hydrogen ion concentration. It isn’t a very useful graph because it is hard to get accurate information for [H 3 O +1 ] below 1x10 -3 M.

In this graph the x-axis is logarithmic. It allows a much greater range of data to be displayed in a readable format.

 Each unit decrease in pH is a 10-fold increase in acidity.  Imagine a soln with a pH of 5. ◦ A soln with a pH of:  4 is 10 times more acidic.  3 is 100 times more acidic.  2 is 1000 times more acidic.  1 is 10,000 times more acidic.  0 is 100,000 times more acidic.

Where does the pH scale come from? pH scale -1… …15 Acidic Basic [H 3 O + ] A lot of H 3 O + Not a lot of H 3 O + Acidic Basic

pOH Scale The pOH scale indicates the hydroxide ion concentration, [OH-] or molarity, of a solution. (In other words how many OH- ions are in the solution. If there are a lot we assume it is a base, if there are very few it is an acid.) Two chemists meet for the first time at a symposium. One is American, one is British. The British chemist asks the American chemist, "So what do you do for research?" The American responds, "Oh, I work with aerosols." The British chemist responds, "Yes, sometimes my colleagues get on my nerves also."

pOH scale -1… …15 Basic Acidic [OH - ] A lot of OH - Not a lot of OH - Basic Acidic

Example: 1.Lemon juice (citric acid) pH = 2, pOH = 12 2.Pure water pH = 7, pOH = 7 3.Milk of magnesia pH = 10, pOH = 4 The last words of a chemist: 1.And now for the taste test. 2. I wonder if this is hot? 3. And now a little bit from this And now shake it a bit.

[H 3 O +1 ] [OH -1 ]

4. Calculations Involving pH, pOH, [H 3 O+], and [OH-] of strong Acids and Bases 1 st: determine which ion will be produced, either OH or H 3 O+ (Acids produce H 3 O+ and bases produce OH-). 2 nd: use formula to determine pH or pOH. 3 rd: check to see if answer is reasonable. pH = -log [H 3 O + ] pOH = -log [OH - ] pOH + pH = 14

 What are the pH values of the following solutions? ◦ 1x10 -1 M H 3 O +1  pH = -log(1x10 -1 M) = 1 ◦ 1x10 -3 M H 3 O +1  pH = -log(1x10 -3 M) = 3 ◦ 1x10 -5 M H 3 O +1  pH = -log(1x10 -5 M) = 5 ◦ 1x10 -1 M OH -1  [H 3 O +1 ] = (1x ) / (1x10 -1 M) = 1x M  pH = -log(1x M) = 13

 Given pH, you can calculate [H 3 O +1 ] and [OH -1 ].  [H 3 O +1 ] = 10 -pH  [H 3 O +1 ] [OH -1 ] = 1x  If pH = 2, ◦ [H 3 O +1 ] = 1x10 -2 M ◦ [OH -1 ] = 1x M

9/4/ Titrations Determining the pH of an unknown solution using the pH of a known solution Titrations take a very long time and you have to have excellent lab technique You add small amounts of known solution until a pre-determined endpoint is reached

#H + a M a V a = #OH - b M b V b #H + in acid formula M= Molarity V= Volume used to neutralize #OH - in base formula

Example You have 50 drops CH 3 COOH & it takes 5 drops 5M NaOH reach the endpoint. What is the molarity of the acetic acid? (1 H + )(M a )(50 drops )=(1 OH - )(5 M)(5 drops ) M a = 0.5 M

Example 2 What is the molarity of sulfuric acid if it takes 12 mL of H 2 SO 4 to neutralize 30 mL of 5 M NaOH. (2 H + )(M a )(12 drops )=(1 OH - )(5 M)(30 drops ) 6 M