Acids and Bases

What do lemon juice, lime juice, sauerkraut, vinegar and sour cream have in common?

Acids colorless Sour Electrolytes Change the color of some indicators litmus paper changes from blue to red phenolphthalein is React with metals to produce hydrogen React with bases to produce water and a salt colorless

Naming acids Binary Ternary two elements hydro-element-ic ex: hydrochloric  HCl hydrosulfuric  H2S Ternary three or more elements oxyacids  polyatomic ions ex: sulfuric  H2SO4 chloric  HClO3

Bases pink Bases have a bitter taste. Bases are slippery. Bases are electrolytes. Bases change the color of some indicators: litmus paper changes from red to blue phenolphthalein is Bases react with acids to form water and a salt. An example of a base is soap Bases are called alkaline Demo: place 10 drops of 1M HCl in 50 mL water in 600mL beaker on overhead. Add 20 drops universal. Drop in one Phillips antacid tablet. Stir slowly for 2 mins. As the pH increases the color changes from orange to blue. pink

Naming Bases Name positive ion – hydroxide ex: NaOH  sodium hydroxide NH4OH  ammonium hydroxide

Bases Families I and II on the periodic table readily form bases when mixed with water. Alkali/alkaline means basic Family I/II + H2O  metal(OH) + H2 Na + H2O  NaOH + H2 Note the base quickly dissociates.

Naming acids and bases Stop and practice by doing worksheet 7

Acid/Base/Salt Acid – hydrogen ions Base –hydroxide ions Salt – metal-nonmetal ionic compound

Bronsted-Lowry acid/base Includes all Arrhenius and more. Acid = hydrogen donor It gives away hydrogen ions Base = hydrogen acceptor It takes hydrogen

Bronsted-Lowry acid/base NH3 + H2O  NH4+ + OH- Base Acid Although we are mixing an acid and a base the solution is not neutral. It is basic, since there is more –OH- on the product side.

Bronsted-Lowry acid/base HF + H2O  H3O+ + F- Acid Base Although we are mixing an acid and a base the solution is not neutral. It is acidic, since there is “extra” H3O+ on the product side.

Bronsted-Lowry acid/base Notice the reverse of this equation. HF + H2O  H3O+ + F- Acid Base In the reverse equation the acid and base changes places, this is called conjugate acid and conjugate base pairs.

Bronsted-Lowry acid/base HF + H2O  H3O+ + F- Acid Base conj. Acid conj. Base Conjugate acid – formed when a base acquires a proton H+ to form the acid. Conjugate base – the particle that remains after the proton has been released by the acid.

Bronsted-Lowry acid/base Identify the acid, base, conj. acid, conj. base for the following reaction. HNO3 + NaOH  NaNO3 + H2O

Bronsted-Lowry acid/base Identify the acid, base, conj. acid, conj. base for the following reaction. NaHCO3 + HCl  NaCl + H2CO3

Amphoteric A substance that can be both an acid and a base. Water has a split personality. It can be both an acid and base.

Summary of Acid/Base Theories Bases Arrhenius releases H+ ions in water. releases OH- ions in water. Bronsted-Lowry donates a proton (H+) accepts protons. (H+) Lewis accepts an electron pair donates an electron pair

Summary of Acid/Base Theories

Hydrogen Ions from Water At times water collides with so much force that one water molecule loses a H+ to another water molecule – this is called self-ionization .. .. .. + .. - H – O : + H – O :  H – O – H + H – O : l l l .. H H H

Hydrogen Ions from Water Simplified Water decomposes into hydrogen ion and hydroxide. .. + .. - H – O :  H + H – O : l .. H In class you may use either interchangeably. You need to recognize both. In reality there is so much water that H+ never occurs it is always H3O+ (hydronium)

Pure water In pure water the [OH-] = [H3O+] Neutral - any aqueous solution where [OH-] = [H3O+]. [H3O+] = 1.0 E -7 mol/L In class you may use either interchangeably. You need to recognize both.

Pure water [OH-] and [H3O+] are inversely related. Therefore, [OH-] [H3O+] = Kw Kw = 1.0E-14 (mol/L)2 Ion-product constant for water In class you may use either interchangeably. You need to recognize both.

Acidic solutions HCl + H2O  H+ + Cl- + H+ + OH- 2H+ and 1 OH- Remember there is so much water in an aqueous solution that the H+s are really H3O+. Therefore, [H3O+] > [OH-] In class you may use either interchangeably. You need to recognize both.

[OH-] vs. [H3O+] [H3O+] > [OH-]  acid [H3O+] < [OH-]  basic Ex: NaOH + H2O  Na+ + OH- + H+ + OH- [H3O+] = [OH-]  neutral In class you may use either interchangeably. You need to recognize both.

[H3O+] vs. [OH-] practice problems Calculate [OH-] if: [H3O+] = 1E-1 M [H3O+] = 1E-12 M Calculate [H3O+] if: [OH-] = 1E-3 M [OH-] = 1E-10 M In class you may use either interchangeably. You need to recognize both.

How strong is your acid?

How strong is your acid? A strong acid completely ionizes (dissociates) in aqueous solution. HCl water H+ + Cl-

How strong is your acid? A weak acid slightly ionizes (dissociates) in aqueous solution. HC2H3O2 water H+ + C2H3O2-

pH and pOH

pH and pOH

pH Indicates the hydronium ion concentration. The negative logarithm of the hydronium ion concnetration. pH = - log [H3O+]

pH scale Always use pH to determine acid/base/neutral 0 1 2 3 4 5 6 7 8 9 1011 12 13 14 strong acid strong base neutral

pH What is the pH of pure water? U: pH K: [H3O+] = 1.0E-7M P: pH = - log [H3O+] S: pH = - log [1.0E-7M] = 7

pH Calculate the pH of a solution with a hydronium ion concentration of 5.0E-6M? U: pH K: [H3O+] = 5.0E-6M P: pH = - log [H3O+] S: pH = - log [5.0E-6M] 5.30

pOH Indicates the hydroxide ion concentration. The negative logarithm of the hydroxide ion concnetration. pOH = - log [OH-] never use pOH to determine acid/base/neutral!! (ZERO CREDIT!!)

Equations Kw = 1.0E-14 = [H3O+][OH-] pH = -log[H3O+] pOH = -log[OH-] pH + pOH = 14

pH and pOH problems Determine the pH and pOH of a solution which contains 0.0035 mole of H3O+ per liter. Is this an acid/base or neutral solution? P: pH = -log [H3O+] ; pH + pOH = 14 S: pH = - log [0.0035M] = 2.46 pOH = 14 – 2.46 = 11.54 acid

pH and pOH Problems What is the pOH of a solution containing 0.042M KOH? Is this a/b/n? since this is a strong base  100% ionizes Therefore: KOH  K+ + OH- before 4.2E-2 0 0 after 0 4.2E-2M 4.2E-2M

pH and pOH Problems (cont’d) What is the pOH of a solution containing 0.042M KOH? Is this a/b/n? u: pOH k: [OH-] = 4.2E-2M p: pOH= -log [OH-] s: pOH = -log [4.2E-2M] pOH = 1.38 pH = 14 – 1.38 = 12.62  base

pH and pOH Problems What is the pH of a solution containing 0.035M H2SO4? Is this a/b/n? since this is a strong acid  100% ionizes Therefore: H2SO4  2H+ + SO4-2 before 3.5E-2 0 0 after 0 7.0E-2M 3.5E-2M

pH and pOH Problems (cont’d) What is the pH of a solution containing 0.035M H2SO4? Is this a/b/n? u: pH k: [H+] = 7.0E-2M p: pH= -log [H+] s: pH = -log [7.0E-2M] pH = 1.15  acid

pH and pOH Problems Calculate the [H3O+] of a solution that has a pH of 3.70. u: [H3O+] K: pH = 3.70 p: pH = -log [H3O+]  10-pH s: [H3O+] = 10-3.70 = 2.0E-4M

Titration

Titration an experimental procedure in which the unknown concentration of a known volume of solution is determined by measuring the volume of a solution of known concentration required to react completely with it.

Titration an analytical method in which a standard solution is used to determine the concentration of another solution. Standard solution – one of known concentration. End point – the point at which neutralization is achieved.

Steps for Titration A measured amount of an acid/base of unknown concentration is added to a flask. An appropriate indicator (phenolphthalein, BB, or methyl orange) is added to the solution. A measured amount of base/acid (known concentration) is mixed into the first solution.

Titration Experiment Preview A ring stand, a clamp, a burette, a 250mL flask, and a white sheet of paper are required materials The materials should be assembled as the picture shows.

Titration Experiment Preview Condition the burette. Add a small amount of NaOH to the burette (about 5 mL). Place the burette almost horizontal and turn. Empty contents through stopcock into sink.

Titration Experiment Preview Be sure the stopcock of the burette is closed! Using the funnel/stirring rod at the top, fill the burette with the NaOH solution. In the flask place the measured HCl solution, and 5 drops of the indicator.

Titration Experiment Preview Begin the titration by opening the stopcock to make the NaOH solution flow dropwise into the flask. Be sure to swirl the flask constantly so the solutions mix thoroughly.

Titration Experiment Preview Continue to add NaOH solution to the flask until a color change begins (colorless to pink) At this point, slow the flow even more. (It is over titrated when the solution turns fuchsia.)

Titration curve Graphical depiction of pH of solution as it is neutralized by a given volume of titrate added.

Titration curve: strong vs. strong

Titration curve: strong acid vs. weak base

Titration curve: weak acid vs. strong base

Titration curve: weak vs. weak

Titration curve summary

Titration problem MaVa = MbVb

Titration problem sample 1 In a laboratory experiment, 20.0 mL of NH3(aq) solution is titrated to the methyl orange endpoint using 15.65mL of a 0.200M HCl solution. What is the concentration of the aqueous solution?

Titration problem sample 1 U:Mb K: vHCl=15.65mL = 0.01565L Ma = 0.200M HCl Vb = 20.0mL = 0.020L P: MaVa = MbVb S: (0.200M)(0.01565L) = (Mb)(0.020L) 0.157M = MNH3