pH Scale Soren Sorensen (1868 - 1939) The pH scale was invented by the Danish chemist Soren Sorensen for a brewery to measure the acidity of beer. Soren Sorensen (1868 - 1939)

pH Scale 7 Acid Base 14 Acidic Neutral Basic [H+] pH 10-14 14 10-13 13 10-14 14 10-13 13 10-12 12 10-11 11 10-10 10 10-9 9 10-8 8 10-7 7 10-6 6 10-5 5 10-4 4 10-3 3 10-2 2 10-1 1 100 0 1 M NaOH Ammonia (household cleaner) Blood Pure wate Milk Vinegar Lemon juice Stomach acid 1 M HCl Acidic Neutral Basic Acid Base 7 14 pH Scale “S.P.L. Sorensen (1868-1939) introduced the pH scale to measure the concentration of hydrogen ions in solution. The more hydrogen ions, the stronger the acid. The amount of hydrogen ions in solution can affect the color of certain dyes found in nature. These dyes can be used as indicators to test for acids and alkalis. An indicator such as litmus (obtained from lichen) is red in acid. If base is slowly added, the litmus will turn blue when the acid has been neutralized, at about 6-7 on the pH scale. Other indicators will change color at different pH’s. A combination of indicators is used to make a universal indicator.” - Eyewitness Science “Chemistry” , Dr. Ann Newmark, DK Publishing, Inc., 1993, pg 42 Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 515

pH of Common Substances Timberlake, Chemistry 7th Edition, page 335

pH of Common Substance 14 1 x 10-14 1 x 10-0 0 13 1 x 10-13 1 x 10-1 1 pH [H1+] [OH1-] pOH 14 1 x 10-14 1 x 10-0 0 13 1 x 10-13 1 x 10-1 1 12 1 x 10-12 1 x 10-2 2 11 1 x 10-11 1 x 10-3 3 10 1 x 10-10 1 x 10-4 4 9 1 x 10-9 1 x 10-5 5 8 1 x 10-8 1 x 10-6 6 6 1 x 10-6 1 x 10-8 8 5 1 x 10-5 1 x 10-9 9 4 1 x 10-4 1 x 10-10 10 3 1 x 10-3 1 x 10-11 11 2 1 x 10-2 1 x 10-12 12 1 1 x 10-1 1 x 10-13 13 0 1 x 100 1 x 10-14 14 Strong Base NaOH, 0.1 M Household bleach Household ammonia Lime water Milk of magnesia Borax Baking soda Egg white, seawater Human blood, tears Milk Saliva Rain Black coffee Banana Tomatoes Wine Cola, vinegar Lemon juice Gastric juice More basic 7 1 x 10-7 1 x 10-7 7 More acidic Strong Acid

Acid – Base Concentrations 10-1 pH = 3 pH = 11 H3O+ OH- pH = 7 10-7 concentration (moles/L) H3O+ OH- OH- H3O+ 10-14 [H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-] acidic solution neutral solution basic solution Timberlake, Chemistry 7th Edition, page 332

pH pH = -log [H+] Kelter, Carr, Scott, Chemistry A World of Choices 1999, page 285

pH Calculations pH pOH [H3O+] [OH-] pH = -log[H3O+] [H3O+] = 10-pH [H3O+] [OH-] = 1 x10-14 pOH [OH-] pOH = -log[OH-] [OH-] = 10-pOH

pH = - log [H+] Given: pH = 4.6 determine the [hydronium ion] choose proper equation 4.6 = - log [H+] substitute pH value in equation - 4.6 = log [H+] multiply both sides by -1 2nd log - 4.6 = log [H+] take antilog of both sides [H+] = 2.51x10-5 mol/L 10x antilog Recall, [H+] = [H3O+] You can check your answer by working backwards. pH = - log [H+] pH = - log [2.51x10-5 mol/L] pH = 4.6

Acid Dissociation pH = ? pH = - log [H+] monoprotic HA(aq) H1+(aq) + A1-(aq) 0.3 mol/L 0.3 mol/L 0.3 mol/L pH = - log [3 mol/L] e.g. HCl, HNO3 pH = - 0.48 pH = - log [H+] diprotic H2A(aq) 2 H1+(aq) + A2-(aq) 0.3 mol/L 0.6 mol/L 0.3 mol/L pH = - log [6 mol/L] e.g. H2SO4 pH = - 0.78 polyprotic 3 H1+(aq) + PO43-(aq) Given: pH = 2.1 H3PO4(aq) e.g. H3PO4 ? mol/L x mol/L find [H3PO4] assume 100% dissociation

3 H1+(aq) + PO43-(aq) Given: pH = 2.1 H3PO4(aq) X mol/L 0.00794 mol/L find [H3PO4] assume 100% dissociation Step 1) Write the dissociation of phosphoric acid Step 2) Calculate the [H+] concentration pH = - log [H+] [H+] = 10-pH 2.1 = - log [H+] [H+] = 10-2.1 - 2.1 = log [H+] [H+] = 0.00794 mol/L 2nd log - 2.1 = log [H+] 7.94 x10-3 mol/L [H+] = 7.94 x10-3 mol/L Step 3) Calculate [H3PO4] concentration Note: coefficients (1:3) for (H3PO4 : H+) 7.94 x10-3 M 3 = 0.00265 mol/L H3PO4

How many grams of magnesium hydroxide are needed to add to 500 mL of H2O to yield a pH of 10.0? Mg2+ Step 1) Write out the dissociation of magnesium hydroxide OH1- Mg(OH)2 Mg(OH)2 (aq) Mg2+ (aq) + 2 OH1-(aq) 0.5 x10-4 mol/L 1 x10-4 mol/L pH + pOH = 14 Step 2) Calculate the pOH 10.0 + pOH = 14 pOH = 4.0 pOH = - log [OH1-] Step 3) Calculate the [OH1-] [OH1-] = 10-OH [OH1-] = 1 x10-4 mol/L Step 4) Solve for moles of Mg(OH)2 n = C V n = 5 × 10-5mol/1 × 0.5 L n = 5 x 10-5 mol Mg(OH)2 Step 5) Solve for grams of Mg(OH)2