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## pH Scale

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**pH Scale**Soren Sorensen (1868 - 1939)**[H+] pH**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) 7 Acid Base 0 14 Blood Pure wate Milk Acidic Neutral Basic Vinegar Lemon juice Stomach acid 1 M HCl pH Scale Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 515**pH of Common Substances**Timberlake, Chemistry 7th Edition, page 335**pH of Common Substance**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 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**Acid – Base Concentrations**10-1 pH = 3 pH = 11 OH- H3O+ 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 [H1+] Kelter, Carr, Scott, Chemistry A World of Choices 1999, page 285**pH Calculations**pH [H3O+] pH = -log[H3O+] [H3O+] = 10-pH pH + pOH = 14 [H3O+] [OH-] = 1 x10-14 pOH [OH-] pOH = -log[OH-] [OH-] = 10-pOH**10x**log 2nd antilog pH = - log [H+] Given: determine the [hydronium ion] pH = 4.6 choose proper equation pH = - log [H+] substitute pH value in equation 4.6 = - log [H+] multiply both sides by -1 - 4.6 = log [H+] take antilog of both sides - 4.6 = log [H+] [H+] = 2.51x10-5M Recall, [H+] = [H3O+] You can check your answer by working backwards. pH = - log [H+] pH = - log [2.51x10-5M] pH = 4.6**pH and pOH Calculations**pH and pOH Calculations pH and pOH Calculations http://www.unit5.org/christjs/tempT27dFields-Jeff/AcidBase1.htm**Acid Dissociation**pH = ? pH = - log [H+] H1+(aq) + A1-(aq) HA(aq) monoprotic 0.3 M 0.3 M 0.3 M pH = - log [3M] e.g. HCl, HNO3 pH = - 0.48 pH = - log [H+] 2 H1+(aq) + A2-(aq) H2A(aq) diprotic 0.3 M 0.6 M 0.3 M pH = - log [6M] e.g. H2SO4 pH = - 0.78 Given: pH = 2.1 3 H1+(aq) + PO43-(aq) polyprotic H3PO4(aq) e.g. H3PO4 ? M x M find [H3PO4] assume 100% dissociation**- 2.1 = log [H+]**2nd log 2nd log 7.94 x10-3M 3 Given: pH = 2.1 3 H1+(aq) + PO43-(aq) H3PO4(aq) XM 0.00794 M 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 M 7.94 x10-3M [H+] = 7.94 x10-3M Step 3) Calculate [H3PO4] concentration Note: coefficients (1:3) for (H3PO4 : H+) = 0.00265 M H3PO4**How many grams of magnesium hydroxide are needed to add to**500 mL of H2O to yield a pH of 10.0? Mg2+ OH1- Step 1) Write out the dissociation of magnesium hydroxide Mg(OH)2 Mg(OH)2(aq) Mg2+(aq) + 2 OH1-(aq) 0.5 x10-4 M 5 x10-5 M 1 x10-4 M 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 M Step 4) Solve for moles of Mg(OH)2 x = 2.5 x 10-5 mol Mg(OH)2 Step 5) Solve for grams of Mg(OH)2**Practice Problems - Key**Practice Problems - Answer Key Practice Problems - Answer Key http://www.unit5.org/christjs/tempT27dFields-Jeff/AcidBase1.htm