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pH Scale Soren Sorensen (1868 - 1939) pH Scale Acid Base 0 7 14 Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 515 [H + ] pH 10 -14 14 10.

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Presentation on theme: "pH Scale Soren Sorensen (1868 - 1939) pH Scale Acid Base 0 7 14 Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 515 [H + ] pH 10 -14 14 10."— Presentation transcript:

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2 pH Scale Soren Sorensen ( )

3 pH Scale Acid Base Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 515 [H + ] pH M NaOH Ammonia (household cleaner) Blood Pure water Milk Vinegar Lemon juice Stomach acid 1 M HCl Acidic Neutral Basic

4 pH of Common Substances Timberlake, Chemistry 7 th Edition, page M HCl 0 gastric juice 1.6 vinegar 2.8 carbonated beverage 3.0 orange 3.5 apple juice 3.8 tomato 4.2 lemon juice 2.2 coffee 5.0 bread 5.5 soil 5.5 potato 5.8 urine 6.0 milk 6.4 water (pure) 7.0 drinking water 7.2 blood 7.4 detergents bile 8.0 seawater 8.5 milk of magnesia 10.5 ammonia 11.0 bleach M NaOH (lye) acidic neutral basic [H + ] = [OH - ]

5 pH of Common Substance 14 1 x x x x x x x x x x x x x x x x x x x x x x x x x x x x 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 More acidic pH [H 1+ ] [OH 1- ] pOH 7 1 x x

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

7 pH pH = -log [H 1+ ] Kelter, Carr, Scott, Chemistry A World of Choices 1999, page 285

8 pH Calculations pH pOH [H 3 O + ] [OH - ] pH + pOH = 14 pH = -log[H 3 O + ] [H 3 O + ] = 10 -pH pOH = -log[OH - ] [OH - ] = 10 -pOH [H 3 O + ] [OH - ] = 1 x10 -14

9 pH = - log [H + ] pH = 4.6 pH = - log [H + ] 4.6 = - log [H + ] = log [H + ] Given: 2 nd log 10 x antilog multiply both sides by -1 substitute pH value in equation take antilog of both sides determine the [hydronium ion] choose proper equation [H + ] = 2.51x10 -5 M You can check your answer by working backwards. pH = - log [H + ] pH = - log [2.51x10 -5 M] pH = 4.6 Recall, [H + ] = [H 3 O + ]

10 pH and pOH Calculations Keys pH and pOH Calculations

11 Acid Dissociation monoprotic diprotic polyprotic HA(aq) H 1+ (aq) + A 1- (aq) 0.03 M pH = - log [H + ] pH = - log [0.03M] pH = 1.52 e.g. HCl, HNO 3 H 2 A(aq) 2 H 1+ (aq) + A 2- (aq) 0.3 M0.6 M0.3 M pH = - log [H + ] pH = - log [0.6M] pH = 0.22 e.g. H 2 SO 4 Given: pH = 2.1 find [H 3 PO 4 ] assume 100% dissociation e.g. H 3 PO 4 H 3 PO 4 (aq) 3 H 1+ (aq) + PO 4 3- (aq) ? Mx M pH = ?

12 Given: pH = 2.1 find [H 3 PO 4 ] assume 100% dissociation H 3 PO 4 (aq) 3 H 1+ (aq) + PO 4 3- (aq) X MX M M Step 1) Write the dissociation of phosphoric acid Step 2) Calculate the [H + ] concentration pH = - log [H + ] 2.1 = - log [H + ] = log [H + ] 2 nd log = log [H + ] 2 nd log [H + ] = 10 -pH [H + ] = [H + ] = M [H + ] = 7.94 x10 -3 M 7.94 x10 -3 M Step 3) Calculate [H 3 PO 4 ] concentration Note: coefficients (1:3) for (H 3 PO 4 : H + ) 7.94 x10 -3 M 3 = M H 3 PO 4

13 How many grams of magnesium hydroxide are needed to add to 500 mL of H 2 O to yield a pH of 10.0? Step 1) Write out the dissociation of magnesium hydroxide Mg 2+ OH 1- Mg(OH) 2 Mg(OH) 2 (aq)Mg 2+ (aq) 2 OH 1- (aq)+ Step 2) Calculate the pOH pH + pOH = pOH = 14 pOH = 4.0 Step 3) Calculate the [OH 1- ] pOH = - log [OH 1- ] [OH 1- ] = 10 -OH [OH 1- ] = 1 x10 -4 M 1 x10 -4 M0.5 x10 -4 M5 x10 -5 M Step 4) Solve for moles of Mg(OH) 2 x = 2.5 x mol Mg(OH) 2 Step 5) Solve for grams of Mg(OH) 2 x g Mg(OH) 2 = 2.5 x mol Mg(OH) 2 1 mol Mg(OH) 2 = g Mg(OH) 2 58 g Mg(OH) 2

14 Practice Problems - Key KeysKeys Practice Problems - Answer Key


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