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Acid-Base Titration & pH Chapter 15B

Acid-Base Titration & pH Chapter 15B. West Valley High School General Chemistry Mr. Mata. Acid Rain and Stream Water pH. Acid rain occurs when SOx and NOx compounds from air pollution react with water to produce acids. H 2 SO 4 ( sulfuric acid ) found in car batteries .

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Acid-Base Titration & pH Chapter 15B

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  1. Acid-Base Titration & pHChapter 15B West Valley High School General Chemistry Mr. Mata

  2. Acid Rain and Stream Water pH • Acid rain occurs when SOx and NOx compounds from air pollution react with water to produce acids. • H2SO4 (sulfuric acid) found in car batteries. • HNO3 (nitric acid) used to clean glass and stone of the most stubborn stains. • The worst reported case of acid rain was in West Virginia • pH measured at pH of 1.5, slightly weaker than stomach acid!

  3. Effects of Acid Rain on structures • The high acidity dissolves the calcium compounds use in bridges, building mortar, pavement. • Weakens the material and puts the structural integrity of the buildings at risk. • Marble and granite used in very old statues is at a greater risk. • Absence of modern compounds used to reduce mineral dissolution. • Thin layers of automotive paint suffer substantial damage from acid rain with only small amount of exposure.

  4. Effects of Acid Rain on Humans • Acidic water vapor in the air we breathe causes inflammation in the lungs and deterioration of lung tissue. • The effect can be fatal for those at risk for respiratory complications. • Acidic particles in the air react to produce smog. • Photochemical smog – a haze that reflects or blocks light like a strong mist on the horizon.

  5. Effects on Forest Ecosystems • The reduced pH of soils: • reduces ability of trees and underbrush to grow. • As pH decreases in aqueous systems: • metals (positive ions) are dissolved more readily and bound up in complex molecules. • As the soils grow more acidic: • valuable minerals are dissolved and bound up. • made unavailable to plants that draw minerals from soils through their roots.

  6. Effects of Acid Rain on Stream water • Acid rain is most devastating to aquatic life. • Small change in our blood pH threatens our lives, • Small changes in stream water pH threatens aquatic life. • Some metal ions will not dissolve into ions above a certain pH. Ex: Aluminum pH is about 5.0. • When pH of a lake drops below this pH, Al lying at the bottom of the lake goes into solution as ions. • Those dissolved ions, as well as other metals acts in two ways to kill aquatic life…

  7. Effects of Acid Rain on Stream water • Elevated levels of Al ions are toxic to many species of aquatic life. • Water entering the gills of lake fish are elevated to a pH of about 7 in the gills. • When the ion-containing acidic water enters the higher-pH gills, the metal ions come out of solution, forming a film on the gills that suffocates the fish.

  8. Reduction of Acid Rain • Acid rain is on the decline globally. • Coal-burning power plants and industrial manufacturing plants must have installed elaborate “scrubbers” to absorbSOxbefore it can be released. • Due to costs of pollutant-reducing equipment, developing nations still release large amounts of SOx pollutants. • Further, basic compounds and acid-base buffers are being added to main lake waters to protect aquatic wildlife.

  9. Titration Sample Problem #1 • Calculate the concentration (molarity) of an HCl solution determined by titration to which 50 mL of the acid is added to a flask. Indicator is added, and 29.6 mL of 0.967 M NaOH are required to reach the end point. MaVa = MbVb -> Ma = MbVb Va Ma = ? M Va = 50 mL Mb = 0.967 M Vb = 29.6 mL Ma = Mb Vb = (0.967 M) (29.6 mL) = 0.57 M Va 50 mL

  10. Titration Sample Problem #2 • How much 3.5 M NaOH is required to neutralize 75 mL of 12 M HCl? MaVa = MbVb -> Vb = MaVa Mb Ma = 12 M Va = 75 mL Mb = 3.5 M Vb = ? mL Vb = Ma Va = (12 M) (75 mL) = 257.1 mL Mb 3.5 M

  11. Titration Sample Problem #3 • An 80 mL sample of KOH was titrated to the endpoint with 39 mL of 6.5 M HCl. What was the molarity of the KOH? MaVa = MbVb -> Mb = MaVa Vb Ma = 6.5 M Va = 39 mL Mb = ? M Vb = 80 mL Mb = Ma Va = (6.5 M) (39 mL) = 3.2 M Vb 80 mL

  12. Titration Sample Problem #4 • How much 2.8 M Ca(OH)2 is necessary to titrate 44 mL of 0.75 M H2SO4 solution to the endpoint? MaVa = MbVb -> Vb = MaVa Mb Ma = 0.75 M Va = 44 mL Mb = 2.8 M Vb = ? mL Vb = Ma Va = (0.75 M) (44 mL) = 11.8 mL Mb 2.8 M

  13. Chapter 15 SUTW Prompt • Describe how titration and indicators are used to find the molarity of unknown solutions. • Complete a 8 -10 sentence paragraph using the SUTW paragraph format. Hilight using green, yellow, and pink. • Due Date: Monday, February 6, 2017 (start of class).

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