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Chapter 16.7 and 16.8

Chapter 16.7 and 16.8. Acid –Base Titrations Buffered Solutions. 16.7 Acid – Base Titrations. Strong acids contain H + Strong bases contain OH - H + + OH -  H 2 O Neutralization reaction = equal amounts of H+ and OH- results in a neutral solution (pH=7).

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Chapter 16.7 and 16.8

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  1. Chapter 16.7 and 16.8 Acid –Base Titrations Buffered Solutions

  2. 16.7 Acid – Base Titrations • Strong acids contain H+ • Strong bases contain OH- • H+ + OH-  H2O • Neutralization reaction = equal amounts of H+ and OH- results in a neutral solution (pH=7). • Titration = involves the delivery of a measured volume of a solution of known concentration (titrant) into the solution being analyzed (analyte)

  3. Titrations • Standard solution = solution of known concentration • Buret = cylindrical device; allows for a measured amount of liquid to be dispensed. • Stochiometry point or equivalence point =Keep adding titrant until all of analyte reacts with titrant, when pH = 7

  4. Titration Curve or pH curve • A titration curve is a plot of the pH of a solution (acid or base) against the volume of titrant added (base or acid).

  5. Curve for the titration of a strong acid by a strong base. Figure 17.11

  6. Titration problems • What volume of .1 M NaOH is needed to titrate (neutralize) 50.0 mL of .2 M HNO3? • We need moles of OH- added to equal moles of H+ present. • M = m / L • Ma x Va = ma and Mb x Vb = mb • So, Ma x Va (mb/ma) = Mb x Vb • mb/ma (mole to mole ratio)

  7. What volume of .1 M NaOH is needed to titrate (neutralize) 50.0 mL of .2 M HNO3? • NaOH + HNO3  NaNO3 + HOH • MaVa (b/a) = MbVb • (.2 M)(50. mL)(1/1) = (.1 M)(V) • V=100 mL of NaOH

  8. Problem • It takes 25mL of H2SO4 to titrtate 30mL of 3 M NaOH. Calculate the molarity of H2SO4. • H2SO4 +2NaOH  Na2SO4 + 2 HOH • MaVa (b/a) = MbVb • Ma (.026L) (2/1) = (3)(.030L) • Ma = 1.7

  9. Problem • When 14.3 mL of 0.573 M Ca(OH)2 are neutralized with 0.426 M HNO2 what volume of acid would be used? • Ca(OH)2 + 2 HNO2  Ca(NO3)2 + 2HOH • MaVa (b/a) = MbVb • (.426)(V) (1/2) = (.573)(14.3 mL) • V acid = 66.67 mL

  10. Buffers Blood is a buffer. Blood’s pH ranges between 7.35 – 7.45. If the pH of blood drops below 6.8 or is higher than 7.8, you can not survive. CO2 + H2O  H2CO3  H+ + HCO3-

  11. 16.8 Buffers • A buffer is a solution characterized by the ability to resist large changes in pH when limited amounts of acid or base are added to it. • Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. • Thus, a buffer contains both an acid species and a base species in equilibrium.

  12. Buffers • Buffering capacity = the amount of H+ or OH- that a buffer system can absorb with out significant changes in pH. • Aspirin is an acid. When someone overdoses, the body can not neutralize the aspirin.

  13. Buffers

  14. Buffers • If you add an acid: H+ + HCO3- H2CO3 (pop) (blood) If you add a base: OH- + H2CO3  H2O + HCO3-

  15. Buffers • Blood buffers: H2CO3 /HCO3- and H2PO4-/HPO4-2 • Kidneys: Excrete acidic/basic solutions and back up the neutralization system • Breathing rate: controls amount of CO2 in body. CO2 + H2O  H2CO3 (carbonic acid) • Acidosis or alkalosis – can cause death.

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