Acid Base Titrations - PowerPoint PPT Presentation

1 / 16
About This Presentation
Title:

Acid Base Titrations

Description:

Neutralization reaction = equal amounts of H and OH- results in a neutral solution (pH=7) ... CO2 H2O H2CO3 (carbonic acid) Acidosis or alkalosis can cause death. ... – PowerPoint PPT presentation

Number of Views:92
Avg rating:3.0/5.0
Slides: 17
Provided by: VHHS
Category:

less

Transcript and Presenter's Notes

Title: Acid Base Titrations


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 (pH7).
  • 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
(No Transcript)
5
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).

6
Figure 17.11
Curve for the titration of a strong acid by
a strong base.
7
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)

8
  • 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)
  • V100 mL of NaOH

9
Problem
  • It takes 25mL of H2SO4 to titrtate 30mL of 3 M
    NaOH. Calculate the molarity of H2SO4.
  • H2SO4 2 NaOH ? Na2SO4 2 HOH
  • MaVa (b/a) MbVb
  • Ma (.026L) (2/1) (3)(.030L)
  • Ma 1.7

10
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

11
Buffers
Blood is a buffer. Bloods 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-
12
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.

13
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.

14
Buffers
15
Buffers
  • If you add an acid
  • H HCO3- ? H2CO3
  • (pop) (blood)
  • If you add a base
  • OH- H2CO3 ? H2O HCO3-

16
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.
Write a Comment
User Comments (0)
About PowerShow.com