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![Page 1: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/1.jpg)
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Introduction to Acids and BasesIntroduction to Acids and Bases
Pure or distilled water undergoes a veryslight ionization as shown below.
H2O(l) H+(aq) + OH-(aq)
The equilibrium constant for water is givenby:
Kw = [H+][OH-] = 1.00 x 10-14
![Page 2: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/2.jpg)
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Note that the product of [H+] and [OH-] is aconstant at 25°C and that [H+] and [OH-]
areinversely proportional to each other.
The pH scale has been devised to determine
the molarity of the hydrogen ion, [H+], in anaqueous solution and is given by:
pH = -log[H+]
Similarly, pOH is defined as:
pOH = -log[OH-]
![Page 3: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/3.jpg)
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Another useful equation is:
pH + pOH = 14.00
It follows that:
[H+] = 10-pH and [OH-] = 10-pOH
![Page 4: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/4.jpg)
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Acidic, basic, or neutral solutions can bedistinguished as shown below:
Solution pH [H+] pOH [OH-]
Acidic < 7.00 > 10-7 > 7.00 < 10-7
Neutral = 7.00 = 10-7 = 7.00 = 10-7
Basic > 7.00 < 10-7 < 7.00 > 10-7
![Page 5: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/5.jpg)
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An alternative approach to the relationshipbetween pH and pOH is shown below.
pOH
14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
pH
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Increasing Acidity Decreasing Acidity
Decreasing AcidityIncreasing AcidityN
eu
tral
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Strong AcidsStrong Acids
There are six strong acids.
HCl, HNO3, HClO4, HI, HBr, and H2SO4
When given any of the above acids
always assume 100% ionization.
100% ionization is shown by using a single arrow in the ionization equation.
The ionization equation for H2SO4 is not
intuitively obvious to the most casual observer.
![Page 7: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/7.jpg)
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All the strong acids are assumed to have
one ionizable hydrogen including H2SO4.
The ionization of H2SO4 must be shown as:
H2SO4(aq) → H+(aq) + HSO4-(aq)
The single arrow as always indicates that the ionization is 100% and there
are no H2SO4 molecules remaining in solution.
![Page 8: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/8.jpg)
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Strong Acid ProblemStrong Acid Problem
Determine the pH and pOH of a 0.20 M HClsolution.
[HCl] = 0.20 M
HCl(aq) → H+(aq) + Cl-(aq)
pH = -log[H+] = -log(0.20) = 0.70
pH + pOH = 14.00
pOH = 14.00 – 0.70 = 13.30
![Page 9: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/9.jpg)
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Another Strong Acid ProblemAnother Strong Acid Problem
Determine the pH and pOH of a 10-10 M HNO3
solution.
[HNO3] = 10-10 M
HNO3(aq) → H+(aq) + NO3-(aq)
pH = -log[H+] = -log(10-10) = 10
What??
![Page 10: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/10.jpg)
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How can a strong acid have a pHcorresponding to a base?
The answer lies with the autoionization ofwater.
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
[ ]I 0 0[ ]c +1.0 x 10-7 +1.0 x
10-7
[ ]e 1.0 x 10-7 1.0 x 10-
7
![Page 11: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/11.jpg)
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Although water ionizes only to a slight extent,there is a dynamic equilibrium which remainsintact.
Distilled or pure water has a pH = 7 becausethe [H+] = [OH-] = 1.0 x 10-7 M.
[H+]T = [H+]water + [H+]nitric acid
[H+]T = 1.0 x 10-7 M + 10-10 M ≈ 1.0 x 10-7 M
pH = -log[H+] = -log(1.0 x 10-7) = 7.00
![Page 12: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/12.jpg)
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Weak Acids Equilibrium ConstantWeak Acids Equilibrium Constant
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
[ ]I 0 0[ ]c +1.0 x 10-7 +1.0 x 10-7
[ ]e 1.0 x 10-7 1.0 x 10-7
Keq = [H+] [OH-]
Kw = Keq = [H+] [OH-] = 1.00 × 10-14
![Page 13: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/13.jpg)
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For the reaction
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
remember the following:
This is an example of a heterogeneous equilibria because more than one
phase (state) is present.
When the reactants/products are solids or liquids, their concentration are
nearly constant and incorporated into the Keq.
![Page 14: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/14.jpg)
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Pure solids and liquids are not incorporated into the equilibrium expression.
The position of the equilibrium is independent of the amount of solid or liquid present.
The solvent for the reaction does not appear in the equilibrium expression.
Ions or molecules appear as their molar
concentrations.
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The effect of temperature on the Keq
depends on which reaction is endothermic.
The inverse relationship between H+ and
OH- in any solution is given by Kw.
Remember the numerical value for Kw is
both reaction and temperature dependent.
![Page 16: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/16.jpg)
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A Weak Acid ProblemA Weak Acid Problem
Acetic acid, HC2H3O2, has a Ka = 1.8 x 10-5.Calculate the hydrogen ion concentration in asolution prepared by adding 2.0 moles ofacetic acid to form one liter of solution.
Ka = 1.8 x 10-5 n = 2.0 mol HC2H3O2 V = 1.0 L
[HC2H3O2] =
[HC2H3O2] =
nV
2.0 mol HC2H3O2
1.0 L= 2.0 M
![Page 17: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/17.jpg)
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HC2H3O2(aq) H+(aq) + C2H3O2-(aq)
[ ]I 2.0 0 0[ ]c -x +x +x[ ]e 2.0 - x x x
Ka =[H+] [C2H3O2
-][HC2H3O2]
1.8 x 10-5 =2.0 - xx x×
![Page 18: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/18.jpg)
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At this point in the problem an approximation
will be made called the 5% rule.
You will note that Ka = 1.8 x 10-5 is a very
small number and that x may be very small compared to the initial concentration of the acid.
This assumption will not always be true and should always be tested after solving for x which is the [H+].
![Page 19: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/19.jpg)
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The approximation made will be 2.0 – x ≈ 2.0
which avoids using the quadratic formula.
x = [H+] = 6.0 x 10-3 M
The 5% rule is given by:
1.8 x 10-5 =2.0 - xx x× ≈ x2
2.0
%ion =[H+]
[HC2H3O2]× 100%
![Page 20: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/20.jpg)
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%ion =
The 0.30% is well within the tolerance of 5%so the assumption is valid.
If the % ionization was more than 5%, then the
assumption would not be valid and you wouldhave to use the quadratic formula.
There are quadratic formula programs for theTI calculator and also the Solver function onthe TI.
6.0 x 10-3 M
2.0 M× 100% = 0.30%
![Page 21: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/21.jpg)
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To learn more about the Solver, visit
http://www2.ohlone.edu/people2/joconnell/ti/solver8384.pdf
Another useful quantity that is used is the pKa
of an acid and is defined as pKa = -log Ka.
So for the current problem, acetic acid,
Ka = 1.8 x 10-5 and pKa = 4.74
pKa’s are more convenient to work with than
Ka’s.
![Page 22: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/22.jpg)
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Combining Acidic SolutionsCombining Acidic Solutions
What is the pH of a solution obtained bymixing 235 mL of 0.0245 M HNO3 with 554 mL of 0.438 M HClO4?
[HNO3] = 0.0245 M [HClO4] = 0.438 MV1 = 235 mL V2 = 554 mL
HNO3(aq) → H+(aq) + NO3-(aq)
HClO4(aq) → H+(aq) + ClO4-(aq)
![Page 23: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/23.jpg)
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[HNO3] = n/V
1 mol HNO3
n1=0.0245 mol HNO3
1.00 L× 235 mL ×
1 L
103 mL×
×1 mol H+
=5.76 × 10-3 mol H+
1 mol HNO3
n2=0.438 mol HClO4
1.00 L× 554 mL ×
1 L
103 mL×
×1 mol H+
=2.42 × 10-1 mol H+
1 mol HClO4
![Page 24: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/24.jpg)
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.
[H+] =nT
VT
[H+] =2.48 × 10-1 mol H+
789 mL × 1 L103 mL
= 0.314 M
![Page 25: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/25.jpg)
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Polyprotic AcidsPolyprotic Acids
Polyprotic acids are those acids, H2CO3 andH2SO4, which have more than one ionizablehydrogen.
The Ka value for each hydrogen is differentand Ka vastly decreases with each
hydrogen.
The first hydrogen is the easiest to removebecause the original molecule or ion is thestrongest acid.
![Page 26: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/26.jpg)
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The reason for the increased difficultly isbecause each successive hydrogen is tryingto be removed from a more negative
species.
Compare the two ionization equations forcarbonic acid.
H2CO3(aq) H+(aq) + HCO3-(aq)
HCO3-(aq) H+(aq) + CO3
2-(aq)
The greater attraction between the HCO3
- and the H+ will make it more difficult to ionize.
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The pH of a diprotic or a triprotic acid isdetermined almost completely by the firstionization.
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Diprotic Acid ProblemDiprotic Acid Problem
Calculate the pH of a 0.0020 M solution ofcarbonic acid.
[H2CO3] = 0.0020 M
Ka1 = 4.4 x 10-7 Ka2 = 4.7 x 10-11
![Page 29: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/29.jpg)
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H2CO3(aq) H+(aq) + HCO3-(aq)
[ ]I 0.0020 0 0[ ]c -x +x +x[ ]e 0.0020 - x x x
4.4 x 10-7 =0.0020 - x
x x×
Ka1 =[H+] [HCO3
-][H2CO3]
4.4 x 10-7 =0.0020 - x
x x×
![Page 30: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/30.jpg)
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We will make the assumption that 0.0020 – x ≈
0.0020 and verify the assumption after thecalculation is made.
x = [H+] = [HCO3-] = 3.0 x 10-5 M
4.4 x 10-7 =0.0020
x2
%ion =[H+]
[HCO3-]
× 100%
%ion =3.0 x 10-5 M
2.0 x 10-3 M×100% =1.5%
![Page 31: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/31.jpg)
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The %ion = 1.5% is well within the 5%,therefore the assumption is valid.
HCO3-(aq) H+(aq) + CO3
2-(aq)
[ ]I 3.0 x 10-5 3.0 x 10-5 0
[ ]c -x +x +x[ ]e 3.0 x 10-5-x 3.0 x 10-5+x x
•
You must use the concentrations of H+ andHCO3
- resulting from the first hydrogen ionizing.
![Page 32: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/32.jpg)
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B
Ka2 =[H+] [CO3
2-][HCO3
-]
4.7 x 10-11=(3.0 x 10-5 + x) x×
(3.0 x 10-5 - x)
The extremely small value of the exponent,10-11, allows for the assumption3.0 x 10-5 ± x ≈ 3.0 x 10-5 which simplifies theabove expression to
![Page 33: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/33.jpg)
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x = [H+] = [CO32-] = 4.7 x 10-11 M
%ion =[H+]
[HCO3-]
× 100%
%ion =4.7 x 10-11 M
3.0 x 10-5 M×100% =1.6 x 10-4%
The %ion = 1.6 x 10-4 % is well within the 5%,therefore the assumption is valid.
![Page 34: 8 - 1 Introduction to Acids and Bases Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l)H + (aq) + OH - (aq) The equilibrium.](https://reader036.fdocuments.us/reader036/viewer/2022062805/5697bff11a28abf838cbb7f6/html5/thumbnails/34.jpg)
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What was the point of doing this rather longand tedious problem?
These steps can be applied to any diprotic or
triprotic acid and what you verify is that the[H+] contribution from the second or thirdhydrogen can be ignored.
So, what is the pH of a 0.0020 M solution ofcarbonic acid?
pH = -log[H+] = -log(3.0 x 10-5) = 4.52
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Common Ion EquilibriaCommon Ion Equilibria
Calculate the pH of a solution containing0.075 M nitrous acid and 0.20 M sodiumnitrite.
Ka = 4.5 x 10-4
[NaNO2] = 0.20 M [HNO2] = 0.075 M
NaNO2(aq) → Na+(aq) + NO2-(aq)
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HNO2(aq) H+(aq) + NO2-(aq)
[ ]I 0.075 0 0.20[ ]c -x +x +x[ ]e 0.075 - x x x
Ka =[H+] [NO2
-][HNO2]
4.5 x 10-4 =0.075 - x
x (0.20 + x)× ≈0.075 0.20x
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[H+] = 1.7 x 10-4 M
%ion =[H+]
[HNO2]× 100%
%ion =1.7 x 10-4 M
× 100%0.075 M
= 0.23%
The 5% rule applies so the previous twoassumptions are valid.
pH = -log[H+] = -log(1.7 x 10-4) = 3.77