Chapter 16
AcidBase Titration and pH

Aqueous Solutions and the Concept of pH
 Hydronium Ions and Hydroxide Ions
Water can form both hydronium and hydroxide ions.
 Self Ionization of Water
Water conducts but very little  pure water i.e. distilled water
Reason is self ionization
Figure 161 page 481
Self ionization  definition
H_{2}O (l) + H_{2}O (l) = H_{3}O^{+1} _{(aq)} + OH^{1} _{(aq)}
The concentration of the hydronium and the concentration of the hydroxide is
1.0 x 10 ^{7} mol/L of water at 25^{o}C
use brackets, [], around formula to indicate “concentration in moles per liter”,
which is the same as molarity.
[H_{3}O^{+1}] x [OH^{1}] = 1 x 10^{14} M since [H_{3}O^{+1}] = 1 x 10^{7} and
[OH^{1}]= 1 x 10^{7}
note that 10^{7} is equivalent to 1 x 10^{7}
The 10^{14} is a constant in water and also dilute aqueous solutions at constant temperature. That means that the [H_{3}O^{+1}] times the
[OH^{1}] must always equal 10^{14} if you have pure water or if you have a solution in which water is the solvent e.g. acids, etc.
ionization constant of water is
Kw = [H_{3}O^{+1}] x [OH^{1}] = 1 x 10^{14} M^{2} = (1.0 x 10^{7}M ) (1.0 x 10^{7} M)
As temp increases the ionization of water increases and so does Kw.
Table 161 page 482
 Neutral, Acidic, and Basic Solutions
Since [H_{3}O^{+1}] = [OH^{1}] in pure water, pure water is neutral.
If [H_{3}O^{+1}] were > [OH^{1}] then the solution would be acidic. The reverse situation would yield a basic (alkaline) solution.
Note: If the [OH^{1}] were greater than 10^{7} the solution would be basic or alkaline i.e. values such as 10^{6}, or 10^{5}.
 Calculating the Concentration of Hydronium and Hydroxide Ions
Table 162 page 483
Example: The dissociation of NaOH can be represented as
NaOH(s) > Na^{+}_{(aq)} + OH^{}_{(aq)}
i.e. one mole of sodium hydroxide yields one mole of sodium ions and one mole of hydroxide ions.
If you had a solution of NaOH whose concentration was 1.0 x 10^{2} M you would have 1.0 x 10^{2} moles of sodium ions in solution and 1.0 x 10^{2} moles of hydroxide ions in solution.
Kw = [H_{3}O^{+1}] x [OH^{1}]
Kw has a value of 1 x 10^{14} M^{2} for aqueous solutions. That means we know two of the three variables in the above equation.
1 x 10^{14} M^{2} = [H_{3}O^{+1}] x (1.0 x 10^{2} M)
1 x 10^{14} M^{2} / 1.0 x 10^{2} M = [H_{3}O^{+1}]
[H_{3}O^{+1}] = 1 x 10^{12} M
Note: Review rules for doing arithmetic operations using scientific notation.
Example: Determine the hydronium and hydroxide ion concentrations in a solution that is 1.0 x 10^{4} M Ca(OH)_{2}.
Need to write the equation to indicate what ions are produced in solution:
Ca(OH)_{2} > Ca^{+2} + 2OH^{1} i.e. one mole of Ca(OH)_{2} produces one mole of Ca^{+2} ions and 2 moles of OH^{1} ions.
The concentration of the solution is 1.0 x 10^{4} M which means that the concentration of the Ca^{+2} ions is the same or 1.0 x 10^{4} M and the concentration of the OH^{1} ions is twice that amount or
2(1.0 x 10^{4}) M.
Kw = [H_{3}O^{+1}] x [OH^{1}]
We know that Kw = 1.0 x 10^{14} M^{2} and from above we indicated that the [OH^{1}] = 2(1.0 x 10^{4}) M
Substituting into the formula we have
1.0 x 10^{14} M^{2} = [H_{3}O^{+1}] x 2(1.0 x 10^{4}M)
1.0 x 10^{14} M^{2} / 2(1.0 x 10^{4} M) = [H_{3}O^{+1}]
1.0 x 10^{14} M^{2} / 2.0 x 10^{4} M = [H_{3}O^{+1}]
0.5 x 10^{10} M = [H_{3}O^{+1}]
5.0 x 10^{11} M = [H_{3}O^{+1}]
Homework: 16.1
 The pH Scale
Way of expressing acidity/alkalinity of a solution.
pouvoir hydrogene  hydrogen power
pH  definition
formula: pH = log [H_{3}O^{+1}]
common log v natural log
purpose of logs
power to which 10 must be raised to equal a number i.e. a log is an exponent
e.g. if the [H_{3}O^{+1}] = 1.0 x 10^{7} M as it does in pur water then the pH would be
pH = log [H_{3}O^{+1}] = log (1.0 x 10^{7}) = (7) = 7 Note that the pH is the same as the exponent of 10 when the number before the x 10 is 1.
Scale: 1>7>14
figure 163 page 485
pOH  definition
pOH = log [OH^{1}]
pH + pOH = 14
table 163 page 486
Example: Calculate the pH of a solution whose [H_{3}O^{+1}] = 1.0 x 10^{6} M.
pH = log [H_{3}O^{+1}] = log (1.0 x 10^{6}) = (6 M) = 6
table 164 page 486
if [H_{3}O^{+1}] > [OH^{1}] then solution is acidic
if [H_{3}O^{+1}] < [OH^{1}] then solution is basic or alkaline
if [H_{3}O^{+1}] = [OH^{1}] then solution is neutral
Kw is temperature dependent.
 Calculations Involving pH
If either [H_{3}O^{+1}] or pH is known we can calculate the other.
pH has unique significant figure
pH is a logarithm or an exponent.
Because of this the number to the left of the decimal only locates the decimal
point. It is not included when counting significant figures. So there must be
a many significant figures to the right of the decimal as there are in the number
whose logarithm was fund.
1 x 10^{7} has one significant figure thus the pH must have one digit to the right of the decimal. pH = 7.0 correctly indicates the proper number of significant figures.
Homework 16.2
 Calculating pH from [H_{3}O^{+1}]
Sample Problem 162 page 487
What is the pH of a 1.0 x 103 M NaOH solution?
Given: pH = 1.0 x 10^{3} M; NaOH solution; pH = ?
[H_{3}O^{+1}] [OH^{1}] = Kw
[H_{3}O^{+1}] = Kw / [OH^{1}]
[H_{3}O^{+1}] = 1.0 x 10^{14} M^{2} / 1.0 x 10^{3}M
[H_{3}O^{+1}] = 1.0 x 10^{11} M
pH = log [H_{3}O^{+1}]
pH =  log (1.0 x 10^{11} M)
pH = 11.00
since pH > 7 the solution is basic or alkaline
Sample Problem 163 page 488
What is the pH of a solution if the [H_{3}O^{+1}] is 3.4 x 10^{5} M?
Given: pH = ?; [H_{3}O^{+1}] = 3.4 x 1^{5} M
pH =  log [H_{3}O^{+1}]
pH =  log (3.4 x 10^{5} M)
pH =  log (3.4) +  log(10^{5})
pH =  .53 + 5
pH = 4.47
Homework: 16.3
 Calculating [H_{3}O^{+1}] and [OH^{1}] from pH
Given pH calculate either [H_{3}O^{+1}] or [OH^{1}]
Note that the base of common logs is 10. Thus the antilog of a common log is 10 raised to that number.
pH =  log[H_{3}O^{+1}]
log[H_{3}O^{+1}] = pH
[H_{3}O^{+1}] = antilog (pH)
[H_{3}O^{+1}] = 10^{pH}
e.g. if the pH = 2 then [H3O^{+1}] = 10^{2} M
e.g. if pH = 0 then [H3O^{+1}] = 1 since 10^{0} = 1
Sample Problem 164 page 489
Determine the hydronium ion concentration of a aqueous solution that has a pH of 4.0.
Given: [H3O^{+1}] = ?; aqueous solution; pH = 4.0
pH =  log [H3O^{+1}]
log [H_{3}O^{+1}] =  pH
[H_{3}O^{+1}] = antilog (pH)
if your calculator does not do antilogs use
[H_{3}O^{+1}] = 1 x 10^{pH}
[H_{3}O^{+1}] = 1 x 10^{pH}
[H_{3}O^{+1}] = 1 x 10^{4} M
Since pH is less than 7 the solution is acidic.
Sample Problem 165 page 490
The pH of a solution is measured and determined to be 7.52.
a) What is the hydronium ion concentration?
b) What is the hydroxide ion concentration?
c) Is the solution acidic or basic?
Given: pH = 7.52; [H_{3}O^{+1}] =?; [OH^{1}] = ?; acidic or basic?
pH =  log [H_{3}O^{+1}]
log [H_{3}O^{+1}] = pH
[H_{3}O^{+1}] = antilog (pH) = antilog (7.52)
if your calculator does not do antilogs use
1.0 x 10^{7.52} =
3.0 x 10^{8} M H_{3}O^{+1
}[H_{3}O^{+1}] [OH^{1}] = Kw = 1.0 x 10^{14}
[OH^{1}] = 1.0 x 10^{14} / [H_{3}O^{+1}]
[OH^{1}] = 1.0 x 10^{14} / 3.0 x 10^{8} M
[OH^{1}] = 3.3 x 10^{7} M OH^{1 }
^{
pH given as 7.52 which is greater than 7, thus the solution is basic.
Homework: 16.4
}
 pH Calculations and the Strength of Acids and Bases
Table 16.5 page 491
Strong acid v weak acid
strong base v weak base
For weak species, you cannot use Molarity to determine the pH since not all of the particles form ions when in solution. Need to measure the pH using instruments.
From the data in the table determine the strong species and the weak species.
Homework 16.5

Determining pH and Titrations

Indicators and pH Meters
Acid
Base indicators  definition
Used to get approximate pH of a solution.
Color change comes from the indicator acting as either a weak acid or a weak base:
HIn = H^{+1} + In^{1
}The H^{+1} is the weak acid part and In^{1} is the anion part of the indicator.
The HIn and the ions on the right of the equation are different colors.
The In ion accepts the H^{+1} from the acid and forms HIn and has its acid color e.g. for litmus is red.
In basic solutions, the OH^{1} ions combine with the H^{+1 }ions of the indicator which leaves mainly In^{1} ions in solution and this produces the base color which for litmus is blue.
Figure 164 page 493
Figure 165 page 494
pH paper  paper soaked in a mixture of indicators.
Many different indicators.
pH range of color change for each is different.
Transition
interval  definition.
Table 166 page 495
Indicators that change color below 7 are stronger acids than the other types of indicators. e.g. methyl orange
pH meter  definition
Figure 166 page 494
 Titration
A neutralization reaction is a reaction between an acid and a base. The acid
provides the hydronium ion and the base provides the hydroxide ion.
Summarized by
H_{3}O^{+1} _{(aq)} + OH^{1} _{(aq)} > 2
H_{2}O _{(l)
}Note: one hydronium ion combines with one hydroxide ion. When equal amounts of hydronium and hydroxide ions are mixed from an acid and a base, the resulting solution is neutral.
One liter of 0.10 M HCl contains 0.10 mol of hydronium ions.
One liter of 0.10 M NaOH contains 0.10 mol of hydroxide ions.
Mix the one liter of each and you get a neutralization reaction in which the products are NaCl + H_{2}O. This solution is neutral.
The progressive addition of an acid to a base (or visa versa) can be used to
compare the concentrations of the acid and the base.
Titration  definition
Used to determine the chemical equivalent volumes of acidic and basic solutions.
Figure 167 page 497
Burets are used to carry out a titration.

Equivalence Point
Equivalence
point  definition
Use either indicators (most common) or pH meters (most accurate) to determine.
pH meter shows large voltage change at equivalence point.
Indicator permanently changes color.
Figure 168 page 498
End point  definition
Different indicators change colors at different pH’s.
Table 166 page 495  note pH (black horizontal line) for each indicator and strength of both acid and base.
No good indicator for weak acid/weak base because the relative strengths of the reactants may vary greatly.
Usually add base to acid i.e. going from low pH to higher pH.
Change in pH is slow then rapid through the equivalence point then slowly after that.
Figure 169 page 499

Molarity and Titration
Note: We use only one buret at our locker and use one buret from the side table  saves solution and possible contamination.
Note: We never use an eye dropper to add phenolphthalein  bottles are special dropper bottles  never put anything into these bottles.
For our purposes, the buret at our place contains the solution of known concentration  the standard solution.
Standard solution  definition.
Usually carry out a titration until you have three results that agree within 0.05 mL.
Need to fill buret to approximately ( not above) the zero mark with the base.
Fill tip with solution.
Read volume of buret.
Record as initial volume of solution.
To a clean dry flask add measured amount of acid from side table (need initial reading and then final reading to get accurate volume).
Take flask to your place and add indicator.
Add base from your buret to flask, with swirling, slowly, until a permanent color change takes place. You know you are getting close to the permanent color change when you see a color change but continued swirling makes the color disappear.
Take a final reading from your buret. Subtracting the initial and final reading tells you the exact volume of base you used.
Do the calculations.
Sample Problem 166 page 502
In a titration, 27.4 mL of a 0.0154 M Ba(OH)_{2} is added to a 20.0 mL sample of HCl solution of unknown concentration. what is the molarity of the acid solution?
Given: 27.4 mL base; 0.0154 M Ba(OH)_{2}; 20.0 mL of HCl used; M of HCl = ?
Balanced equation:
Ba(OH)_{2} + 2HCl > BaCl_{2} + 2 H_{2}O
1 mol + 2 mol > 1 mol + 2 mol
a) for Barium hydroxide: (molarity) x (volume) x (1 L / 1000 mL) = mol
(0.0154 mol Ba(OH)_{2} / L) x (27.4 mL Ba(OH)_{2}) x (1 L / 1000 mL) = 4.22 x 10^{4} mol Ba(OH)_{2}
b) (molar ratio from balanced equation) x (mol Ba(OH)_{2} from a)) = mol HCl
(2 mol HCl / 1 mol Ba(OH)_{2}) x (4.22 x 10^{4} mol Ba(OH)_{2}) = 8.44 x 10^{4} mol HCl
c) (mol HCl / vol HCl) x (1000 mL / 1 L) = (mol HCl / 1 L) = M
(8.44 x 10^{4} mol HCl / 20.0 mL) x (1000 mL / 1 L) = (4.22 x 10^{2} mol HCl / 1 L) = 4.22 x 10^{2} M HCl
Homework: 16.6
end of notes
Titration is the controlled addition and measurement of the amount of a solution
of known concentration required to react completely with a measured amount of
a solution of unknown concentration.
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Equivalence point is the point at which the two solutions used in a titration
are present in chemically equivalent amounts .
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End point of an indicator is the point in a titration at which an indicator changes color.
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Standard solution is the solution that contains
the precisely known concentration of a solute.
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Self ionization of water involves the transfer of a proton (H^{+1})
from one water molecule to another to form hydronium ion and a hydroxide
ion.
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pH of a solution is defined as the negative of the common logarithm of
the hydronium ion concentration.
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pOH of a solution is defined as the negative of the common logarithm of
the hydroxide ion concentration.
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Acid base indicators are compounds whose colors are sensitive to pH.
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The transition interval is the pH range over which an indicator changed
color.
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pH meter determines the pH of a solution by measuring the voltage between
the two electrodes that are placed in the solution.
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