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Quantitative Problem Solving of Acids and Bases Notes
(pages 126 – 144 in your textbook)
(E1) analyse the equilibrium that exists in water
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write equations representing the ionization of water using either H3O+ or H+ and OHwrite the equilibrium expression for the ion product constant of water (Kw)
o when a strong acid and base react, we get the following double replacement
reaction (balanced chemical equation):
HCl(aq) +NaOH(aq) ↔ NaCl(aq) + H2O(l) + 59 kJ
o the complete ionic equation is:
o and the net ionic equation is:
o the reverse of this reaction is called the self – ionization of water:
o the equilibrium expression for this reaction is:
o this is called the ion product constant of water (kw)
Note: The self – ionization of water can also be written as:
The ion product constant of water (kw) would then be:
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state the value of Kw at 25°C
describe and explain the variation in the value of Kw with temperature
o In the Dynamic Equilibrium unit, we learnt that the equilibrium expression (keq)
is a constant for a certain temperature. Therefore, the keq value does not change
unless the temperature changes.
o The ion product constant of water (kw), works in the same way. It is a constant
for a certain temperature.
o Kw at 25°C is:
o However, if this temperature changes, then the value of Kw changes
Determine how Kw changes when water is heated.
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calculate the concentration of H3O+ (or OH-) given the other, using Kw
Determine [H3O+] and [OH-] in pure, neutral water.
Do you think there might be some correlation between these concentrations
and the pH scale? If so, explain.
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What is the concentration of [H3O+] and [OH-] in 0.0010 M HCl(aq)?
What is the concentration of [H3O+] and [OH-] in 0.5 M NaOH(aq)?
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predict the effect of the addition of an acid or base to the equilibrium system
o As we learnt in the Dynamic Equilibrium unit, if we act to increase the
concentration of a substance, the reaction will respond by decreasing it.
Determine what happens when we add an acid to the equilibrium system.
Determine what happens when we add a base to the equilibrium system.
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state the relative concentrations of H3O+ and OH- in acid, base, and neutral solutions
In a neutral solution,
In an acidic solution,
In a basic solution,
[H3O+] = [OH-]
[H3O+] > [OH-]
[H3O+] < [OH-]
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(E3) explain the significance of the Ka and Kb equilibrium expressions
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write Ka and Kb equilibrium expressions for weak acids or weak bases
o So far, we have dealt with strong acids and bases with 100% dissociation, where
we only consider [H3O+] and [OH-]. This is not the case for weak acids and bases.
o The equilibrium expression for a weak acid is the acid ionization constant (ka).
o The equilibrium expression for a weak base is the base ionization constant (kb).
o ka found in “Relative Strength of Brönsted-Lowry Acids and Bases” Table
Determine the equilibrium expression for CH3COOH(aq).
Determine the equilibrium expression for NH3(aq).
How can you find the Kb value?
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relate the magnitude of Ka (the acid ionization constant) or Kb (the base ionization
constant) to the strength of the acid or base
o the greater the ka value, the stronger the acid
o the greater the kb value, the stronger the base
What happens when you mix acids and bases? Consider H2CO3(aq) mixed with SO32-(aq).
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(E2) perform calculations relating pH, pOH, [H3O+], and [OH-]
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define pH and pOH
pH = – log10[H3O+]
pOH = – log10[OH-]
Logarithms and What They Means
1. The expression log10[x] means the logarithm to the base 10 of x
Examples:
log10[0.01] = – 2
Note: 0.01 = 10-2 (log10 just gets rid of the base 10)
log10[1000] =
log10[8.76 x 10-9] =
log10[2.43 x 10-4] =
2. The opposite of log is antilog. Antilog[x] means 10 to the power of x.
Examples:
antilog[4] = 104 = 10 000
Note: antilog just puts it to the power of 10
antilog[–7] =
antilog[5.12] =
antilog[11.4] =
3. Combining logaritms.
log[A x B] = log[A] + log[B]
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define pKw , give its value at 25°C, and its relation to pH and pOH
pKw = – log10[kw]
= – log10[1.00 x 10-14] =
Note: This can also be expressed as
pKw = pH + pOH
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calculate [H3O+] or [OH-] from pH and pOH
Note: In pH (pOH, or pKw), only the figures after the decimal are significant. The
figure before the decimal comes from the nonsignificant figure in the power of 10.
1. If [H3O+] = 3.94 x 10-4, what is pH?
7. If pH is 6.330, what is [OH-]?
2. If pH is 4.798, what is the [H3O+]?
8. If pH is 4.44, what is [OH-]?
3. If [OH-] = 9.51 x 10-12, what is pOH?
9. If pOH is 8.09, what is [H3O+]?
4. If pOH is 9.106, what is [OH-]?
10. If pOH is 10.645, what is [H3O+]?
5. If pH is 9.355, what is pOH?
6. If pH is 6.577, what is pOH?
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Use the following schematic to converting from pH and pOH to [H3O+] and [OH-]
Note: When the pH is increase by 1, the [H3O+] is decreased by 10
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describe the pH scale with reference to everyday solutions
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(E4) perform calculations involving Ka and Kb
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calculate the value of Kb for a base given the value of Ka for its conjugate acid (and vice
versa)
o refer to outcome E3 on page #4 on this notebook
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calculate the value of Ka or Kb given the pH and initial concentration
given the Ka, Kb, and initial concentration, calculate [H3O+], [OH-], pH, or pOH
calculate the initial concentration of an acid or base, given the appropriate Ka, Kb, pH, or
pOH values
If 10.0 mL of 0.100 M HCl is added to 90.0 mL of 0.100 M NaOH, what is the pH of
the mixture?
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How many moles of HCl must be added to 40.0 mL of 0.180 M NaOH to produce a
solution having a pH = 12.500, if it is assumed that there is no change in volume
when the HCl is added?