Molarity and pH
plus Titration
Thursday, October 14, 2010
1
Pop Quiz Take-Home
a. Balanced Equation with products
b. Net Ionic Equation and spectator ions
c. Name of products
1. Solutions of sodium carbonate and copper (II) nitrate
2. Solutions of nitric acid and potassium hydroxide
3. Sol’ns of nitrous acid and magnesium hydroxide
4. Sol’ns of calcium chloride and sodium sulfate
5. Sol’ns of lead (IV) chlorate & calcium hydroxide
6. Sol’ns of lead (IV) hydroxide & hydrofluoric acid
Thursday, October 14, 2010
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Solutions of sodium carbonate and copper (II) nitrate
Na2CO3 + Cu(NO3)2 -> CuCO3 + 2NaNO3
Cu2+ + CO32– -> CuCO3 (s) ; Sp Ions: Na+, NO3–
Solutions of nitric acid and potassium hydroxide
HNO3 + KOH -> H2O + KNO3
H+ + OH– -> H2O Sp Ions: K+ ; NO3–
Sol’ns of nitrous acid and magnesium hydroxide
2HNO2 + Mg(OH)2 -> Mg(NO2)2 (aq) + 2H2O
HNO2 + OH– -> NO2– + H2O Sp Ions: Mg2+
or 2HNO2 + Mg(OH)2 -> Mg2+ + 2NO2– + 2H2O
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Sol’ns of calcium chloride and sodium sulfate
CaCl2 + Na2SO4 -> CaSO4 (s) + 2NaCl
Cu2+ + SO44– -> CaSO4 (s) ; Sp Ions: Na+, Cl–
Sol’ns of lead (IV) chlorate & calcium hydroxide
Pb(ClO3)4 + 2 Ca(OH)2 -> Pb(OH)4 + 2 Ca(ClO3)2
Pb4+ + 4 OH– -> Pb(OH)4 (s) ; Sp Ions: Ca2+ ClO3–
Sol’ns of lead (IV) hydroxide & hydrofluoric acid
Pb(OH)4 + 4 HF -> PbF4 (s) + 4H2O
Pb4+ + 4OH– + 4 HF -> PbF4 (s) + 4H2O (l)
Thursday, October 14, 2010
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Concentration
Concentration is an amount per unit volume
0.5 mol solute / 1 L solution = molarity
0.5 mol solute / 1 kg solvent = molaity
0.5 g solute / 1 g mixture X 100 = % by mass
0.05 mL solute / 1 mL solution X 100 = % by volume
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Molarity & Molar Mass
Molarity changes the volume to moles (or moles to
volume)
JUST LIKE
Molar mass changes mass to moles (or moles to mass)
For : Pb(NO3)2 + 2NaCl ➔ PbCl2 + 2NaNO3
Determine the mass of PbCl2 formed from 50.0 mL of
0.25 M Pb(NO3)2 solution.
0.050 L Pb(NO3)2 • 0.25 mol • 1PbCl2 •215.1 g PbCl2 = 1.1 g
1 L Pb(NO3)2 1Pb(NO3)2 1 mol PbCl2
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So:
2NaOH + H2SO4 ➔2H2O + Na2SO4
{0.12 M NaOH and 0.40 M H2SO4}
Starting Convert
Amount to Moles
(given)
(given)
1 mol NaOH
40.0 g NaOH
1 mol NaOH
5.0 g NaOH
40.0 g NaOH
0.12 mol NaOH
0.05 L NaOH
1L NaOH
0.12 mol NaOH
0.05 L NaOH
1L NaOH
5.0 g NaOH
Thursday, October 14, 2010
Mole
Ratio
Convert
to Ans.
(bal. eq.)
(given)
1 H2SO4
2 NaOH
1 H2SO4
2 NaOH
1 H2SO4
2 NaOH
1 H2SO4
2 NaOH
98 g H2SO4
1 mol H2SO4
1 L H2SO4
0.4 mol H2SO4
1 L H2SO4
0.4 mol H2SO4
98 g H2SO4
1 mol H2SO4
Answer
x g H2SO4
x L H2SO4
x L H2SO4
x g H2SO4
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Dilution Equation
Dilution is the act of changing a concentrated solution
into a less concentrated solution (like adding water to
frozen concentrated orange juice)
Determine total moles in the Known solution (usually
the dilute solution) and then divide by the known
amount of the unknown solution:
M1V1 = M2V2
Find the volume in mL of concentrated 6.0 M H2SO4
needed to make 500 mL of 0.25 M H2SO4.
0.25 mol/L • 0.500L = 6.0 mol/L •V => 20.8 mL 6M
H2SO4
Thursday, October 14, 2010
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Additive Concentrations
A solution is made by adding 40 mL of 0.50 M CaCl2
and 100 ml of 0.20 M NaCl. Determine the
concentration of Cl– in solution.
Find the total moles of Cl– and then divide by the total
volume.
Thursday, October 14, 2010
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pH and
+
[H3O ]
The amount of H+ or H3O+ in solution is important for
determining the vigorousness of the reaction of an
acid.
{Equal concentrations of strong and weak acid will
react completely with the same amount of base, but
the strong acid will react immediately since it is
already dissociated into the reactive H+ or H3O+.}
pH is a measure of the concentration of H+
pH = -log[H3O+]
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pH by example
pH
[H+] or [H3O+]
pH = -log[H+]
[H+] = 10–pH
12
1 X 10–12 M
3
1 X 10–3 M
3.50 (closer to 4)
10–3.50 = 3.2 X 10–4
2.300
5.01 X 10–3 M
-1
10 M
Thursday, October 14, 2010
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pH and precipitation
30 mL of a HCl solution with a pH of 2.35 is added to
a 50 mL saturated solution of lead (II) acetate.
Assuming all the Cl– is recovered as solid lead (II)
chloride, what is the mass of lead (II) chloride?
[H+] = [HCl] b/c it is a strong acid
0.030 L HCl • 10–2.35 M HCl • 1PbCl2 • 278.1 g PbCl2 =
1 L HCl 2 HCl
1 mol PbCl2
0.019 g PbCl2
Thursday, October 14, 2010
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Acid Base Reaction
Find the molarity of an HCl solution if
23.2 mL of acid reacts completely with
40.0 mL of 0.25 M Ca(OH)2.
Find the concentration 200 mL NaOH that
reacts with 45 mL of 1.00 M H2SO4.
Find the pH of a solution where 200 mL of
2.00 M HCl reacts with only 100 mL of 0.85 M
NaOH.
Thursday, October 14, 2010
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More about pH
Distilled (Pure Water) has a pH of 7.0000000
Distilled water is neutral this means that
there is an equal amount of H3O+ and OH– (the
reactive species of an acid and a base)
pH less than 7 are acidic and pH > 7 are basic
-1 0 1 2 ... 7 ... 12 13 14 15
A
Thursday, October 14, 2010
...
B
14
And More
Neutral pH means [H3O+] = [OH–]
10–7 = 10–7
So Water dissociation Constant
Kw = [H3O+]•[OH–] ALWAYS!!!
So 10-14 = 10–7•10–7
Given a pH of 3: the [H+] = 10–3 and [OH–] = 10–11
pH = -log[H3O+] and pOH = -log[OH–]
So pKw = pH + pOH (yes plus) 14 = pH + pOH
Thursday, October 14, 2010
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Weak Acids
Weak acids do not dissociate completely so the
[H+] ≠ [weak acid]
For example a 1.0 M acetic acid solution has a
pH of 0.0042M H+.
So the pH of weak acid solutions is usually
smaller than the pH of a strong acid, but if
the strong acid is dilute enough (e.g.,
0.00000001 M ) then the pH of the strong acid
can be smaller)
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Thursday, October 14, 2010
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Titration
Thursday, October 14, 2010
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