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CHM135H1 (333)
Lecture

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Department
Chemistry
Course Code
CHM135H1
Professor
Kris Quinlan

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Slide No. Notes
13 -half equivalence point – half of acid added
-equivalent concentration of HA and A-
-at equivalent point, pH not 7 but greater than 7, because
OH- present
-EX. 40.00mL of 0.1000M HPr is titrated with 0.100 M
NaOH. What is the pH after 24.00mL NaOH is added?
-Ka(HPr) = 1.3 x 10^-5
-Same concentration so look at volume – more acid,
therefore acidic
-Calculate [HPr] & [NaOH] after dilution/mixing up
each other
- [HPr] = 0.1000M/L times 40.00mL divided by 65.00
mL = 0.06154
- [NaOH] = (0.100M/L)(25.00mL) / 65.00mL = 0.03846
-HPr + OH- -> water + Pr-
-0.06154M 0.03846M 0M
-0.02308M 0M 0.03846M
-HPr + water -> Pr- + H3O+
-I: 0.02308M 0.03846M 0M
-C: -x +x +x
-E: 0.02308M-x 0.03846M+x x
-Ka = [products]/[reactants] = (0.03846+x)(x)/(0.02308-
x) = 1.3 x 10^-5
-Assume x << 0.02308 M
-X = 7.8 x 10 ^ -6 M
-pH = 5.11
14 -titration curves vary depending on acid strength
-different shapes – buffer formed so need to add more
NaOH
-Pka = pH at half equivalence point
-The equivalence point is same for all curves
15 -physiological pH is at 7.4
-Ans: B
-Want buffer as close to desired pH as possible
16 -as OH- is added, first half equivalence point is reached
-2 steep regions
17 -PbSO4 sparingly soluble, not very soluble
-1.6x10^-6 very small – not very soluble
-Equal amount of each concentration, so that to find one
concentration, take square root of one Ksp
-Ans: C
18 -EX1.
-CaF2 -> Ca2+ + 2F-
-I: 0M 0M
-C: +x +2x
-E: x 2x
-Ksp = [Ca2+][F-]squared
-X = 1.5 x 10^-4 g CaF2 divided by 100.0mL times
1000mL/L times mol/78.1g = 1.92x10^-5 M of what
dissolves
-Ksp = x times 2x squared = 2.8x10^-14
-EX2. want to find the amount that dissolves
-Solubility = mol/L of Mg(OH)2 that dissolves
-Mg(OH)2 -> Mg2+ + 2OH-
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Description
Slide No. Notes 13 - half equivalence point half of acid added - equivalent concentration of HA and A- - at equivalent point, pH not 7 but greater than 7, because OH- present - EX. 40.00mL of 0.1000M HPr is titrated with 0.100 M NaOH. What is the pH after 24.00mL NaOH is added? - Ka(HPr) = 1.3 x 10^-5 - Same concentration so look at volume more acid, therefore acidic - Calculate [HPr] & [NaOH] after dilutionmixing up each other - [HPr] = 0.1000ML times 40.00mL divided by 65.00 mL = 0.06154 - [NaOH] = (0.100ML)(25.00mL) 65.00mL = 0.03846 - HPr + OH- -> water + Pr- - 0.06154M 0.03846M 0M - 0.02308M 0M 0.03846M - HPr + water -> Pr- + H3O+ - I: 0.02308M 0.03846M 0M - C: -x +x +x - E: 0.02308M-x 0.03846M+x x - Ka = [products][reactants] = (0.03846+x)(x)(0.02308- x) = 1.3 x 10^-5 - Assume x << 0.02308 M - X = 7.8 x 10 ^ -6 M - pH = 5.11
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