Notes taken during lecture

27 views2 pages
user avatar
Published on 31 Dec 2010
School
UTSG
Department
Chemistry
Course
CHM135H1
Professor
Page:
of 2
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-
www.notesolution.com
LECTURE 11
www.notesolution.com

Document Summary

Notes half equivalence point half of acid added equivalent concentration of ha and a- at equivalent point, ph not 7 but greater than 7, because. 40. 00ml of 0. 1000m hpr is titrated with 0. 100 m. 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. Hpr + oh- -> water + pr- Hpr + water -> pr- + h3o+ Ka = [products]/[reactants] = (0. 03846+x)(x)/(0. 02308: = 1. 3 x 10^-5. X = 7. 8 x 10 ^ -6 m ph = 5. 11 titration curves vary depending on acid strength different shapes buffer formed so need to add more. Pka = ph at half equivalence point physiological ph is at 7. 4. The equivalence point is same for all curves. Want buffer as close to desired ph as possible.