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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LECTURE 11

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## 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.