TITRATION CURVE WEAK ACID WITH STRONG BASE MG-KP

TITRATION CURVE WEAK ACID WITH STRONG BASE MG-KP

TITRATION CURVE WEAK ACID WITH STRONG BASE MG-KP 2014 EQUIVALENCE POINT HA moles original =OH- added moles T ACID IONIZATION IN PURE WATER Ka= [H+] [A-]/[HA] HA At equivalence HA initial = OH- added = A- produced The A- produced is the major pH determining ion through A HYDROLYSIS REACTION: A- + H2O HA + OHKa Point Need V and M Ka=[H+][OH-] [HA] BuFFER REGION pH = pKa + LOG [A-]/[HA} Kb = [HA] [OH-] [A-] AT EQUIVILENCE THEN HYDROLYSIS RECOGNIZING AREAS OF THE TITRATION CURVE 1) THE a)

b) c) Ka POINT: NO BASE HAS BEEN ADDED, BASE ADDED IS (0.0 mL). THE pH CALCULATED WITH THE Ka MASS ACTION. ASSUMES HA IN PURE WATER. 2) THE BUFFER ZONE a) SOME MOLES OF WEAK ACID (REMMAINNING) b) BEFORE EQUIVALENCE POINT c) THE BASE IS THE LIMMITING REAGENT IN NEUTRALIZATION ICE CHART. d) AFTER YOU FIND HA AND A- IN THE ICE CHART, USE THEM IN HENDERSON-HASSLEBACH. 3) EQUIVALENCE a) ALL OF THE ACID HAS BEEN NEUTRALIZED EXACTLY (STOICHIOMECTRIC ). b) MOLES OF HA NEUTRALIZED =MOLES OF BASE ADDED = MOLES OF CONJUGATE PRODUCED. HA + OH- H2O + Ac) THE BASE AND ACID ARE 0.0 IN THE END BAR OF THE ICE CHART. d) MaVa = MbVb 4) HYDROLYSIS OF THE CONJUGAT E BASE. a) AFTER NEUTRALIZATION THEN HYDROLYSIS b) USE THE Kb: Kb * Ka = Kw c) Kb = [OH-] [HA]/[A-] A- + H20 OH- + HA EX) Calculate the pH of a solution of 100.0 mL of 1.0M HA (pKa = 6.2)that has had 0.0 mL of 1 M NaOH added in a titration. STEP ONE: - you are at the Ka point, the titration has not begun and no base (OH-) has been added as of now. Simply do a Ka ice chart working in molarities.

HA A- 1 molar -x 1 M -x Approximate this x, 5% rule STEP TWO: Ka = 6.3 * 10-7 Ka = [A- ][H+]/[HA] Ka = 6.3 * 10-7 = x2 /1.0 M X = [H+] = 0.00079 M pH = - LOG 0.00079 = 3.1 = answer 0.0 M H+ 0.0M x x x x EX) Calculate the pH of a solution of 100.0 mL of 1.0M HA (pKa = 6.2)that has had 40 mL of 1 M NaOH added in a titration. Moles of HA from volume Moles of OH- from volume and STEP ONE: - you are below equivalence as the OH- added is less than the moles of and molarity: molarity: HA initial. You are in the buffer range so evaluate the neutralization stoic first to get M=mole/L M=mole/L

A- and HA remaining. 1.0 M = x/0.1 L 1.0 M = x/.040 L X = 0.10 mole HA initial./#L X = 0.040 mole OH- added./#L HA OHAH2O 0.1 mole -x 0.04 mole Limiting Rx. -x 0.06 mol 0 0.0 mol x 0.04 mol STEP TWO: Use the Henderson-Hasslebach equation to find the pH using the A- and HA remaining from step one. pH = pKa + log [A-]/[HA] pH= 6.2 + log (.04)/(.06) pH = 6.2 + log 0.666 pH = 6.2 -0.176 pH = 6.04 EX) Calculate the pH of a solution of 100.0 mL of 1.0M HA (pKa = 6.2)that has had Moles100 ofmL HAmL from of aOHfrom volume and of volume 1 M NaOH Moles added in titration. and molarity:

molarity: M=mole/L M=mole/L STEP ONE: - you are below equivalence as the OH- added is less than the moles of 1.0 MHA = x/0.1 1.0 M =range x/0.10 initial.L You are in the buffer so Levaluate the neutralization stoic first to get X = 0.10 moleHA HAremaining. initial./#L X = 0.10 mole OH- added./#L A- and HA OH- 0.01 mole A- 0.01 mole H2O 0.0 mol -x -x x 0 0

0 .01 mole Equivalenceall consumed Convert moles of A- -to molarity for the next step which is the hydrolysis equilibrium, remember you should work in molarity for any equilibrium calculation.: M=0.010 mole / 0.2 L (total volume HA + OH-)=.050 molar ASEE NEXT SLIDE FOR HYDROLYSIS Eq The molarity of the conjugate from last slide is 0.05 molar A- H 2O 0.050 M HA OH0.0 0.0 -x x x 0.050M -x x M xM

Now solve for [OH-] in the hydrolysis (Kb) mass action. Kb = [OH-] [ HA] / [A-] 5.9 x 10-11 = (x)(x)/0.05, x = [OH-] = 1.7 x10-6 , - LOG(1.7 x 10-6 ) = pOH- = 5.7 14=pH + pOH , 14 = pH + 5.7, pH 8.3 Kb = Kw/Ka Kb=1 x 10-14/1.8 x 10-4 Kb= 5.9 x 10-11 MolesEX)ofCalculate HA from the volume pH of a solution Moles of OH100.0 from mL volume of 1.0Mand HA that has had 150 mL mL and molarity: of 1 M NaOH added in a titration. molarity: M=mole/L M=mole/L 1.0 MSTEP = x/0.1 ONE: L - you are beyond 1.0equivalence M = x/0.15 as L the OH- added is more than the moles X = 0.10 of HA mole initial. HA initial./#L You are in theXovershoot = 0.15 mole range OH-so added./#L evaluate the neutralization stoic first

to get OH- remaining. HA 0.01 mole OH- A- 0.015 mole H2O 0.0 mol -x -x x 0 .005 0 .01 mole Base overshootpH is from [OH-] Convert moles of OH- -to molarity for the next step which is the pOH-, remember you should work in molarity for any equilibrium calculation.: M=0.005 mole / 0.25 L (total volume HA + OH-)=.020 molar OHpOH- = - log [.025] = 2.0 and 14 = pH + pOH so 14 = pH + 2.0 pH =13

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