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TITRATION CURVE WEAK ACID WITH STRONG BASE MG-KP 2014

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

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TITRATION CURVE WEAK ACID WITH STRONG BASE MG-KP 2014

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  1. 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]

  2. 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 + OH- Kb = [HA] [OH-] [A-] AT EQUIVILENCE THEN HYDROLYSIS Ka Point Need V and M Ka=[H+][OH-] [HA] BuFFER REGION pH = pKa + LOG [A-]/[HA}

  3. RECOGNIZING AREAS OF THE TITRATION CURVE • THE 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. • 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. • 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 + A- • c) THE BASE AND ACID ARE 0.0 IN THE END BAR OF THE ICE CHART. • d) MaVa = MbVb • HYDROLYSIS OF THE CONJUGAT E BASE. • a) AFTER NEUTRALIZATION THEN HYDROLYSIS A- + H20  OH- + HA • b) USE THE Kb: Kb * Ka = Kw • c) • Kb = [OH-] [HA]/[A-]

  4. 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. 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 2-(

  5. 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. STEP ONE: - you are below equivalence as the OH- added is less than the moles of HA initial. You are in the buffer range so evaluate the neutralization stoic first to get A- and HA remaining. Moles of HA from volume and molarity: M=mole/L 1.0M = x/0.1 L X = 0.10 mole HA initial./#L Moles of OH- from volume and molarity: M=mole/L 1.0M = x/.040L X = 0.040 mole OH- added./#L 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

  6. EX) Calculate the pH of a solution of 100.0 mL of 1.0M HA (pKa = 6.2)that has had 100 mL mL of 1 M NaOH added in a titration. STEP ONE: - you are below equivalence as the OH- added is less than the moles of HA initial. You are in the buffer range so evaluate the neutralization stoic first to get A- and HA remaining. Moles of HA from volume and molarity: M=mole/L 1.0M = x/0.1 L X = 0.10 mole HA initial./#L Moles of OH- from volume and molarity: M=mole/L 1.0M = x/0.10 L X = 0.10 mole OH- added./#L Equivalence…all 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 A- SEE NEXT SLIDE FOR HYDROLYSIS Eq

  7. The molarity of the conjugate from last slide is 0.05 molar 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

  8. Moles of HA from volume and molarity: M=mole/L 1.0M = x/0.1 L X = 0.10 mole HA initial./#L EX) Calculate the pH of a solution of 100.0 mL of 1.0M HA that has had 150 mL mL of 1 M NaOH added in a titration. STEP ONE: - you are beyond equivalence as the OH- added is more than the moles of HA initial. You are in the overshoot range so evaluate the neutralization stoic first to get OH- remaining. Moles of OH- from volume and molarity: M=mole/L 1.0M = x/0.15 L X = 0.15 mole OH- added./#L Base overshoot…pH 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 OH- pOH- = - log [.025] = 2.0and 14 = pH + pOH so 14 = pH + 2.0 pH =13

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