Neutralization Titration PDF

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Summary

Indicators used, example of generating the pH at various points along the sigmoidal curve
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Neutralization titration- titrations involving reactions of acids and bases The standard solutions are always strong acids or strong bases - Most common acid HCl - Cheap - Strong acid - Has a relatively redox neutral conjugate base/anion - Most common bases are NaOH and KOH Indicators for these titrations are weak organic acids or weak organic bases - Have differing colors in their protonated and deprotonated forms

Weak base: In + H2O ⇋ HIn+ + OH↓ ↓ Base acid Color color

Weak acid: HIn + H2O ⇋ In- + H3O+ ↓ ↓ Acid base Color color

Kb = [HIn+] [OH-] / [In]

Ka = [In-] [H3O+] / [HIn]

The Ka and Kb for these species are typically VERY small

Ka = [In-] [H3O+] / [HIn] [H3O+] = Ka [HIn] / [In-] -

Shows that [H3O+] controls the ratio of [HIn] to [In-] in the solution, thus controlling the color of the solution

The eye can generally detect color changes when [HIn] / [In-] is between 0.1 and 10. - This means that color change happens when pH = pKa ∓ 1 - So if an indicator has a pKa= 8 the color change will occur between pH 7 & 9 xX + yY → products At the equivalence point the amounts of each species present is exactly that required by the

stoichiometry of the reaction: (1/x) MxVx = (1/y) MyVy Before and after equivalence (where X = species in excess): [X] = MxVx - (x/y) MyVy / Vx + Vy Generate sigmoidal curve for strong acid and strong base: 50 ml 0f a 0.0500 M solution of a strong diprotic acid is titrated with a 0.200 M solution of a strong monobasic base H2A + 2 BOH → B2A + 2 H2O Determine the volume of the titrant (base) used to reach the equivalence point: (½) [BOH] VBOH = (1/1) [H2A] VH2A VBOH = 2 [H2A] VH2A / [BOH]

→

2 (0.05 M) (50ml) / (0.02 M) = VBOH= 25.0 ml

EXAMPLE: Generating the pH at various points along the curve: 1) Initial pH (no BOH added) H2A + 2 H2O → A2- + 2H3O+ [H3O+] = 2[H2A] → pH = -log [2*0.0500] = 1.00

2) After 10 mL of BOH added (H2A still in excess) [H2A] = [H2A]o VH2A - (½) [BOH]o VBOH / VH2A + VBOH [H2A] = (0.05)(50 mL) - (½) (0.20 M)(10mL) / (50 mL + 10 mL) = 0.0250 M pH= -log (2 * 0.0250 M) = 1.301 3) Near equivalence → at 24.90 mL of BOH added [H2A] = (0.0500M) (50.00 mL) - (½) (0.200 M) (24.90 mL) / (50 mL) + (24.90 mL) = 1.34 * 10-4 M pH = -log (2* (1.34 * 10-4 M)) = 3.573 4) At equivalence, all of the H2A is consumed and only B+ and A2- are in solution - Since initial species were a strong acid and a strong base, their conjugate species are

-

not strong enough acids or bases to change the pH at all The only H3O+ present is that from neutral, pure water itself - pH=7.0

5) Just past the equivalence point → 25.10 mL of BOH added [BOH] = (0.200 M) (25.10 mL) - (2/1) (0.0500 M) (50.00 mL) / (25.10 mL + 50.00 mL) = 2.66 * 10-4 M pOH = -log (2.66 * 10-4 M) = 3.575 pH= 10.425

6) Well after the equivalence point 35 mL BOH → pH= 12.372 50 mL BOH → pH= 12.699

For weak acids, the calculations are slightly more complicated: EXAMPLE: 50.00 mL of 0.1000 M acetic acid (HOAc) is being titrated with 0.1000 M NaOH - Acetic acid is weak and does not ionize completely HOAc + H2O ⇋ OAc- + H3O+

Ka = 1.75 * 10-5 = [H3O+] [OAc-] / [HOAc]

1) Initial conditions If HOAc is only significant source of [H3O+] [H3O+] = [OAc-] → Ka = [H3O+]2 / [HOAc] - Since Ka is very small, assume [HOAc]eq ≅ [HOAc]o [H3O+] = √Ka[HOAc]o = √(1.75 * 10-5) (0.1000) = 1.32 * 10-3 M pH= 2.88

At any conditions up to very near the equivalence point, it behaves as a buffer, so you can calculate the pH using the Henderson-Hasselbach equation:

pH = pKa + log [Base] / [Acid]

In our case pH = pKa + log [OAc-] / [HOAc] We first calculate the amount of HOAc left after adding the NaOH and how much OAc- is formed and plug into this equation: 2) After adding 10.00 mL of the NaOH: [HOAc] = (0.1000 M) (50.00 mL) - (0.1000 M) (10.00 mL) / (50.00 mL - 10.00 mL) = 0.06667 M How much OAc- is present? M1V1 = M2V2 → M2 = M1V1 / V2 M2 = (0.1000 M) (10.00 mL) / (60.00 mL) = 0.01667 M OAc- or OH- ion pH = -log (1.75 * 10-5) + log (0.01667 M / 0.06667 M) = 4.16

3) At exactly halfway to the equivalence point: Half titration point- at this point the amount of titrant added is exactly half that required to get to the equivalence point LOOK AT THIS SHIT FOR LAB :) -

At this point [Base] = [Acid] - The Henderson-Hasselbach equation pH=pKa (log([base]/[acid]) = log(1) = 0 - At this point the buffering capacity of the solution is at its maximum - This is an inflection point in the curve where the slope of the plot reaches a minimum - Can find the pKa for an acid in this type of titration by looking for this point on the titration curve

At 25 mL pH = pKa = -log (1.75 * 10-5) = 4.76

4) At equivalence, we have another interesting situation. Here all of the HOAc has been converted to OAc-, which is itself a weak base OAc- + H2O ⇋ HOAc + OH-

Kb = [HOAc] [OH-] / [OAc-]

If OAc- is the only significant source of OH- then [OH-] = [HOAc] Kb = [OH-]2 / [OAc-] …. If Kb is small, then equilibrium - [OAc-] initial. How can we find Kb? For conjugate acid/base pair, KaKb=Kw Kb = Kw / Ka = 1.00 * 10-14 / 1.75 * 10-5 = 5.71 * 10-10 M1V1 = M2V2 → M2 = (0.1000 M)(50.00 mL) / (100.00 mL) = 0.05000 M [OH-] = √ Kb[OAc]o = √ (5.71 * 10-10)(0.05000 M) = 5.34 * 10-6 M pOH= -log (5.34 * 10-6) = 5.27 pH= 14 - pOH → 14 - 5.27 = 8.73

5) Beyond the equivalence point: at 50.10 mL of NaOH added [NaOH] = [OH-] = (0.1000 M) (50.10 mL) - (0.1000 M) (50.00 mL) / (50.10 mL + 50 mL) = 9.99 * 10-5 M

NOTE: - Very early in the titration and very close to the equivalence point, we can’t use the Henderson-Hasselbach equation to calculate pH - You would use an ICE table for the HOAc equilibrium and potentially solve the resulting quadratic equations

Note that the lower Ka is (the higher pKa is), the harder it is to detect the equivalence point in the titration:

This also affects - Dealing with the three will work in the

your choice of indicator a strong acid, really any of indicators listed in the figure fine, as they all change color “steep portion” of the plot

However, as you can also see in this figure, as Ka gets smaller, the choice of indicator becomes more and more crucial

IF you know Ka for the acid analyzed (or Kb for base analyzed) and you generally have a guess for an approximate concentration of the species at equivalence, you can use the pKa values for indicator in question to choose the one that is the most appropriate for the titration

Ex: Doing titration for acetic acid with a strong base At equivalence, you estimate that [OAc-] will be approximately 0.100 M OAc- + H2O ⇋ HOAc + OHBromocresol green - changes from pH = 4 to pH = 6 (pKa = 4.7) Bromothymol blue - changes from pH = 6 to pH = 8 (pKa = 7.3) Phenolphthalein - changes from pH = 8 to pH = 1o (pKa = 9.5) [OH-] = √ Kb[OAc] = √ (5.71 * 10-10) (0.100 M) = 7.56 * 10-6 M -log (7.56 * 10-6 M) 14 - 5.12 = 8.88

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