Titrations Acid-Base Part 2.

Titrations Acid-Base Part 2

Acid-Base Titration The process of adding a strong acid or a strong base to a titrate of acid or base with the purpose of neutralizing the substance in the beaker Looking to reach equivalence point  mol H+ = mol OH-

Acid-Base Titration Generally done with a pH probe, which is used to create a graph of pH vs. volume added pH Volume added

Determination of the Equivalence Point
Ways to find the equivalence point Can find it stoichiometrically using stoichiometry for the neutralization reaction mol H+ = mol OH- stoichiometrically

Determination of the Equivalence Point
Ways to find the equivalence point Can find it graphically based on pH vs. volume graph When the titrate is neutralized completely by the titrant, the pH changes dramatically The midpoint of this dramatic change is the equivalence point – the middle of the area where neutralization is happening

Use of an Acid-Base Indicator
If the pH at the equivalence point is known, an indicator can be used Want the pKa of the indicator to match the pH at the equivalence point Indicator will then change color as pH changes dramatically, giving visual end to the titration

Things to KNOW about Titrations
Need to be able to EXPLAIN the shape of the curve at ALL points through the titration What happens initially after a small amount of acid/base added When excess of the titrant still remains What happens near the equivalence point What happens beyond the equivalence point – when excess of the strong acid or strong base used as a titrate is added

Things to CALCULATE about Titrations
Calculate the initial pH of acid/base in the beaker Normal acid/base calculation for strong or weak acid/base Calculate the pH at the halfway point of the titration Or use pH at halfway point to find Ka or Kb of substance titrated Calculation of the pH at the equivalence point OR using the equivalence point to find the concentration of unknown acid/base you are titrating Calculation of the pH at a point beyond the equivalence point

Five Types of Titrations
Strong acid titrated with a strong base Strong base titrated with a strong acid Weak acid titrated with a strong base Weak base titrated with a strong acid Polyprotic acid titrated with a strong base

Strong Acid titrated with Strong Base
Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find initial pH

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find initial pH pH = - log [H1+] pH = - log (0.200 M) = 0.699

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH pH Volume added Starts at LOW pH (strong acid in beaker)

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH at halfway point

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH at halfway point L H1+ x (0.200 mol H1+ / 1 L) x (1 mol OH1- / 1 mol H1+) x (1 L / mol OH1-) = L OH1- (Needed to Reach Equivalence Point) = L OH1- / 2 = L OH1- (Needed for Halfway Point)

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH at halfway point 0.050 L OH1- x (0.100 mol OH1- / 1 L) x (1 mol H1+ / 1 mol OH1-) = mol H1+ 0.010 mol H1+ initially – mol H1+ consumed = mol H1+ remain mol H1+ / 0.10 L total volume = M H1+

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH at halfway point 0.050 M H1+ pH = - log [H1+] = - log (0.050 M H1+) = 1.3

Upon adding strong base, there is a very SMALL change as long as there is still excess acid. pH Volume added A very small change with excess strong acid still present

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH at equivalence point

As the equivalence point is reached, there begins to be a dramatic jump in pH as moles acid = moles base pH Volume added Dramatic change

Even just beyond the equivalence point, the addition of excess base leads to a basic pH pH Volume added Excess base now = jump to high pH Midpoint = equivalence point = 7

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH after mL of NaOH has been added

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH after mL of NaOH has been added 0.150 L OH1- x (0.100 mol OH1- / 1L) x (1 mol H1+ / 1 mol OH1-) = mol H1+ Excess of mol OH1- mol OH1- / 0.2 L = M OH1-

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH after mL of NaOH has been added 0.025 M OH1- pOH = - log [OH1-] = - log (0.025 M OH1-) = 1.6 14 – pOH = pH = 14 – 1.6 = 12.4

As more excess strong base is added, there is NOT a dramatic change in pH pH Volume added Same excess base = no dramatic change in pH

Strong Base titrated with Strong Acid
OPPOSITE!!!!! To find initial pH – Just based on [OH-] since it dissociates completely To find pH at halfway point – Find moles of both, the acid added will be limiting Moles of excess base / total Liters = M of [OH-]  use to get pH To find pH at equivalence point – Since mol [OH-] = mol [H+] and no significant conjugates – pH = 7 To find pH beyond equivalence point – Find moles of excess acid added / total L = M of [H+]  can get pH

Strong Base titrated with Strong Acid
OPPOSITE!!!!!

LAB Create a titration curve
If molarity of original acid isn’t known, you can use the curve to find its molarity. HOW??? That’s what you’re designing!!!

WEAK Acid titrated with Strong Base
Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find initial pH

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find initial pH HC2H3O2 + H2O   H3O1+ + C2H3O21- I M C - x x x E M – x x x

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find initial pH 1.8 x 10-5 / 0.1 < 5% x2 / 0.1 = 1.8 x 10-5 X = [H1+] = M pH = - log ( M) = 2.89

Still starts at a low pH (though not quite as low) with weak acid in beaker pH Volume added Starts at LOW pH (weak acid in beaker)

The REACTION that occurs is now slightly different: HC2H3O2 + OH-  C2H3O H2O Makes a GOOD conjugate base now that also influences the pH and

The REACTION that occurs is now slightly different: HC2H3O2 + OH-  C2H3O H2O With not all weak acid reacted and good conjugate base also formed, it creates a BUFFER

Initial spike in pH due to formation of the conjugate base. Then mostly levels off as with both weak acid and conjugate base you have a BUFFER. pH Volume added Jump = conjugate base made, then buffer region

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH at halfway point

HALFWAY POINT Ratio of weak acid to conjugate base is the same
Half of weak acid used up, equal amount of conjugate base formed pH = pKa!!!

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH at equivalence point

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH at equivalence point C2H3O21- + H2O   OH1- + HC2H3O2 I M C x x x E M – x x x

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH at equivalence point kb = 5.6 x 10-10 5.6 x / < 5%

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH at equivalence point x2 / = 5.6 x 10-10 x = 5.3 x 10-6 M = [OH1-] pOH = - log (5.3 x 10-6) = 5.3  pH = 8.7

As the equivalence point is reached, there again is the point where the base neutralizes all of the acid and there is a significant jump in pH. The pH at the equivalence point is GREATER than 7 since the GOOD conjugate base remains. pH Volume added Dramatic jump up as equivalence point is reached again Halfway point: pH = pKa

Even just beyond the equivalence point, the addition of excess base leads to a basic pH pH Volume added Excess base now = jump to high pH Midpoint = equivalence point = GREATER THAN 7

As more excess strong base is added, there is NOT a dramatic change in pH pH Volume added Same excess base = no dramatic change in pH

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH after mL NaOH added

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH after mL NaOH added 50.0 mL of NaOH is excess As NaOH is a stronger base than acetate and is a strong base, the pH will be solely based on the NaOH concentration.

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH after mL NaOH added L OH1- excess x M OH1- = mol OH1- mol OH1- / L total volume = M OH1- pOH = - log (0.033 M) = 1.48  pH = 12.52

WEAK Base titrated with Strong Acid
OPPOSITE!!!!! To find initial pH – To find pH at halfway point – To find pH at equivalence point – To find pH after equivalence point -

WEAK Base titrated with Strong Acid
OPPOSITE!!!!!

LAB Create a titration curve
If the identity and the Ka / Kb of a weak acid or a weak base are not known, you can determine it. HOW??? That’s what you’re designing!!!

Polyprotic Acids Acids that have more than one H+
That means there will be more than one equivalence point!! Each H+ reacted must be neutralized with a stoichiometric amount of base Example – H2CO3 with NaOH

LAB Titration of a polyprotic acid with a strong base.
What would this look like? Why would it look that way?

Titrations Acid-Base Part 2.

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