Acid Base Titration Notes – link to pdf

SCROLL DOWN TO SEE THE 5 POINT NOTES

Stoichiometry (mole ratio) continues with Acid/Base Reactions.  The connection we need to make is that salts that are soluble and that produce a hydroxide ion (OH) into an aqueous solution are bases.  These are important ions because they can upset the balance of H+ and  OH in water.  Remember that these are naturally occurring ions that occur in water.  This called auto-ionization of water.

H2(l)  →    H+ (aq)   +   OH (aq)             Kw =  1 x 10-14
Proton                Hydroxide

or

H2(l)  +  H2(l)  →    H3O+ (aq)   +   OH (aq)        Kw =  1 x 10-14
Hydronium         Hydroxide

From 2 reactions above which are really the same you can see ONE H+ or H3O+ for every ONE OH.

Acids will increase the Hor the  H3O+ concentration above normal* conditions.

Bases will increase the OH concentration above normal* conditions.

*Normal conditions refer to water that is NEUTRAL : (  Hor the  H3O=   OH– )  = pH = 7

H2(l)  →    H+ (aq)        +   OH (aq)
H2(l)  +  H2(l)  →    H3O+ (aq)   +   OH (aq)

Keq =   [Products]   =  Kw = [H+ H3O+][OH]
[Reactants]

*We can use H3Oand H+ interchangeably.  I will write Hgoing forward but remember I can use H3Oat any time and it would mean the same thing.  (Whether you have a free proton or a a proton held by water does not matter)

Now the value of Kw (equilibrium constant for the auto-ionization of water) equals =  Kw = 1 x 10-14
Kw = [H+ ]    x     [OH]

1 x 10-14  = [ H+]     x    [OH]

If one Hdissociates for every ONE OH–  H2(l)  →    H+ (aq)        +   OH (aq)    then:

1 x 10-14    = [ H+]          x        [OH]

1 x 10-14    = [1 x 10-7]     x      [1 x 10-7 ]
As you can see the concentration (molarity, M) of Hand  OH–  ARE THE SAME IN PURE WATER!

They must be the same concentration to equal 1 x 10-14 !
Remember when you multiply exponents you add!

*THIS is WATER AT NORMAL CONDITIONS.

Now we use a convention to measure this balance or unbalance of ions in water.  It is called pH.

pH = – log (base 10) of the H+ or  H3Oconcentration.

Pure water at 25 ºC the [H+] or  [H3O+]  = 1 x 10-7

if you put  -log (1 x 10-7) in your calculator you will get :  pH  = 7

So basically pH is a numerical system that measures “stuffs” the entire concentration (small value) into a exponent (base 10) AND MAKES IT positive for Ease

pH = – log  [H+

**ACID/BASE TITRATION Curve NOTES:

The new skills that we learned from Monday’s/ Tuesday’s Titrations are identifying the parts of a Titration curve.  The striking part of the Titration graph is the asymptotic line that occurs as the pH changes becomes exponential changes as the pH nears the equivalence point.

We learned the endpoint in an Acid/Base titration is really an approximation of the equivalence point.  The endpoint refers to the volume of titrant added (Standard Base Yesterday) and pH that results when the chemical indicator changes color.  This color change will never be exactly at the equivalence point but it approximates the volume of titrant added if the indicator has a color change on the asymptotic line.

Consider 2 titrations like the 2 we did yesterday.  We used phenolphthalein which has a color change (turns from colorless to pink at pH changes of 8 – 9).  What if we used Thymol blue or Methy red?

 Table M Do not forget that these indicators are themselves conjugate acid base pairs    in equilibrium.      In the case of Thymol blue: Notice in the first Titration which you should recognize as a strong acid / strong base titration, with a pH of the equivalence pH of 7, can utilize both Thymol blue and Methyl red as indicators as the endpoints will be on the asymptotic line.  They would have endpoints with approximately the samevolume as the equivalence point.

In the second titration you should recognize as as weak acid / strong base titration because the starting pH is greater and the equivalence point pH is NOT 7!  In this titration the asymptotic line is not quite as long thus Methy Red would no longer be appropriate to use because its color change would occur at MUCH different volume than the equivalence point volume resulting in a very poor approximation of the volume needed by the titrant to neutralize the acid.  It would lead to undervalued concentration of the acid in this case.  THymol blue however would however lead to very good approximation of the equivalence volume because its color change is closer to the equivalence point.

Notice both the Titrations have an equivalence volume of 50 ml of base added, illustrating that both acids are at the same concentration even though the acid in the titration on the right is a weaker acid.  The Strong Base DRIVES the weak acid to COMPLETION like the Strong Acid on the Left does naturally.

Remember the Volume at the equivalence point is what is normally needed to attain the concentration of the solution being titrated  (acid in both cases above and in our lab 10 and 11). This volume of added titrant (chemical with known concentration) is used to determine the concentration of the solution in the beaker under the buret.

The new skill that we added in our graphical analysis of acid base titrations allows us to determine the Ka of acid, or conjugate acid (produced from a base that is titrated).

Remember Ka is really a Keq which is a equilibrium constant that expresses whether the products or reactants are favored.  I have been surprised how many students have asked what this in the past few days?  We have discussed that Keq in terms of thermodynamics when we talked about spontaneity which if you remember is pathway that the universe supports.  A favorable pathway always is supported by a dispersion of energy from a concentrated source (increase in entropy).

Reactants <—>  Products

Since Keq generally = [Products] / [Reactants

Then a Keq greater than  1 is a reaction that has more Products than Reactants at equilibrium.
If there are more products present at equilibrium then THE FORWARD DIRECTION is more favorable pathway then the reverse reaction (which would build up reactants if more favorable).
The forward reaction then is more spontaneous!

Strong Acids have an incredibly large Keq because virtually all of the acid dissociates to leave virtually zero reactants which would drive the Keq to a VERY LARGE NUMBER! Strong Acidsdissociate very spontaneously!

HCl   –>  H+   +    Cl

Keq  = [H+][Cl] /  [HCl]

Keq = VERY LARGE

Weak acids have an incredibly small Keq because virtually all of the acid remains undissociated and thus there are less products and more reactants. Weak acids dissociate very UNspontaneously!

HC2H3O2   < – >   H+   +   C2H3O2

Keq   =    [H+][C2H3O2] / HC2H3O2

Keq  =   Very Small

Now because these reactions are very similar  (HA   — >   H+  +   A– we call them Ka!
Keq  = Ka when we deal with acids dissociating.

We can find the Ka graphically because of the relationship between Ka and what it equals: Using this relationship we know that when we have equal amounts of conjugate Base (A) and conjugate acid (HA)

the log [A] / [HA] will go to zero because the log of 1 = 0 !!

Thus when [A]  =  [HA]   pH = pKa

and we can get the pKa of acid which is just the -log of the Ka.  Once we have the pKa we just

perform the 10-x calculation to attain the Ka  of the acid.

How do we find the point where the conjugate Base (A) and conjugate acid (HA) are equal
We find the half equivalence point!!! So once we establish the equivalence point which is in the middle of the asymptote we attain the volume at that point which in the example to the left is 20 ml.  Now we take the equivalence volume and halve it: 20 ml / 2 = 10 ml 10 ml represents the volume of titrant (base) added to neutralize half of the acid.  This point is where half of the weak acid has been converted to the conjugate base.  So if we had 1 mole of weak acid, .5 moles of the weak acid remains and  .5 moles of conjugate base has been created. This is the point where the acid = conjugate base !                                            pH of this point = pKA
So in the above example the pH at the Half – equivalence point is equal to 3.  Pka of the acid = 3.

The Ka of the acid is thus:    10-3 =    1 x 10-3

This is a Ka of a weak acid because it is less than 1 which means the reactants or the undissociated acid ( HA ) is a larger quantity because the reverse reaction is more spontaneous.

Every ACID / BASE equilibrium calculation is centered around 5 points of a titration curve.

1.  Initial pH of the analyte (chemical being tested)
2.  Half – equivalence point (max buffering position)
3.  Any point in the Buffering region
4.  Equivalence point
5. Final pH of the Titration (extra titrant added beyond eqivalence)

If you can verify OR predict (going backwards) the 5 points on a titration curve you will be ABLE TO COMPLETE ANY ACID BASE EQUILIBRIUM problem.

More Importantly if you can IDENTIFY what point a particular question is asking about you will able to easily identify and solve what are notoriously the most difficult problems on the AP.

So lets take our Weak acid/Strong Base Titration that we were/are working on
to verify these 5 points:

In the titration 25.0 ml of HC2H3O2 was titrated with 0.10M NaOH:

weak acid                                            strong base Skill 1: Determine the net ion reaction

Many students have trouble with the net ion reaction with the acid/base reactions. So many students think that that the net ion reactions of these type of reactions is:  H+  +   OH  —-> H2
But that reaction is only for Strong Acid/Strong Base reactions!  These are very uncommon as most acid/base reactions as are Weak acid / Strong base (like the example above) or Strong acid / Weak Base.

The reason that this is so is because Acid/Base Titrations are reactions that are completion reactions that are driven stoichiometrically by the chemical that is Strong and that most acids or bases that are tested or titrated are weak.  Only a few bases or acids are strong (completely dissociate) and these are used to drive the majority of the acids and bases that are weak to completely dissociate.

In the titration above, a weak acid, acetic acid (HC2H3O2) was titrated with a strong base (NaOH). The strong base FORCED the weak acid to give up its proton (H+) with as many of OH- ions that are added. This is what stoichiometric means and titrations are meant to determine the concentration of the analtye (the weak acid in this case) from the titrant (the strong base in this case).

The reaction is written:
HC2H3O2 (aq)   +   NaOH (aq)  —–>    NaC2H3O2 (aq)   +   HOH (l)

But we need view the reaction as a net ion reaction so lets dissociate the chemicals that are electrolytes.

HC2H3O2 (aq) +  Na+  +  OH (aq)  —–>    Na+ +  C2H3O2 (aq)   +   HOH (l)

Now lets cancel out the ions that are spectating and not part of the overall reaction.

HC2H3O2 (aq) +  Na+  +  OH (aq)  —–>    Na+ +  C2H3O2 (aq)   +   HOH (l)

and we get the net ion reaction:

HC2H3O(aq)  +   OH (aq)  —–>    C2H3O2 (aq)   +   HOH (l)

Skill 2: Determine the concentration of the titrant – chemical that was titrated
The main reason that we do acid/base titrations is to calculate the concentration of the titrant (acetic acid in his case).  Using the titration graph above we must first identify the equivalence point and the volume of base added at this point. Point 4 represents the equivalence point as it is in the  middle of the asymptote. This point represents the volume of   base needed to completely to consume the weak acid. This means stoichiometrically:                moles OH–  = moles of H+ (from weak acid)   To calculate the molarity of the weak acid we need the moles of the H+  and we get it from the moles of OH–  at the equivalence point. Let’s calculate the moles of (OH-):                                   M        x        V           =      moles                             0.10 mol    x   .014 L   =   .0014 mol                                         LThis value was given.      This value wason the graph
.0014 mol of OH = mol of H at the equivalence point

Lets calculate the molarity of the acid with the given volume of weak acid (25 ml) that we started the titration with.

M   =         mole
L

M  =    .0014 mol
.025 L

M  =        .056     =    [HC2H3O2]

Point 1: Initial pH of the Titrant (weak acid in the beaker)-

To verify point 1 we need to understand that the chemical reaction that is occurring in the beaker is different than the net ion reaction we determined in skill 1 above BECAUSE NO Strong Base WAS ADDED YET!

So all we have is weak acid in water:

HC2H3O2 (aq)  +  H2O (aq)   <—–>    H3O(l)    +   C2H3O2 (aq)

Now all we need to verify point 1 on the graph is to find the  [H3O+]
and then -log that value to obtain the pH of the initial solution.

BUT we need to perform an equilibrium table to solve for how much [H3O+]  was produced at equilibrium. Remember acetic acid,  HC2H3O2  , is a weak acid and will not completely dissociate (like a strong acid).
We need to use the weak acids Ka (equilibrium constant for acids) to help solve for how much this acid moves forward.
The larger the Ka, the more products or [H3O+]  will be present!

So lets write the equilibrium table:

We need molarities for equilibrium tables and we need the initial concentration of the weak acid that we solved for in Skill 2.
HC2H3O        +       H2O         <—–>      H3O+       +       C2H3O2

Initial:          .056   M

Change:              – x                                                                 +x                             +x

at Equilibrium:        .056 – x                                                             x                                x

Lets solve for x =  [H3O+ ]

Ka    =    x2 /  .056 – x

approximate:                                    Ka    =    x2 /  .056 – x   ——> WE ARE approximating here!!!
very small Ka for weak acids!!

Ka    =    x2 /  .056

Ok all we need is the Ka of the acid and we are on are way but we need to look at point 2 to determine the Ka of the weak acid and then solve for x.

From point 2 we solved for the Ka = 3.16 x 10-5

3.16 x 10-5    =    x2 /  .056

x   = .00133  = [ H3O+ ]

– log [.00133] = pH = 2.9

The pH of the initial concentration of weak acid in water = 2.9

Point 2:  Half Equivalence Point – (Determine the Ka of the acid)

At point 2, which is the the half equivalence point, we can determine the Ka of the acid. Remember that equilibrium constants have a lot of names and for reactions of acids dissociating we call it an Acid Dissociation Constant, Ka.  It is still an equilibrium constant which indicated how much the products or reactants are favored at equilibrium.  Since this is a weak acid, we expect that our Ka to be very small as weak acid barely dissociate (that is why we need a strong base to force them.)
Our titration graph provides us a way to obtain the Ka from point 2.

Before we can attain the Ka of out acid we need to identify the half-equivalence point, which represents the point where half of the acid has been consumed by the strong acid OR half the amount of base needed to titrate (drive the reaction forward) the weak acid.

When we look at the titration graph above we always determine the equivalence point (point 4) first and then determine the equivalence volume at that point.  That is the volume of base needed to completely titrate ( consume or neutralize) the weak acid.  In that equivalence volume we determined the moles of base needed to completely titrate the weak acid in skill 1. The amount of moles of base that would half-titrate the moles of weak acid would be the amount of volume of base THAT IS HALF the volume of base added at the equivalence point. The amount of moles of base that would half-titrate the moles of weak acid would be the amount of volume of base THAT IS HALF the volume of base added at the equivalence point. So from our titration graph, 14 ml of base was needed to be added to the weak acid to completely neutralize the weak acid AND thus point 2 is the half – equivalence point as its volume of 7 ml is half the volume of the equivalence point (14 ml). I call this point the max buffer position as it represents the point where equal amounts of acid and conjugate base are present.   In order to understand how point 2 can provide us with the Ka of the weak acid we need to understand how the weak acid and it conjugate acid are both present in equal amount.
Let’s look at why the amount of the conjugate acid and base are present in equal amounts at the half -equivalence point. We need to view a stoichiometry table to understand this.

From skill 2 we calculated that .0014 moles of weak acid was initially present .

If we have half the amount of volume needed to consume the weak acid then we would have half the moles of base to match the weak acid and thus the amount of moles of base must be .0007 mole at the half-equivalence point.

THIS is a STOICH TABLE !!  NO Equilibrium!!! REACTION IS DRIVEN FORWARD by the STRONG BASE!

conjugate acid                                                 conjugate base
HC2H3O     +      OH       —–>      C2H3O2       +      H2O

Initial:           .0014 mol            .0007  mol                       0

Change:          .0007 mol        .0007 mol                 + .0007 mol

End:            .0007 mol                 0                          .0007  mol

From the stoichiometry table above we can see that we will have the same amount of moles of conjugate base as conjugate acid IF WE ADD HALF THE NEEDED BASE to drive the reaction to its half-way point = half equivalence point.
This what happens at point 2!

So how does this give us the Ka?  If we use the Henderson- Hasselbach equation that I derived earlier we can see how.

conjugate base     conjugate acid
pH = pKa + log [ C2H3O2 ] /  [HC2H3O2 ]
If we plug in the moles into the formula for the conjugate base and acid we will see how the Ka is determined.  (If you notice there are brackets in the equation which require a molarity but because both concentrations would have the same total volume, the volumes would cancel out, leaving moles.)
conjugate base     conjugate acid
pH = pKa + log [ .0007  mol ] /  [ .0007 mol ]

Equal moles of conjugate base and conjugate acid would make the equation have a log of 1.

pH = pKa + log  [   1   ]

A log of 1 = 0 !

pH = pKa           (at half – equivalence point!)

So from our graph point 2 reveals that the pH at the half – equivalent point is approximately 4.5 thus the pKa is 4.5!

All we have to do now is convert the pKa into the Ka (without the little p!).  The little p = – log and thus a pKa is really an exponent base 10:
1 x 10 4.5

Lets convert this to a non exponent (Ka) by using the 10 key on our calculator.

We must type:  10 -4.5  because the pKa is the NEGATIVE log of the Ka .

pKa = 4.5   —–>    10 -4.5    =   Ka = 3.16 x 10-5

This makes sense as the weak acid should have a equilibrium constant that is less than 1 as more reactants are present than products.

You can now return to complete point 1 as we now have the Ka we need to complete the problem/

Point 3: (on the buffer region)

Previously we determined the initial concentration of the analyte (weak acid: HC2H3O2 )
to be 0.056 M. To solve for this point we need to realize that we are using the DRIVEN reaction:

HC2H3O     +      OH       —–>      C2H3O2       +      H2O

This is a driven stoichiometric reaction because the weak acid would not give up its proton without the strong base ( OH)  pulling it off as many protons as there are weak acid molecules ( HC2H3O2 ).
Because there is no equilibrium here we need to figure out how much weak acid is left if add only 11 ml of base.

Remember that at equivalence we needed 14 ml to completely neutralize the weak acid. At point 3 we are below that value thus we have more weak acid than strong base.  This means that there will be extra acid left over.  What is the pH with the remaining acid that was not neutralized ?

To answer this we need to use a stoichiometric table THAT ALWAYS uses moles!!

Before we complete we have get the moles of the acid in the beaker.  We obtained this in another step when we calculated the original concentration of the acid (0.056 M)

M      x    V             =   moles
(0.056 mol/L) (.025 L)      =  .0014 mole  HC2H3O

At point 3,  11 ml of base (OH-) was added:

(0.10 mol/L) (.011 L)  =  .0011 mole  OH
given

HC2H3O     +      OH       —–>      C2H3O2       +      H2O

Initial:         .0014 mol          .0011 mol                           0

Change:        .0011 mol         – .0011 mol                    + .0011 mol

End:         .0003 mol                 0                              .0011 . mol

Lets convert these values to Molarities so that we can now complete a equilibrium ICE table because
now we have just weak acid and weak conjugate base in water!

At point 3, 11 ml was added to the original 25 ml so :

11 ml +  25.0 ml = 36 ml is the total volume of the solution at point 3.

.0003 mol HC2H3O2 / .036 L = .0083 M

.0011  mol  C2H3O2 / .036 L =  .0306 M

New reaction in water: (all the  OH– the base has been consumed!)
This is an equilibrium reaction!

HC2H3O        +       H2O         <—–>      H3O+       +       C2H3O2

Initial:         .0083 M                                                                                      .133 M

Change:              – x                                                                 +x                          +x

at Equilibrium:          .0083 – x                                                              x                        .133 + x

In step 2 we calculated the Ka of the weak acid and determined it to be: Ka = 3.2 x 10-5

[H3O+ ] [ C2H3O2 ]
Ka = 3.2 x 10-5    =     ________________
[HC2H3O2 ]
plug in values from table and approximate to prevent the quadratic catastrophe!

[x ] [ .0306 + x]
Ka = 3.2 x 10-5    =   ________________
[.oo83 – x]

8.68 x 10-6  =    [ x ]  =  [H3O+

[H3O+ ]  =  8.68 x 10-6 ——-> -log  8.68 x 10-6  pH = 5.07

Is there an easier way?

Yes Henderson – Hasselbach!
If I use the molarities that I just calculated in the the above equilibrium table I get:

pH = pka + log [ C2H3O2 ] /  [HC2H3O2 ]

pH = 4.5 + log [ .0306 ] /  [.0083 ]

pH =   4.5 + .57 = 5.07

If I just use the moles of the HC2H3O (conjugate acid) and that of  C2H3O2  (conjugate base)  that we calculated in the stoichiometry table that began the problem then this problem is even easier.

pH = pka + log [ C2H3O2 ] /  [HC2H3O2 ]

pH = 4.5 + log [ .0011 . mol ]  /  [ .0003 mol ]

pH = 4.5 + log [ 3.66]

pH = 4.5 + .56

pH = 5.06
**So the quickest way to solve for any pH in the buffer region (just after the initial pH and just before the equivalence point) is to use moles in the Henderson-Hasselbach equation.

We can use moles because the molarities of both of the conjugate acid and conjugate base are in the same volume and that volume will cancel.

Point 4:  Equivalence pH

Lets verify the pH at the equivalence point.  ONLY in a Strong Acid / Strong Base Titration will the final pH be 7 and we should understand why! Strong acids and Strong Bases produce conjugate bases and acids that DO NOT IONIZE water. That is they do not have the ability give or take a proton from water.

Since almost all titrations deal with one weak component then most titrations have a pH at the equivalence point THAT is above or below a pH of 7.

In our example, with acetic acid (a weak acid)  being titrated with a strong base we would expect the equivalence pH to be higher than 7 because at the equivalence point 100% of the acid has been converted into the conjugative base.

The driven reaction when we are added base to the weak acid is:

HC2H3O     +      OH       —–>      C2H3O2       +      H2O

But at the equivalence point there is no more weak acid or strong base as both have been consumed and thus the reaction is just the conjugate base in water that reaches an equilibrium:

C2H3O2       +      H2O     <—–>     HC2H3O     +      OH
conjugative base                                                                  conjugative acid

And so the conjugate base reacts with water to reform the conjugate acid AND hydroxide ion until it reaches equilibrium (notice the double arrow above).  The production of hydroxide ion will cause the pH to be greater (basic) than 7.

*Keep in mind if we were titrating a weak base with a strong acid, the equivalence pH will be below 7 (acidic) due to the production of conjugative base and hydronium ion (H3O+).

conjugative acid     +      H2O   <——>     conjugative acid     +     H3O+

Okay back to the problem and reaction at the equivalence point:

C2H3O2       +      H2O     <—–>     HC2H3O     +      OH
conjugative base                                                                  conjugative acid

So now that we understand that we will have 100% conjugate base reacting with water to reach equilibrium we should be able to understand that we need to solve for how much hydroxide is being formed at equilibrium.

We need an equilibrium table and remember that we need Molarity in our equilibrium tables and we need the initial concentration of the conjugative base at the equivalence point:

If you remember from Skill 2 , when we determined the initial concentration of of the weak acid we determined that the moles of the weak acid:

.0014 mol of OH = mol of H+ (weak acid) at the equivalence point

Given the stiochiometry (the ratios) in the driven reaction:    HC2H3O+  OH  —–>   C2H3O2  +   H2O
1         :                 1          :         1

The moles of the weak acid ( HC2H3O2) initially will equal the moles of the conjugate base ( C2H3O2 ) produced at the equivalence point because Strong Base (OH)  drove the reaction to completion!

So  .0014 mol of (weak acid) = .0014 mol of C2H3O2 )

Now we need to convert this to a molarity to complete the equilibrium table and we have to remember that at the equivilance point 14 ml of Base was added to the analyte (beaker with 25 ml of weak acid).  Thus the total volume of solution is:

25 ml initially + 14 ml of base added = 39 ml of total solution

solve:                                                        M = .0014/.039 L  = .036M

C2H3O2       +      H2O     <——>     HC2H3O     +      OH

Initial:                 .036M                                                                    0                             0

Change:                       – x                                                                              +x                           +x

Equilibrium:               .036M – x                                                                  x                              x

Now we solve using the

The Keq (equilibrium constant) for the reaction will tell us how forward the reaction moves forward to reach equilibrium.  Now the Keq will have to be the Kb because this reaction is the conjugate base reacting in water, which is the reverse of the original reaction.

We already have the Ka from point 2.                     Ka = 3.16 x 10-5

We use the formula:
Ka   x    Kb  =   Kw

3.16 x 10-5    x    Kb  =   1 x 10-14

Kb    1 x 10-14
3.16 x 10-5

Kb  =   3.16 x 10-10

___________

There are important implications of this formula.  Remember the stronger the Acid (higher the Ka) the weaker the conjugate base OR the weaker the acid (smaller Ka) the stronger the conjugate base (as in this case):

Table 1.2
conjugate acid                                            conjugate  base
Ka = 3.2 x 109                          HCl (aq)    —-> H+ (aq)     +            Cl  (aq)                      Kb = 3.2 x 10-24
Ka = 3.16 x 10-5                     HC2H3O2 (aq)    —-> H+ (aq)    +   C2H3O2 (aq)               Kb =  3.16 x 10-10
Ka =   5.6 x 10-10                     NH4+ (aq)    —->  H+ (aq)   +           NH3 (aq)                      Kb =  1.79 x 10-5

As the acid gets weaker (smaller Ka) moving down the list its conjugate base gets stronger (larger Kb).

So the weakest acid in the group above (NH4+) has the strongest conjugate base (NH3) meaning that conjugate acid/base pairs have acid/base properties that are inverse to each other.  This means that conjugate acid base pairs do not act on each other AND this means that they can be added to a solution and remain as an available acid and base in the same solution.  We call these solutions buffer solutions.  They are able to absorb H+ (protons) form an added base and also absorb OH- from added bases.  They prevent pH changes or they Buffer the solution from changing their pH.  Point 2, in our graphs is the max buffering position BECAUSE it has equal amount of acid and base.

So a buffer is identified as conjugate / acid base pairs ONLY !
And they do not include Strong Acids as their Conjugate Bases because they are too weak to act as bases. That is why strong acids ARE Strong because their conjugate base has no ability to act as a base and reform the acid.

Buffers from table 1.2 above include:
log                                                    pKa
HC2H3O2 , C2H3O2             Ka = 3.16 x 10-5        ———>     4.5
NH4+ , NH3                     Ka =   5.6 x 10-10     ———>     9.3
The pKa (-log of the Ka) gives us the pH of the max buffering position THUS the HC2H3O2 / C2H3O2 buffer will best buffer a solution at a pH of 4.5 while the NH4/ NH3  buffer will best buffer a solution at a pH of 9.3

Remember when we did acid / base titrations in class and the pH probes needed to be calibrated. I calibrated them with a buffer solution that was buffered at a pH of 7: This buffer was a pre- made tablet that I dissolved in 100 ml of distilled water. That tablet was made from:                                                         Sodium Phosphate – Dibasic                                         Na2HPO4   → 2Na+   + HPO4-2                                                         Potassium Phosphate – monobasic                                            KH2PO4   → K+   + H2PO4-1
So the buffer is made from H2PO4-1 / HPO4-2

A conjugate acid / base pair!!!!

H2PO4-1 (aq)  ——>   H+(aq)   +   HPO4-2  (aq)

Can you guess what the Ka and pKa of the acid is?                         pka = 7 silly!
__________

Okay back to the problem of point 4 (equivalence pH)

We calculated the Kb of the conjugate base to be:  Kb  =   3.16 x 10-10
And we had the following Equilibrium Ice Table for the reaction that occurs at the equivalence point due to 100% conjugate base being present when all of the weak acid (HC2H3O2)was driven to completion by the Strong Base (OH): This conjugate base will now react with the water (most predominant molecule) to reform some conjugate acid and SOME OH- (which will make the pH at equivalence basic).
C2H3O2       +      H2O     <——>                  HC2H3O     +      OH

Initial:                 .036M                                                                    0                             0

Change:                       – x                                                                              +x                           +x

Equilibrium:               .036M – x                                                                 x                             x

Write the equilibrium expression:            Kb =    [HC2H3O2] [OH]
[.036M – x ]                 approximate the x away!

3.16 x 10-10 =                [X]2
[.036M

3.37 x 10-6     =     [X]     =     [OH]

-log  3.37 x 10-6    =    5.47   =      pOH

pKw =   pH  +   pOH

14  =   pH  +   5.47 pH = 8.53  (verified pH at point 4!) So the equivalence point with a weak acid / strong base titration results in a pH of 7 due to production of conjugate base THAT has some ability to remove protons form water. That some ability is measured by its Kb! We predicted that the equivalence point would be Basic in November but now we can calculate the actual pH with our new equilibrium skills.

Point 5:  Final pH at the end of the titration.

Okay the final point that we will verify from our titration does not include any equilibrium problems. It was a point that we were able to do early in the year. It is pure stoichiometry.  The key with determining this point is understanding that the final pH represents just the “extra” titrant (strong base in this case = OH). Remember that at point 4 (equivalence point) we drove away all the moles of weak acid with exactly the moles of OH added. So at point 4 we have no weak acid or strong base left and since there is no acid remaining, then that volume of OH– added from point 4 to point 5 is now just remaining in the total solution added.So from the graph a total of 22 ml of extra OH– was added.                       Point 5          –           Point 4        =  22 ml                    36 ml                           14 ml

Lets calculate the [OH-] :  Remember that we used 0.10 M of OH–

M =    mol of OH
Liter of solution

We need moles of OH- that was “extra” (22 ml):   (0.10M)(0.022L) = 0.0022 mol OH

We need the Total volume of the final solution:  Remember that 25 ml of HC2H3Owas titrated.

25 ml of  HC2H3O2  +  36 ml of total OH- added   =   61 ml of final solution

Lets calculate the [OH-] :

[OH-]   =  M =     0.0022 mol of OH
0.061 L

[OH-]   =   0.036 M

Calculate the – log of 0.036 = 1.44 = pOH

pKw =   pH   +   pOH

14   =    pH    +    1.44

The verified pH at point 5 = 12.66