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# 0.1000 M We need to convert it to the p scale for use in our titration curve

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(1)

As with other types of reactions, the formation of a precipitate can be used as the basis of a titration.

analyte + titrant precipitate The approach assumes that under the

experimental conditions used, the product is virtually insoluble.

pre-equivalence region

over titration region equivalence

point

panalyte

ml titrant

At this point no chloride has been added so the calculation is pretty straightforward.

[Cl-] = 0.1000 M

We need to convert it to the p scale for use in our titration curve.

pCl = -log[Cl-] = 1.000

(2)

This would be any point after you have started to add some reagent but before reaching the equivalence point.

> 0ml and < 50 ml of titrant for this example.

Obviously, we could calculate an infinite number of points. We’ll look from what happens at 5 ml intervals (5 - 45 ml).

Starting moles Cl- - moles AgCl precipitated total solution volume

[Cl-] =

Since the quantities are small and we’ll be using ml quantities of titrants, its best to use mmoles Starting mmoles Cl- = 50.00ml x 0.1000 M

= 5.000 mmol

Starting moles Cl- - moles AgCl precipitated total solution volume

[Cl-] =

[Cl-] = 5.000 mmol - 0.5 mmol

55 ml = 0.08181 M

We can also calculate the [Ag+] using our KSP expression.

[Ag+] = KSP / [Cl-] = 1.8x10-10 / 0.08181 = 2.20x 10-9

Finally, we need to convert to p values.

pCl = -log(0.08181) = 1.09 pAg = -log(2.20x 10-9) = 8.66 To save time, let’s list the other values.

ml

titrant total [Cl-] pCl [Ag+] pAg 0 50 0.1000 1.00 N/A N/A 5 55 0.0818 1.09 2.20x10-9 8.66 10 60 0.0667 1.18 2.69x10-9 8.57 15 65 0.0539 1.27 3.34x10-9 8.48 20 70 0.0429 1.37 4.20x10-9 8.38 25 75 0.0333 1.48 5.41x10-9 8.27 30 80 0.0250 1.60 7.20x10-9 8.14 35 85 0.0176 1.75 1.02x10-8 7.99 40 90 0.0111 1.95 1.62x10-8 7.79 45 95 0.0053 2.27 3.40x10-8 7.47 Note: in all cases [Cl-] >> [Ag+]

(3)

8 6 4 2

0 0 10 20 30 40 50 60 70

Ag+

Cl-

During most of the titration, there is little change in concentration.

p [ ]

ml titrant

The point where sufficient titrant has been added to be stoichiometrically equivalent to the amount of analyte.

Just enough titrant has been added to react with all of our analyte.

Now, neither the amount of Cl- or Ag+ is in excess.

At this point, [Cl-] = [Ag+].

KSPAgCl = 1.8x10-10 = [ Ag+ ] [ Cl- ] and [ Ag+ ] = [ Cl- ]

so KSPAgCl = 1.8x10-10 = [ Ag+ ]2 [ Ag+ ] = [ Cl- ] = 1.34 x 10-5 pCl = pAg = 4.87

8 6 4 2

0 0 10 20 30 40 50 60 70

In this case, both points have the same value.

This would not be the case if our species did not combine 1:1

In this region, we are adding excess Ag+. Compared to the titrant, our solution contains

essentially no silver so:

[Ag+] = ml excess titrant added total solution volume M titrant x

[Cl-] = KSP / [Ag+]

ml total

excess ml [Ag+] pAg pCl 5 105 0.0048 2.32 7.42 10 110 0.0091 2.04 7.70 15 115 0.0130 1.89 7.85 20 120 0.0167 1.78 7.96 25 125 0.0200 1.70 8.04

Note: in all cases [Cl-] << [Ag+]

(4)

8 6 4 2

0 0 10 20 30 40 50 60 70

Electrodes exist that give a response that is proportional to the concentration of an ion in solution - ion selective electrodes.

The classic example is a pH electrode.

We’ll devote an entire unit to this method later.

For now, we’ll just give a summary.

saturated AgCl/KCl

Ag wire

AgCl

(5)

This approach relies on KSP differences for two insoluble silver salts.

Ag+ + Cl- AgCl (titration reaction) 2 Ag+ + CrO42- Ag2CrO4 (at endpoint) AgCl is much less soluble than Ag2CrO4 so it

will precipitate first.

Ag2CrO4 is brick-red in color so a color change is observed at the endpoint

You can’t simply dump in some Na2CrO4 and expect to get good results.

The point that starts to precipitate is dependent on both the silver and chromate concentration.

The solution must be buffered to prevent formation of dichromate - silver dichromate is quite soluble.

Determine the maximum amount Na2CrO4 to use.

KSP AgCl = 1.8 x10-10 KSP Ag2CrO4 = 1.1 x10-12

We don’t want any to precipitate until after we reach the AgCl equivalence point.

At the equivalence point, [Ag+] = [Cl-] so [Ag+] = KSP AgCl = 1.34x10-5 M

At higher concentrations, Ag2CrO4 starts to form.

KSP Ag2CrO4 = 1.1 x10-12

The maximum chromate concentration that we can have with out forming a precipitate when [Ag+] = 1.34x10-5 M is:

[CrO42-] = KSP / [Ag+]2 = 6.1 x 10-3 M

By trial and error, 2.5 x 10-3 M has been found to work best.

(6)

Ag+ + Cl- AgCl(s) + Excess Ag+ + SCN-

AgSCN (s)

Fe3+

Fe(SCN)2+

(red)

endpoint

The endpoint is not very sharp but gives good results.

AgCl

- - - - - - - - - - -

- - - - - - - - -

Until we the equivalence point is reached, chloride is in excess.

It is our primary absorbed ion.

The outer surface is negative which acts to repel the indicator.

After the equivalence point, silver ion is in excess.

It becomes our primary absorbed ion.

Our indicator can now be attracted to the surface.

AgCl

+ + + + + + + + + + +

+ + + + + + + + +

(7)

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