Volumetric Analysis (4) A Silver Nitrate Precipitation Titration.

A common precipitation titration is that involving silver nitrate solution. 

The indicator would be yellow potassium chromate(VI) solution. 

The essence of this titration is to titrate chloride ions with silver ions forming the white precipitate of silver chloride. 

Ag⁺ (aq)    +     Cl^– (aq)    =    AgCl(s)

When all the chloride ions have reacted, any further silver ions will precipitate with the chromate(VI) ions.

A red precipitate of silver chromate (VI) forms and the white precipitate of silver chloride turns red/brown in colour.  

Ag⁺(aq) +    CrO₄ ^2-(aq)    =    Ag₂CrO₄ (s)

This happens of course because silver chloride is less soluble than silver chromate(VI).

So what you are always looking for in this titration is one drop of the silver ions solution producing the first pink tinge to the white silver chloride precipitate.

There are several experiments in the repertoire of college and Advanced level experiments to illustrate this titration technique

e.g. you’ll find it used to determine the moles of water of crystallisation in say barium chloride that is y in BaSO₄.yH₂O.

This particular experiment is delightful because it involves two of the most toxic chemicals available to us in the Chemistry Laboratory viz: silver nitrate solution and hydrated barium chloride solid!!

Furthermore, the calculation of the resulting mole ratio of barium salt to water is a challenge too.

But hey–ho, this is what Chemistry is all about and it all adds to the fun of studying this science.  Would you want it any other way?

Let’s start by working out how to go about this experiment.

We are going to have to weigh out a given mass of hydrated barium chloride, say somewhere in the region of 1.50g

We then take this accurately weighed mass and prepare a standard solution of it dissolving it in distilled water and making up to the mark in a volumetric flask.

We titrate 10mls of this solution with 0.050M silver nitrate solution.

We add about 1g sodium sulphate to the flask before the titration.

We also add the indicator potassium chromate(VI) 2-3drops.

We take particular safety precautions with the materials and we do not wash toxic solutions down the sink but retain them in residue bottles for appropriate disposal. 

Here are the instructions for a typical precipitation titration to estimate chloride ion concentration.

You can download a set of instructions from the University of Canterbury New Zealand site here and watch a youtube video here with the grooviest soundtrack ever to a titration. 

In the video you’ll see how the titration is carried out with a safe silver nitrate burette assembly—interesting.

Here is typical set of results:

First, the weighing of the barium chloride solid:

Mass of weighing bottle and BaCl2 before transfer to flask: (g)	11.79
Mass of weighing bottle  after transfer: (g)	10.21
Mass of BaCl2 sample: (g)	1.58
Mass of BaCl2 in 10ml solution: (g)	0.0632

Second, the titration results:

Pipette solution	Barium Chloride			? mol/dm3		10.0ml
Burette solution	Silver nitrate			0.050mol/dm3
Indicator	Potassium chromate(VI)
Burette rdgs	Rangefinder	1	2	3
Final rdg (ml)	10.40	20.70	30.90	41.20
First rdg (ml)	0.00	10.40	20.70	30.90
Volume used (ml)	10.40	10.30	10.20	10.30
Mean titre (ml)	10.2(7)

Calculation

Well, I’m going to take you through one way of working out the BaCl₂ : water ratio, but please note that this is not the only way of getting to the answer. 

Here we go then:

First, use the titration results to calculate the number of moles of silver ion used in the mean titre.

Use the equation n= cV

where n = number of moles of silver ion

c is the concentration of the silver ion solution and

V is the volume (in litres) of the silver ion solution used (i.e. the mean titre)

So n(moles Ag+ ion) =  0.050  × 10.27  =  0.0005135moles  Ag+ ion

                                                           1000

Second, apply the equation for the precipitation reaction which is:

Ag⁺ (aq)    +     Cl^– (aq)    =    AgCl(s)

Therefore the number of moles of chloride ion in 10ml of the solution from the volumetric flask is also 0.00051moles since 1 mole of silver ion reacts with one mole of chloride ion.

Third, we’ll now try and calculate how many moles of chloride ion were in the volumetric flask altogether. 

If 0.0005135 moles Cl– are in 10ml then 0.0005135 × 25 are in 250ml

0.0005135 × 25   =   0.01284moles  Cl– ion. 

(be careful with the number of zeros you add here!!)

Fourth, we can now calculate how many moles of barium chloride were in the volumetric flask since for every mole of barium chloride two moles of chloride ions are produced.

Therefore moles barium chloride =  0.01284/2  =  0.00642moles

Fifth, if we work out what this number of moles barium chloride weigh then we can get at the weight of water in the barium chloride we added to the flask at first. 

BaCl₂ has the molar mass:   137 + 35.5 + 35.5 = 208g/mole.

Therefore the mass of barium chloride added to the flask was

0.00642  × 208  = 1.335g

Sixth, we can now calculate how much water the sample of hydrated barium chloride contained. 

Hydrated mass – anhydrous mass = mass of water

1.58g  –  1.335g  =   0.244 g  water

Seventh, now we can get at the moles water and moles barium chloride which will give use the ratio we are looking for.

Moles water  =  0.244/18  =  0.0136moles

Moles barium chloride =  0.00642

And we hope that 0.00642   :   0.0136 is about 1 : 2  !!

Which it is roughly!!  Yes y = 2

Here are a couple of questions:

1.    Why was sodium sulphate added to the flask?

2.    What happens to the result if we overshoot the endpoint?

3.    Why is potassium chromate(VI) yellow?

4.    What happens to the white AgCl precipitate on standing in the light?  Why is this what is happening?