Precipitation Titration Calculations Video Example Here we’ll be given some data from a titration...

Precipitation Titration

CalculationsVideo Example

Here we’ll be given some data from a titration and asked to use this data to calculate the concentration of an ion in a sample.

nA

We’ll start by looking at how we handle titration calculations in general. Titration calculations in Chem 12 involve the reaction between two reactants, which we’ll call A and B here.

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

n molesc molar concentration (mol/L)m mass (g)V volume (L)

nA

In the center of all titration calculations are the moles of reactant A and the moles of reactant B. We represent number of moles in chemistry by the letter n.

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB


nA

Reactant A represents the reactant that we are given enough information to find the number of moles of.

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB


Given enough

information to find

nA

To convert from moles of A to moles of B, (click) we Always use the mole ratio, or coefficient ratio of B to A in the balanced equation for the titration reaction.

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


nA

The information we are given about A could be the molar concentration of A ( represented by the letter c)

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A

n molesc molar concentration (mol/L)V volume (L)m mass (g)

nA

and the volume of A in Litres, represented by the letter V

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


nA

or it could be the mass of A in grams, represented by the letter m.

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


nA

Whatever we’re given, step 1 of a titration calculation is to convert what we’re given to moles of reactant A, OR N (A)

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


1

1

nA

Step 2 of any titration calculation is to convert moles of reactant A to moles of reactant B

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


2

nA

This is done using the mole ratio or coefficient ratio of B to A in the balanced equation.

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


2

nA

We could be asked one of 3 different things for reactant B. We could be given the volume of B and asked to find its molar concentration c (B).

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


?

nA

We could be given the concentration of B and asked to find its volume V (B).

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


?

nA

Or we could be asked to find the mass of B in grams.

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


?

nA

Step 3 in any titration calculation is to convert moles of B to whatever we’re asked for: concentration of B, volume of B, or mass of B.

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


3

3

nA

So here is a generic diagram that outlines the possible steps to take in most titration calculation problems. Remember the first step is always to find moles of what we can.

nB

mA

cA & VAcB (given

VB)VB (given

cB)mB

Mole ratio:B to A


Let’s do an example precipitation titration question.

Example of a Precipitation Titration Calculation Problem

A 50.0 mL sample of a solution known to contain chloride (Cl–) ions is titrated with 0.100 M AgNO3 solution. A small amount of sodium chromate is added to the sample as an indicator.

0.100 M AgNO3

A few drops of

Na2CrO4(aq)

A 50.0 mL sample of a solution known to contain chloride (Cl–) ions is titrated with0.100 M AgNO3 solution. A small amount of sodium chromate (Na2CrO4) indicator is added to the sample.

50.00 mL of a solution containing Cl–

Three separate trials are done.

0.100 M AgNO3


A few drops of

Na2CrO4(aq)

A 50.0 mL sample of a solution known to contain chloride (Cl–) ions is titrated with0.100 M AgNO3 solution. A small amount of sodium chromate (Na2CrO4) indicator is added to the sample. Three separate trials are done.

We’re asked to find the concentration of Cl minus in the original sample. The results are recorded in a table.

0.100 M AgNO3


A few drops of

Na2CrO4(aq)

A 50.0 mL sample of a solution known to contain chloride (Cl–) ions is titrated with0.100 M AgNO3 solution. A small amount of sodium chromate (Na2CrO4) indicator is added to the sample. Three separate trials are done.

Find the [Cl–] in the original sample

Like this. The first thing we need to do is calculate the volume of AgNO3 solution used in each trial.

A 50.0 mL sample of a solution containing Cl– ions is titrated with 0.100 M AgNO3. The net ionic equation for the precipitation reaction is: Find the [Cl–] in the original sample.

(aq) (aq) (s)Ag Cl AgCl

[AgNO3]=0.100 M

Volume of Cl– solution = 50.0 mL

Trial 1 Trial 2 Trial 3

Initial buret reading (mL) 0.95 4.45 7.65

Final buret reading (mL) 4.46 7.65 10.87

Volume of AgNO3 used (mL)

We do that by subtracting the initial buret reading from the final buret reading. So in Trial 1, it is 4.46 minus 0.95

[AgNO3]=0.100 M





Volume of AgNO3 used (mL)



4.46 – 0.95 = 3.51

Which is 3.51 mL

[AgNO3]=0.100 M





Volume of AgNO3 used (mL) 3.51



4.46 – 0.95 = 3.51

For trial 2, the volume is 7.65 minus 4.45

[AgNO3]=0.100 M





Volume of AgNO3 used (mL) 3.51



7.65 – 4.45 = 3.20

Which is 3.20 mL

[AgNO3]=0.100 M





Volume of AgNO3 used (mL) 3.51 3.20



7.65 – 4.45 = 3.20

And in Trial 3, the volume used is 10.87 minus 7.65

[AgNO3]=0.100 M





Volume of AgNO3 used (mL) 3.51 3.20



10.87 – 7.65 = 3.22

Which is 3.22 mL

[AgNO3]=0.100 M





Volume of AgNO3 used (mL) 3.51 3.20 3.22



10.87 – 7.65 = 3.22

Taking a look at these three results, we see that the volume used in Trial 1, 3.51 mL, is considerably higher than the 3.20 and 3.22 used in trials 2 and 3, respectively.

[AgNO3]=0.100 M








Considerably higher than 3.20 and 3.22

For that reason, we just discard the value of 3.51

[AgNO3]=0.100 M








DISCARD

We calculate the best average volume of AgNO3, by taking 3.20 plus 3.22 and dividing by 2, which gives us

[AgNO3]=0.100 M








Best Average = 3.20 3.2

22

23. 1

3.21 mL

[AgNO3]=0.100 M









3.212

We’ll make a note of the average volume of 3.21 mL up here in the table.

[AgNO3]=0.100 M


Volume of AgNO3 used:3.21 mL








22

23. 1

We’ll convert the 3.21 mL to 0.00321 L

[AgNO3]=0.100 M


Volume of AgNO3 used:3.21 mL = 0.00321 L



At this point, let’s dissociate the AgNO3 here

[AgNO3]=0.100 M





And we get Ag+ and NO3 minus

[Ag+ NO3–] = 0.100 M





And we’ll also dissociate the AgNO3 here…

[Ag+ NO3–] = 0.100 M





Giving us Ag+ and NO3 minus

[Ag+ NO3–] = 0.100 M


Volume of Ag+ NO3– used:

3.21 mL = 0.00321 L



The nitrate ion, NO3 minus is a spectator ion. It does not form any precipitates.

[Ag+ NO3–] = 0.100 M



3.21 mL = 0.00321 L



So we’ll just discard it.

[Ag+ NO3–] = 0.100 M



3.21 mL = 0.00321 L



So we can simply say that the concentration of Ag+ is 0.100 molar

[Ag+] = 0.100 M


Volume of Ag+ used:3.21 mL = 0.00321 L



And the volume of Ag+ solution used is

[Ag+] = 0.100 M





Equal to 0.00321 L

[Ag+] = 0.100 M





So we have all the information we need up here now. We’ll just rearrange it a bit…

A 50.0 mL sample of a solution containing Cl– ions is titrated with 0.100 M AgNO3. The net ionic equation for the precipitation reaction is: Find [Cl–] in the original sample.


[Ag+] = 0.100 MVolume of Cl– solution = 50.0 mL

Volume of Ag+ used: 3.21 mL = 0.00321 L

So it looks like this.

[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L




We have the concentration of Ag+ and the volume of Ag+ here.



Ag Ag & c V



And the volume of Cl minus here.



Ag Ag & c V Cl

V



At this point, we’ll convert the 50.0 mL of Cl minus solution


Volume of Cl– solution = 50.0 mL = 0.0500 L

Ag Ag & c V

ClV



To 0.0500 litres


Volume of Cl– solution = 50.0 mL = 0.0500 L

Ag Ag & c V

ClV



And leave it like this.


Volume of Cl– solution = 0.0500 L



Here is the generic diagram we came up with for titrations.





We are given the concentration and volume of Ag+





And we’re asked for the concentration of the other reactant, Cl minus





And we’re given its volume.





We’ll start with the concentration and volume of Ag+





Ag Ag Ag Cl Cl& c V n n c

Convert that to moles of Ag+





AgA Cl Cg Ag l& nc cV n

Then find moles of Cl minus





Ag Ag Ag Cl Cl& c cV n n

And use that and the given volume, to calculate the concentration of Cl minus





Ag Ag Ag l CC l& c cV n n

We can do the first 2 steps using conversion factors.





Ag Ag Ag l CC l& c cV n n

We take the concentration of Ag+, which is 0.100 mol Ag+ per 1 litre Ag+





Ag Ag Cl ClAg& Vc n n c

Cl

1 mol Cl0.00321 L Ag

0.100 mol Ag

1 L Ag0.000321 mol Cl

1 mol Agn

and multiply it by the volume of Ag+ used, which is 0.00321 L of Ag+





Ag Ag Cl CAg l& c nV n c

Cl0.00321 L Ag

0.100 mol Ag 1 mol Cl0.000321 mol Cl

1 L Ag 1 mol Agn

If we cancel out the unit, litres of Ag+






ClL Ag

0.100 mol Ag 1 mol Cl0.00321 0.000321 mol Cl

1 1 mL Ag ol Agn

we’re left with moles of Ag+






Cl

0.100 1 mol Cl0.00321 L Ag 0.000321 mol Cl

1 L Ag 1 mo

mol

l A

A

g

gn

so evaluating this would give us moles of Ag+, or n Ag+





AgAg Ag Cl Cl& V nc n c

Cl

1 mol ClL Ag

0.100 mol Ag0.000321 mol Cl

1 L Ag 1 mol A0.00321

gn

However, rather than stopping here, we’ll add another conversion factor to get the moles of Cl minus, or n Cl minus





AgAg Ag CCl l& c V cn n

Cl

0.100 mol Ag 1 mol Cl0.00321 L Ag 0.000321 mol Cl

1 L Ag 1 mol Agn

Looking at the coefficients of Ag+ and Cl minus in the balanced net ionic equation, the mole ratio is 1 mole of Cl minus to 1 mole of Ag+.





AgAg Ag CCl l& c V cn n

Cl

0.100 mol Ag0.00321 L Ag

1 mol Cl

1 mol Ag0.000321 mol Cl

1 L Agn

We’ll cancel the unit moles of Ag+






Cl

0.100 1 mol Cl0.00321 L Ag

mol Ag

mol Ag0.000321 mol Cl

1 L Ag 1n

And we’re left with the unit moles of Cl minus






Cl

0.100 mol Ag 10.00321 L Ag 0.000321 mol Cl

1 L Ag 1 mo

mo

l A

l C

g

ln

So evaluating this expression will give us the moles of Cl minus, or n Cl minus





Ag Ag Ag l CC l& c V n cn

Cl

0.100 mol Ag 1 mol Cl0.00321 L Ag

1 L Ag 1 mol A0.0

g00321 mol Cln

0.100 times 0.00321 times 1 over 1 gives us 0.000321 moles of Cl minus.





Ag Ag Ag l CC l& c V n cn

Cl

0.100 mol Ag 1 mol Cl0.00321 L Ag

1 L Ag 1 mol A0.0

g00321 mol Cln

To convert moles of Cl minus to concentration of Cl minus, we’ll use the formula c equals n over V





Ag Ag Ag Cl Cl& c cV n n

Cl


1 L Ag 1 mol Agn

ncV

n Cl minus is 0.000321 moles of Cl minus






Cl

0.100 mol Ag 1 mol Cl0.00321 L Ag

1 L Ag 1 mol A0.0

g00321 mol Cln

0.00642 mol/L or 0.00642 M0.0

0.000321 mol Cl

500 L ClClC

l

l

Ccn

V

And the volume of the Cl minus solution, V Cl minus, is 0.0500 L






Cl


1 L Ag 1 mol Agn

0.0500

0.000321 mol Cl0.00642 mol/L or 0.0

L l6

C0 42 MCl

ClCl

nc

V

So the concentration of Cl minus in the original sample, c Cl minus, is






Cl


1 L Ag 1 mol Agn

0.000321 mol Cl0.00642 mol/L or 0.00642 M

0.0500 L ClCl

ll

C

C

cn

V

0.000321 mol of Cl minus over 0.0500 L of Cl minus






Cl


1 L Ag 1 mol Agn

0.000321 mol Cl

0.0500 L Cl0.00642 mol/L or 0.00642 MCl

ClCl

nc

V

which equals 0.00642 moles per litre.






Cl


1 L Ag 1 mol Agn

0.000321 Clor 0.00642 M

0.0

mol0.00642 mol

500 C/L

lLCl

ClCl

nc

V

or 0.00642 molar






Cl


1 L Ag 1 mol Agn

0.000321 mol Cl0.00642 mol/L or

0.00.00

500 L Cl642 MCl

ClCl

nc

V

You may want to pause the video at this point, so you can take a screen shot and use this as a copy of the whole solution to this problem.






Cl


1 L Ag 1 mol Agn

0.000321 mol Cl0.00642 mol/L or 0.00642 M

0.0500 L ClCl

ClCl

nc

V

So we can summarize by stating that the concentration of Cl minus in the original chloride solution was 0.00642 molar.





The [Cl–] in the originalsample was 0.00642 M


Precipitation Titration Calculations Video Example Here we’ll be given some data from a titration...

Documents

Transcript of Precipitation Titration Calculations Video Example Here we’ll be given some data from a titration...