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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
n molesc molar concentration (mol/L)m mass (g)V volume (L)
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
n molesc molar concentration (mol/L)m mass (g)V volume (L)
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
n molesc molar concentration (mol/L)m mass (g)V volume (L)
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
n molesc molar concentration (mol/L)V volume (L)m mass (g)
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
n molesc molar concentration (mol/L)V volume (L)m mass (g)
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
n molesc molar concentration (mol/L)V volume (L)m mass (g)
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
n molesc molar concentration (mol/L)V volume (L)m mass (g)
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
n molesc molar concentration (mol/L)V volume (L)m mass (g)
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
n molesc molar concentration (mol/L)V volume (L)m mass (g)
?
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
n molesc molar concentration (mol/L)V volume (L)m mass (g)
?
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
n molesc molar concentration (mol/L)V volume (L)m mass (g)
?
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
n molesc molar concentration (mol/L)V volume (L)m mass (g)
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
n molesc molar concentration (mol/L)V volume (L)m mass (g)
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
50.00 mL of a solution containing Cl–
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
50.00 mL of a solution containing Cl–
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 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)
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
4.46 – 0.95 = 3.51
Which is 3.51 mL
[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) 3.51
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
4.46 – 0.95 = 3.51
For trial 2, the volume is 7.65 minus 4.45
[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) 3.51
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
7.65 – 4.45 = 3.20
Which is 3.20 mL
[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) 3.51 3.20
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
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 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) 3.51 3.20
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
10.87 – 7.65 = 3.22
Which is 3.22 mL
[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) 3.51 3.20 3.22
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
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
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) 3.51 3.20 3.22
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
Considerably higher than 3.20 and 3.22
For that reason, we just discard the value of 3.51
[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) 3.51 3.20 3.22
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
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
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) 3.51 3.20 3.22
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
Best Average = 3.20 3.2
22
23. 1
3.21 mL
[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) 3.51 3.20 3.22
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
Best Average = 3.20 3.22
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 Cl– solution = 50.0 mL
Volume of AgNO3 used:3.21 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) 3.51 3.20 3.22
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
Best Average = 3.20 3.2
22
23. 1
We’ll convert the 3.21 mL to 0.00321 L
[AgNO3]=0.100 M
Volume of Cl– solution = 50.0 mL
Volume of AgNO3 used:3.21 mL = 0.00321 L
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
At this point, let’s dissociate the AgNO3 here
[AgNO3]=0.100 M
Volume of Cl– solution = 50.0 mL
Volume of AgNO3 used:3.21 mL = 0.00321 L
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
And we get Ag+ and NO3 minus
[Ag+ NO3–] = 0.100 M
Volume of Cl– solution = 50.0 mL
Volume of AgNO3 used:3.21 mL = 0.00321 L
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
And we’ll also dissociate the AgNO3 here…
[Ag+ NO3–] = 0.100 M
Volume of Cl– solution = 50.0 mL
Volume of AgNO3 used:3.21 mL = 0.00321 L
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
Giving us Ag+ and NO3 minus
[Ag+ NO3–] = 0.100 M
Volume of Cl– solution = 50.0 mL
Volume of Ag+ NO3– used:
3.21 mL = 0.00321 L
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
The nitrate ion, NO3 minus is a spectator ion. It does not form any precipitates.
[Ag+ NO3–] = 0.100 M
Volume of Cl– solution = 50.0 mL
Volume of Ag+ NO3– used:
3.21 mL = 0.00321 L
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
So we’ll just discard it.
[Ag+ NO3–] = 0.100 M
Volume of Cl– solution = 50.0 mL
Volume of Ag+ NO3– used:
3.21 mL = 0.00321 L
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
So we can simply say that the concentration of Ag+ is 0.100 molar
[Ag+] = 0.100 M
Volume of Cl– solution = 50.0 mL
Volume of Ag+ used:3.21 mL = 0.00321 L
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
And the volume of Ag+ solution used is
[Ag+] = 0.100 M
Volume of Cl– solution = 50.0 mL
Volume of Ag+ used:3.21 mL = 0.00321 L
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
Equal to 0.00321 L
[Ag+] = 0.100 M
Volume of Cl– solution = 50.0 mL
Volume of Ag+ used:3.21 mL = 0.00321 L
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
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.
(aq) (aq) (s)Ag Cl AgCl
[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
Volume of Cl– solution = 50.0 mL
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.
(aq) (aq) (s)Ag Cl AgCl
We have the concentration of Ag+ and the volume of Ag+ here.
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 50.0 mL
Ag Ag & c V
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.
(aq) (aq) (s)Ag Cl AgCl
And the volume of Cl minus here.
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 50.0 mL
Ag Ag & c V Cl
V
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.
(aq) (aq) (s)Ag Cl AgCl
At this point, we’ll convert the 50.0 mL of Cl minus solution
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 50.0 mL = 0.0500 L
Ag Ag & c V
ClV
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.
(aq) (aq) (s)Ag Cl AgCl
To 0.0500 litres
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 50.0 mL = 0.0500 L
Ag Ag & c V
ClV
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.
(aq) (aq) (s)Ag Cl AgCl
And leave it like this.
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Here is the generic diagram we came up with for titrations.
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
We are given the concentration and volume of Ag+
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
And we’re asked for the concentration of the other reactant, Cl minus
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
And we’re given its volume.
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
We’ll start with the concentration and volume of Ag+
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
Convert that to moles of Ag+
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
AgA Cl Cg Ag l& nc cV n
Then find moles of Cl minus
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c cV n n
And use that and the given volume, to calculate the concentration of Cl minus
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag l CC l& c cV n n
We can do the first 2 steps using conversion factors.
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
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+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
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+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
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+
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
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+
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
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+
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
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
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
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+.
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
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+
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
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
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
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+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
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+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
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+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c cV n n
Cl
0.100 mol Ag 1 mol Cl0.00321 L Ag 0.000321 mol Cl
1 L Ag 1 mol Agn
ncV
n Cl minus is 0.000321 moles of Cl minus
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
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
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
Cl
0.100 mol Ag 1 mol Cl0.00321 L Ag 0.000321 mol 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
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
Cl
0.100 mol Ag 1 mol Cl0.00321 L Ag 0.000321 mol 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
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
Cl
0.100 mol Ag 1 mol Cl0.00321 L Ag 0.000321 mol 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.
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
Cl
0.100 mol Ag 1 mol Cl0.00321 L Ag 0.000321 mol 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
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
Cl
0.100 mol Ag 1 mol Cl0.00321 L Ag 0.000321 mol 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.
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
Ag Ag Ag Cl Cl& c V n n c
Cl
0.100 mol Ag 1 mol Cl0.00321 L Ag 0.000321 mol 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.
[Ag+] = 0.100 MVolume of Ag+ used = 0.00321 L
Volume of Cl– solution = 0.0500 L
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.
(aq) (aq) (s)Ag Cl AgCl
The [Cl–] in the originalsample was 0.00642 M
50.00 mL of a solution containing Cl–