M l i off ions
Molarity
i
Early
E
l in
i this
thi chapter
h t you learned
l
d that
th t when
h placed
l d in
i water,
t
strong electrolytes ionize, or separate into their component ions.
In molarity problems sometimes you will be asked to calculate
the molarity of an ion instead of the molarity of the complete
molecule. For simple
p binaryy compounds
p
with onlyy one anion
and cation this is trivial, because each molecule gives you one of
each ion, so there is a simple 1:1 correspondence between the
molarity of the molecule and the molarity of the component ions.
In other compounds it’s a bit trickier, as you will see here.
M l i off ions
Molarity
i
Problem: I have
h
di l d 15 grams off CaCll2 in
dissolved
i enoughh water to
make 100 mls of solution. What are the molar concentrations of
CaCl2, Ca2+ and Cl- in this solution?
First let’s focus on the chemistry. What is the chemical reaction
that occurs when this ionic compound ionizes?
CaCl2(s) = Ca2+ (aq) + 2 Cl-(aq)
For now we will assume that this reaction goes to 100%
completion so no CaCl2(s) remains and only the ions Ca2+ and
Cl- exist in solution. Next semester we will tackle problems where
this isn’t true.
I have dissolved 15 grams of CaCl2 in enough water to
make 100 mls of solution.
solution What are the molar concentrations
of CaCl2, Ca2+ and Cl- in this solution?
CaCll2(s)
( ) = Ca2+ (aq)
( ) + 2 Cll-(aq)
( )
Let s tackle the easy one first
Let’s
first. If the reaction goes to
100% completion, then CaCl2 has been completely used up
in the reaction, and the concentration of CaCl2 in solution is 0,
so [CaCl2]=0.0 M.
I have dissolved 15 grams of CaCl2 in enough water to
make 100 mls of solution.
solution What are the molar concentrations
of CaCl2, Ca2+ and Cl- in this solution?
To tackle the concentration of the ions,
ions we will start by
calculating what the moles of CaCl2. Then we will use the same
stoichiometry math that you learned earlier to convert from
moles of reactant to moles of product. Once we have the moles
of the appropriate product, we divide by the volume of
the solution to get the final molarity.
I have dissolved 15 grams of CaCl2 in enough water to
make 100 mls of solution.
solution What are the molar concentrations
of CaCl2, Ca2+ and Cl- in this solution?
The moles of CaCl2, the reactant in this reaction is:
Moles of CaCl2 = 15 g /[40.08
[
+ 2(35.45)]
(
)] =.13516 moles
Now let’s convert from moles of reactant to moles of Ca2+ product.
CaCl2(s) = Ca2+ (aq) + 2 Cl-(aq)
So 1 mole CaCl2 = 1 Mole of Ca2+ = 2 moles of Cl1 mole Ca 2 +
.13516 moles CaCl2 ×
= .13516 moles Ca 2 +
1 mole CaCl2
I have dissolved 15 grams of CaCl2 in enough water to
make 100 mls of solution.
solution What are the molar concentrations
of CaCl2, Ca2+ and Cl- in this solution?
The molarity of Ca2+ product is then:
.13516 moles Ca2+ /.100 liters = 1.3516 molar
[Ca2+] = 1.4M
I have dissolved 15 grams of CaCl2 in enough water to
make 100 mls of solution.
solution What are the molar concentrations
of CaCl2, Ca2+ and Cl- in this solution?
And convert from moles of reactant to moles of Cl- product.
CaCl2(s) = Ca2+ (aq) + 2 Cl-(aq)
So 1 mole CaCl2 = 1 Mole of Ca2+ = 2 moles of Cl2 mole Cl −
.13516 M CaCl
C Cl2 ×
= .27032 moles
l Cl −
1 mole CaCl2
And [Cl-] = .27032 moles /.100 liters = 2.7M
So our final answer is:
0.0 M CaCl2, 1.4M Ca2+, 2.7M Cl-
M l i off ions
Molarity
i
If you think
thi k about
b t it,
it these
th
problems
bl
are actually
t ll pretty
tt easy, just
j t
look at the stoichiometry and multiply by the appropriate
coefficients Let
coefficients.
Let’ss try one more problem just to make sure you
have it.
I am going to put 5.00 grams of Ca3(PO4)2 in a 250. ml volumetric
flask and fill it to the mark. What is the concentration of
Ca3(PO4)2 , Ca2+ and PO43- in the resulting solution.
I am going to put 5.00 grams of Ca3(PO4)2 in a 250. ml volumetric
flask and fill it to the mark.
mark What is the concentration of
Ca3(PO4)2 , Ca2+ and PO43- in the resulting solution.
Let’s start by predicting the chemical reaction.
Ca3(PO4)2(s) = 3Ca2+ (aq) + 2 PO43- (aq)
So our molar conversions are:
1 Ca3(PO4)2(s) = 3Ca2+ (aq) = 2 PO433- (aq)
Again let
let’ss start with the easy one; assuming the reaction goes
to completion, the concentration of Ca3(PO4)2(s) = 0.00M
I am going to put 5.00 grams of Ca3(PO4)2 in a 250. ml volumetric
flask and fill it to the mark.
mark What is the concentration of
Ca3(PO4)2 , Ca2+ and PO43- in the resulting solution.
Ca3(PO4)2(s) = 3Ca2+ (aq) + 2 PO43- (aq)
1 Ca3(PO4)2(s) = 3 Ca2+ (aq) = 2 PO43- (aq)
Molar mass of Ca3(PO4)2 = 3(40.08) + 2(30.97) + 8(16.00)
=310.18 g
M l Ca
Moles
C 3(PO4)2 = 5 g x (1 mole/310.18g)
l /310 18 ) = 0.01612
0 01612 moles
l
Moles Ca2+ = .01612 x [3 moles Ca2+ / 1 mole Ca3(PO4)2 ]
= .04837
04837
Liters of solution = 250 mls x (1x10-3 l/ml) = .250 liters
Molarityy Ca2+ = .04836 moles Ca2+ /.250 liters= .1935 M
[Ca2+] = .194 M
I am going to put 5.00 grams of Ca3(PO4)2 in a 250. ml volumetric
flask and fill it to the mark.
mark What is the concentration of
Ca3(PO4)2 , Ca2+ and PO43- in the resulting solution.
Ca3(PO4)2(s) = 3Ca2+ (aq) + 2 PO43- (aq)
1 Ca3(PO4)2(s) = 3 Ca2+ (aq) = 2 PO43- (aq)
Molar mass of Ca3(PO4)2 = 3(40.08) + 2(30.97) + 8(16.00)
=310.18 g
M l Ca
Moles
C 3(PO4)2 = 5 g x (1 mole/310.18g)
l /310 18 ) = 0.01612
0 01612 moles
l
Moles PO43- = .01612 x [2 moles PO43- / 1 mole Ca3(PO4)2 ]
= .03224
03224
Liters of solution = 250 mls x (1x10-3 l/ml) = .250 liters
Molarityy PO4-3= .03224 moles PO43- /.250 liters= .12896 M
[PO4-3] = .129 M
P i problems
Practice
bl
11. II’m
m going to dissolve 50.
50 g of aluminum nitrate in a 500.
500 ml
solution. What is the [Al+3] and [Cl-] in this solution?
2. I want a solution that is 0.50 M in nitrate. How many grams
of Calcium nitrate do need to dissolve in 1liter of water to make
this solution?
As usual
usual, find you own answers pause the video if you need to.
to
I m going to dissolve 50.
I’m
50 g of aluminum nitrate in a 500.
500 ml
solution. What is the [Al+3] and [Cl-] in this solution?
Aluminum nitrate has the formula Al(NO3)3
Molar mass of Al(NO3)3= 26.98 + 2(14.01) + 9(16.00) = 199.00
Moles
o es oof Al(NO
(NO3)3 = 50g x (1
( mole/199.00g)
o e/ 99.00g) = .25126
. 5 6 moles
o es
Liters of solution = 500 mls x (1x10-3 liters/ml) = .500 liters
Dissolution reaction = Al(NO3)3(s) = Al3+(aq) + 3 Cl-(aq)
Conversion factors: 1 Al(NO3)3= 1Al3+ = 3 Cl-(aq)
[Al+3] = [(moles Al(NO3)3 x (1Al3+/1 Al(NO3)3)]/.500 liters
=[.25126 x(1/1)]/.500 = .50 M
[Cl-] = [(moles Al(NO3)3 x (3Cl-/1 Al(NO3)3)]/.500
)]/ 500 liters
=[.25126 x(3/1)]/.500 = 1.50 M
I want a solution that is 0.50 M in nitrate. How many grams
of Calcium nitrate do need to dissolve in 1liter of water to
make this solution?
Molarityy = Moles/volume;; 1.00M = X moles/1 l
.50M = X moles/1 liter
X moles = .50M x 1 Liter
so I need .50 moles of NO3Calcium nitrate = Ca(NO3)2
2 (aq)+ 2NO -(aq)
Dissolution reaction: Ca(NO3)2(s) = Ca2+
3
Conversion factors: 1 Ca(NO3)2 = 1 Ca2+ = 2NO3.50
50 moles NO3- x (1 mole Ca(NO3)2/2 mole NO3-)
=.25 moles Ca(NO3)2
Molar mass of Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00)
= 124.02g
.25 moles x 124.02 g/mol = 31g Ca(NO3)2