Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Solution concentration and Stoichiometry

Molar Concentration of Solutions
The concentration of a solution is defined as the amount of solute dissolved in a given quantity of
solvent or solution
...
This material focuses on Molarity
...

Mathematically,
Molarity =

𝑚𝑜𝑙𝑒 𝑠𝑜𝑙𝑢𝑡𝑒
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝐿

The unit for Molarity is mole/L
...
We then simply
determine the ratio of the number of moles of solute to the volume of the solution
...
For example, if the concentration of a Sulfuric acid (H 2SO4) solution
is 0
...
32M in H2SO4, 0
...
32M in SO42-
...
50g of Potassium nitrate (KNO3) is mixed with enough water to create a 500ml solution
...
Determine the number of moles of solute in the solution
...

Calculate the molar mass
...
10g/mol
1
39
...
01g/mol
1
14
...
00 g/mol
3
48
...
11g/mol
Next, we determine the number of moles using the molar mass of the substance
...
11𝑔/𝑚𝑜𝑙 𝐾𝑁𝑂3

) = 0
...
Determine the volume of the solution in liters
...
We need to convert this
quantity to liters before plugging it in the molar concentration formula
...
Dela Cruz

1𝐿
1000𝑚𝑙

1

) = 0
...
Determine the Molarity of the solution by applying the formula
...
495 𝑚𝑜𝑙 𝐾𝑁𝑂3
0
...
99 mol/L
We can say that the solution has a concentration of 0
...
99M (molar) solution
...
Determine the concentration of each ion in the solution
...

KNO3(aq) → K+(aq) + NO3-(aq)
Then we multiply the molar concentration to the number of moles of each ion in the dissociation
equation
...

Hence, the concentration of K+ is 0
...

Example 2
...
3L solution of 125g of glucose (C 6H12O6) and
water? What is the molar concentration of each ion in the solution?
Step 1
...

We are given the mass of the solute so we simply use the molar mass of the solute to get the number
of moles
...

Element in the
compound

Molar mass of the
element

Total mass
contributed by the
element
Carbon
12
...
06g/mol
Hydrogen
1
...
00g/mol
Oxygen
16
...
00g/mol
Molar mass of C6H12O6
180
...

(125g C6H12O6) (

Subscript

1𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6
180
...
692 mol C6H12O6

Step 2
...

We are given the volume of the solution and it is already expressed in liters (2
...

Step 3
...

Molarity =
Molarity =

Reymon T
...
692 𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6
2
...
30 mol/L
We can say that the solution has a concentration of 0
...
30M (molar) solution
...
Determine the concentration of each ion in the solution
...

Example 3
...
5L of a
0
...

Step 1
...

Molarity =

𝑚𝑜𝑙𝑒 𝑠𝑜𝑙𝑢𝑡𝑒
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝐿)

mole solute = Molarity X liter solution
mole NaCH3COO = (0
...
5L)
mole NaCH3COO = 1
...
Convert the number of moles to mass (since we are asked to determine the mass of the
solute) using the molar mass of the solute
...

Element in the
compound

Total mass
contributed by the
element
Sodium
23
...
00g/mol
Carbon
12
...
02g/mol
Hydrogen
1
...
00g/mol
Oxygen
16
...
00g/mol
Molar mass of NaCH3COO
82
...
5L solution of 0
...

(1
...
02𝑔 𝑁𝑎𝐶𝐻3 𝐶𝑂𝑂
1𝑚𝑜𝑙 𝑁𝑎𝐶𝐻3 𝐶𝑂𝑂

) = 89
...
81 g of NaCH3COO to make 1
...
73M solution
...
Determine the concentration of each ion in the solution
...

NaCH3COO (aq) → Na+(aq) + CH3COO-(aq)
Then we multiply the molar concentration to the number of moles of each ion in the dissociation
equation
...

Hence, the concentration of Na+ is 0
...


Reymon T
...

The key point to remember is that in dilution, the number of moles of solute remains the same since
only the quantity of the solvent is increased
...

V(conc) is the volume of the stock solution in liters
M(dil) is the molar concentration of the diluted solution
V(dil) is the volume of the diluted solution in liters

Example 1
...
2M stock solution of HI is needed to prepare 150ml of a
0
...

M(conc) x V(conc) = M(dil) x V(dil)
V(conc) =

𝑀(dil) 𝑥 𝑉(dil)
𝑀(𝑐𝑜𝑛𝑐)

V(conc) =

(
...
150𝐿)
1
...
045𝐿
1
...
0375 L
The question asks for the amount in milliliters so the answer is 37
...

Example 2
...
01M solution of NaOH from a 1
...
Determine the volume of the stock solution needed for the dilution
...
01𝑀 𝑥
...
3𝑀

V(conc) =

Reymon T
...
032𝐿
1
...
0246 L
Step 2
...
The difference is the
amount of volume of water that has to be added
...
6ml
V(water) = 295
...
4ml of water should be added to the stock to prepare the diluted solution
...
10ml of a certain stock solution was used to prepare 750ml of a 0
...

What is the molar concentration of the stock solution?
Solution
We determine the concentration of the solution using our dilution formula
...
015𝑀 𝑥 0
...
01𝐿

M(conc) =

0
...
01𝐿

M(conc) = 1
...
13M
...
We can
determine the amounts of reactants and products used or produced, respectively, using:
Mass-mole relationships
Mole ratios in a balanced chemical equation
Molarity
Density
Example 1
...
4M
Potassium iodide (KI) solution and 60ml of a 0
...
Determine the number of moles of solute
...
4M)(0
...
024mol KI

For Pb(NO3)2 – Since the solutes have the same concentration and volume, the number of moles
are also the same
...
024mol

Reymon T
...
Write the balanced chemical reaction equation
...
Determine the limiting reactant
...

For KI
(0
...
012mol PbI2

For Pb(NO3)2
(0
...
024mol PbI2

The theoretical yield based on the quantity of KI is smaller therefore, it is the limiting reactant
...
Determine the amount of PbI2 formed in the reaction
...
012mol PbI2)
...

Element in the
compound

Molar mass of the
element

Pb
I

207
...
9g/mol
Molar mass of PbI2

(0
...
2g/mol
253
...
53g PbI2

If we combine 60 ml of 0
...
4molar Pb(NO3)2
solution, 5
...

Example 2
...
5M HCl solution in milliliters is needed to neutralize 110ml of a
0
...
Determine the number of moles of NaOH in the base solution
...
75 x 0
...
0825mol
Step2
...

HCl(aq) + NaOH(aq)

Reymon T
...
Calculate the number of moles of HCl needed to completely neutralize NaOH
...
0825 mol HCl to completely neutralize
0
...

Step 3
...

Molarity =

𝑚𝑜𝑙𝑒 𝑠𝑜𝑙𝑢𝑡𝑒
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛(𝐿)
𝑚𝑜𝑙𝑒 𝑠𝑜𝑙𝑢𝑡𝑒

Volume of solution (L) =

𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦

Volume of solution (L) =

0
...
5𝑀

Volume of solution (L) =
...
Convert the volume in liters to milliliters
...
165L) (

1000𝑚𝑙
1𝐿

) = 165ml

It will take 165ml of a 0
...
75M NaOH
solution
...
We want to precipitate all Ag+ out of 100ml of a 2
...
5M Na2SO4 solution
...
Calculate the number of moles of Ag+ in the solution using the formula for molarity
...
1M) (0
...
21mol
Step 2
...

2AgNO3 + Na2SO4 → Ag2SO4 + 2NaNO3
Step 3
...

From the balanced equation above, we can see that for every 2 moles of AgNO3, we need 1 mole
of Na2SO4
...
21mol AgNO3) (

1𝑚𝑜𝑙𝑁𝑎2 𝑆𝑂4
2 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3

) = 0
...
Calculate the volume of 1
...

Molarity =

𝑚𝑜𝑙𝑒 𝑠𝑜𝑙𝑢𝑡𝑒
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛(𝐿)

Volume of solution (L) =

Reymon T
...
105 𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
1
...
07 L
We now convert our answer to milliliters
...
07L) (

1000𝑚𝑙
1𝐿

) = 70ml

The AgNO3 solution should be combined with 70ml 1
...


Reymon T

---