Solubility and Net Reactions

Solubility and Net Reactions


Net Reactions or Net Ionic Equations are just a simplified version of a reaction involving the forming of a precipitate. Note: A net Ionic Equation can only occur if one of the products in a double replacement reaction is a solid.

A Net Ionic Equation is written using the following steps:

1. Write, predict and balance the complete reaction with state subscripts written in.
2. If a solid is formed, write out the complete equation separating all ions from one another except the ones forming the solid with the charges and state subscripts written in.
3. Cross out any ions found on both sides of the equation.
4. Write out the remaining ions and compounds in a reaction.

Example:

Step 1:

___BaCl2 (aq)

+

___Na2CO3 (aq)

___BaCO3 (s)

+

_2__NaCl (aq)

Step 2:

_Ba2+(aq)

+

2Cl - (aq)

+

2Na +(aq)

+

CO3 2-(aq)

___BaCO3 (s)

+

2Na +(aq)

+

2Cl- (aq)

Step 3 and 4:

Ba 2+ (aq)

+

CO32- (aq)

BaCO3(s)

In order to tell whether or not a chemical reaction can occur, one must determine the solubility of the products.

Use the following link to determine whether or not a compound or soluble or not / barely soluble.

http://dl.clackamas.edu/ch105-03/solubili.htm

If one of the products is not or barely soluble then a precipitate or solid forms meaning that a reaction has occurred.

If both the products are soluble meaning they are aqueous then a reaction has not occurred.

Examples:

NaOH	Fe2S3	MgSO4	PbCl2	Ba(NO3)2	MgCO3
soluble	insoluble	soluble	insoluble	soluble	insoluble

Double Replacement, Combustion and Neutralization Reactions

Double Replacement, Combustion and Neutralization Reactions


Double Replacement Reactions: When two compounds swap components in one of two ways... The anion of one compound may re- attach to the other compound and vice versa OR the cation of one compound may re-attach to the other compound and vice versa.

This reaction can be further simplified using a dancing analogy:

Two couples, comprised of two partners each (a man and a woman), are taking ballroom lessons.  The instructor later calls "switch" and the female partners switch and begin dancing with the other male partner.

A double replacement reaction can be modeled by this formula:

AB + XY ---> AY + XB

Examples:

KOH + H₂SO₄ ---> K₂SO₄ + H₂O
FeS + HCl ---> FeCl₂ + H₂S
NaCl + H₂SO₄ ---> Na₂SO₄ + HCl
AgNO₃ + NaCl ---> AgCl + NaNO₃




Combustion Reactions: Are essentially a kind of single replacement reaction which involves something burning in air.

It can be modeled by the following formula:

AB + O2  ---> CO2 + H2O

Examples:

CH_4(g)  +  2O_2(g)  →    CO_2(g)  +  2H₂O_(g)

2C₄H_10(g)  +  13O_2(g)  →    8CO_2(g)  +  10H₂O_(g)

_{Neutralization Reactions: Are essentially a kind of double replacement reaction which includes acids reacting with bases to produce H2o and an ionic salt.}

<sub>It can be modeled by this formula:


HA + BOH ---> H2O + BA


Examples:</sub>


HCl + NaOH --> NaCl + HOH

H₂SO₄ + 2 NH₄OH --> (NH₄)₂SO₄ + 2 HOH

2 NaOH + H₂CO₃ --> N₂CO₃ + 2 NaOH

Now, you must look at the "Activity Series". If the compound replacing the previous one is higher on the list, the reaction will occur. If it is lower, nothing will happen.

Here's a video that demonstrates how to do it:

Word Equations and Naming Coumpounds

We're going back to the very beginning; when we first learned to name compounds. Remember? Well, here's a little refresher...

For example,
If a word equations is:
Aluminium + Hydrochloric Acid = Aluminium Chloride + Hydrogen
The chemical formula is:
Al + HCl = AlCl3 + H2
BUT WAIT! We still have to balance it! So, the proper equation is:
2Al + 6HCl = 2AlCl3 + 3H2


Now that was easy, wasn't it? Here's a chart reminding you of all the prefixes for covalent compounds:

Don't forget about naming acids! To help remember, I "ate" something "ic"y. If you just say that, the other ending of "ous" for "ite" just falls into place.

Here is a little video to clear up any other confusion:

And one more to help with balancing:

Molar Volume of Gas (STP)

STP, or Standard Temperature and Pressure is a set of data we use to compare our measurements to the gas' properties under the temperature of 0 degrees (273 K)

Key Formula: (one mole of gas occupies 22.4 L of gas)

1 mole of gas 
22.4L of gas

Putting it Into Use:


Calculate the volume occupied by 4.6g of C (carbon)

4.6 x 1 mole  = 0.38moles
         12.0g C

This is where the Key formula is plugged into the equation.

molar volume = 22.4 L  x 0.38moles = 6.5 L
                         1 mole

And there you go! A few easy steps.

Solution Preparation

Solutions are shipped in concentrated form to minimize shipping costs. Therefore, they must be prepared before they can be used.

The formula M1L1 = M2L2 is used to prepare solutions.
The most important thing to remember is that the MOLES on one side MUST EQUAL the MOLES on the other side!


Example:
I have 2.00 L of 16.0 M of HCl. I need 0.800 L of 2.00 M HCl.
Solve for L1.
(16.0M1)(L1) = (0.800L2)(2.00M2)
L1 = 0.100 L

Let's do one more example:
You have 11.6 mol/L of HCl. How would you make up 250 mL of 0.500 mol/L?
Solve for L1.
(11.6M1)(L1) = (0.25L2)(0.500M2)
L1 = 0.011 L acid
Then, you subtract the original volume by the amount of acid required.
0.25 - 0.011 = 0.239 L
Therefore, you would need 0.011 L of acid and 0.239 L of water.

Here's a video that explains it all step by step...

Dilution.

Remembering this simple formula will save your troubles.
M₁L₁ = M₂L₂ 
M stands for MOLARITY.
L stands for LITERS 


Example: 


You have 2.00L of 15.4M HCL and you need 500.0 mL of 3.00M HCl, how much of the original (HCl) concentration do you ned, and how much H20 do you need? Answer in mL


Now plug in the numbers into the formula.


15.4M HCl x L1 = 3.00M HCl x 0.500L


Solve what side has most info (right side in this case)


15.4M HCl x L1 = 1.5 mol HCl


Like math we're trying to find L1


1.5 mol HCl = 100 mL
1.5M


Then find the water needed.


Simple subtract L2 qith L1.


0.500L - 0.100L = 0.400L or 400mL of water is needed


Quick video on how to dilute :)