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Hey guys,

ok so today in class we learned how to Write Ionic and Net Ionic Equations.

Q: why is a "total ionic equation"?
A: Because every spieces which breaks down into it's ions when dissolved in water has been written in ionic form.

Q: What is a "net ionic eaquation"?:
A: an equation that includes only those compounds and ions that undergo a chemical change in a reation which occurs in aqueous solution; shows the reaction that occurs, omitting the spectator ions.

Vocbaulary:
Spectator: ions found in solutions both before and after a chemical reaction; they "watch" the reaction occur.
Salt: a compound composed of the negative ion of an acid and the positive ion of the bases.

so off the sheet Ms. Kozoriz gave us today titled:

"Summary of Writing Total Ionic and Net Ionic Equations"

it shows the 7 steps in writing these equations. These steps are:

STEP 1: identify type of reaction and possible products.

STEP 2: look up solubility of both products.

STEP 3: Indicate states of reactants and products.

STEP 4: Write chemical equations for reaction.

STEP 5: balance the equation.

STEP 6: Write total ionic equation.

STEP 7: Write Net Ionic Equation.
(on that sheet we are given two different examples which follow these steps.)


WHAT IS NEUTRALIZATION?
Neutralization: a neutralization reaction is one whose products are a salt and a water.

aN equation such as:
ACID + BASE -----> SALT + WATER

CALCULATING AN UNKNOWN CONCENTRATION:
We would follow the 4 steps:

1) write the balanced neutralization reaction

2) calculate the number of moles of the solution. (stoichiometry)

3) calculate the moles of acid/base used.

4) calculate the concentration of the acid/base.

Using the formula:

concentration = n / v (moles / volume)

We can solve questions to determine the amount of acid of base needed to neutralize the solution:

** calculate the concentration of hydrochloric acid, if 25.00 mL is just neutralized by 40.00 mL of a 0.150 mol/L sodium hydroxide solution.



1) HCl(aq) + NaOH -----> H2O(l) + NaCL(aq)

2) (moles = c x v)
= 0.150 mol/L x 0.040L
= 0.006 mol NaOH

3) moles of HCl = moles of NaOH

4) (c = n / V)
= 25 mL x (1 L / 1000 mL) = 0.025 L
= 0.006 mol HCl / 0.025 L = 0.24 mol/L HCL

ok that's about it. i was working on the lab during class so i'm still reading over todays notes. so for anyone whose having problems if you don't get it try reading the notes it can be found in the yellow booklet (pages 4 - 6) and i believe i recall Ms. Kozoriz assigning page 7 for homework so just incase might wanna work on that!

the next scribe will be the_bdl