March 13th - A Whole Lotta Nothing

Our class was cut short today, so we spent our time reorganizing the lab draws. What fun! And next class is the test, of course!

In the meantime, here's something for chuckles:

Posted by Michael.

March 9th - Time to Review

As one would assume, a test is upcoming. Will it be on Tuesday or Thursday? We don’t know! Therefore, one would be wise to be prepared for Tuesday.

Be sure to know the following topics:

Molarity

Stoichiometry and Molarity

Titrations

Dilutions

Ion Concentration

Mixing Acids and Bases

Remember, stay concentrated. 

Posted by Andrew.

March 7th - Was Today’s Lesson Sour or Bitter?

Actually, today’s lesson was very interesting! We were able to determine the pH of the a lake before and after acid rain fell on it. To do this, we needed the following equation:

pH = - log [H⁺]

pOH = - log [OH^-]

Other equations include:

pH + pOH = 14

[H⁺][OH^-] = 10^-14

http://www.sciencegeek.net/Chemistry/taters/graphics/pHSchematic.gif

With this knowledge, we were able ‘to do work’. If 150mL of 0.30M HCl fell on the lake, determine its pH.

Ex. Simple. HCl à H⁺ + Cl^-. Therefore, 0.150L (0.30mol/1L)(1/1)(1/0.150L) = 0.30 M H⁺

pH = - log [0.30M] = 0.52!

Ex. After this amount falls on a lake of 2000mL of 0.02M NaOH, what is the pH of the lake?

Since we have two reactants, we first need to find the limiting reactant.

There are 0.045 mol HCl present. There are 0.040 mol NaOH present.

HCl + NaOH à NaCl + H₂O

There are 0.045 mol NaOH needed. Therefore, NaOH is the limiting reactant.

There are 0.040 mol HCl needed. After the reaction, we are left with 0.005 mol HCl.

Dissociate it! HCl à H⁺ + Cl^-

There will be 0.005 mol H⁺, and since our solution has 2.150L, [H⁺] is 0.0023255814.

pH = 2.6! That’s pretty acidic.

And that’s it. So remember, when performing acid-base reaction with two reactants, ‘follow the following’ steps.

Find the LR

Determine how many moles of excess reactants are left

Finding out number of moles of H⁺ or OH^-.

Finding the concentration by dividing by volume of [H⁺ or OH^-].

Using -log.

Fun fact: K_a is a ratio of dissociated to un-dissociated acid. Ie. [H⁺][F^-] / [HF]

Posted by Andrew.

March 5th - Ion Concentrations

Today, the topic was ion concentration. We’ve already learned about the concentration of substances and molarity in this unit, so we’re going to move onto the concentration of specific ions.

As we’ve learned before, ionic compounds are made up of two parts. The first part is the cation, which is the positively charged particle. The second part is the anion, which is the negatively charged particle. When dissolved in water, these two particles separate from each other. This process is called dissociation.

The top equation is dissociation, while the bottom is the opposite. 

When writing dissociation equations, the atoms and charges must both be balanced.

Ex.) Write the dissociation of Sodium chloride.

NaCl_(s) à Na⁺_(aq) + Cl^-_(aq)

Ex.) Write the dissociation for Na₃PO₄.

Na₃PO₄ à 3Na⁺ + PO₄^3-

If the volume of the solution does not change, then the concentration of individual ions depends on the balanced coefficients in the dissociation equation.

Ex.) Determine [Na⁺] and [PO₄^3-] in a 1.65 M solution of Na₃PO₄.

Na₃PO₄ à 3Na⁺+ PO₄^3-

As always, we must start with a balanced equation. Next, we convert 1.65 M to moles, and then use our molar ratios to find the number of moles of both ions. Finally, we divide by the volume. When it doesn’t tell us, we can assume it’s 1L.

1.65 M   x   1 L/mol   =   1.65 mol

1.65 mol   x   3/1   =   4.95 mol   x   1/L   =   5.0 M [Na⁺]

1.65 mol   x   1/1   =   1.65 mol   x   1/L   =   1.7 M [PO₄^3-]

Done!

Ex.) A 0.100 L solution of 0.500 M PbCl₂ is added to 0.200 L solution of 0.100 M NaOH. Determine the final [Cl^-].

PbCl₂ + NaOH à Pb⁺ + 2Cl^- + Na⁺ + OH^-

Just like we did before, a balanced equation is necessary. In this equation, we need to add the volumes to find the total volume. The total volume is 0.300 L. We only need to calculate the concentration of PbCl₂, so we can just disregard the other parts.

0.500 M   x   0.100 L/mol   x   2/1   x   1/0.300L   =   0.333 M [Cl^-]

And that’s it! Great job!

Here’s a great video on the subject:

Posted by Michael.

March 1st - A Big Blue Lab

Today’s class was rather blue. That is, the solutions were. But, they had different shades, and of course, this was no coincidence. We were told these were solutions of copper (II) chloride, or cupric chloride.

Our task was to figure out which of the solutions corresponded to 0.1 M. No problem, we said! We decided to quickstyle this lab.

The items we worked with included the following:

Copper (II) chloride

Scoopula

Weight

Plastic container

Beaker

Erlenmeyer flask

Graduated cylinder

Here is the procedure:

1. Obtain your needed materials listed above.

2. Weight out 1.345g of CuCl2.

3. Measure 100mL of water in a graduated cylinder.

4. Mix these two substances in an Erlenmeyer flask (it’s fun to spin).

5. Pour into test tube and compare with test tube species.

How we obtained 1.345g?

Simple. To receive a molarity of 1.0, we know that there are 1.0 moles of CuCl2 in 1L. To receive our desired molarity, 0.1, we have 0.1 in 1L. Since one liter is a bit too much, we simply keep our ratio the same, of moles and liters, but decrease the numbers by 10. Now we have 0.01 moles in 100 mL. How many grams is this? Simply molar mass of CuCl2 will tell us.

0.01mol CuCl2  (134.5g / 1 mol) = 1.345g Tada :)

Posted by Andrew.