Hydrogen Concentration, Hydroxide Concentration, pH, and pOH

Today we learned how to convert between hydrogen concentration, hydroxide concentration, pH and pOH. We were able to convert between all of these using one original given value.
 This picture shows how to get from each value to the next in simple problems.

A sample problem from our notes looked like this:
What is the pH of a solution that is 12.5 M HCL?
pH= -log [H+]
     = -log(12.5)
     = -1.097
In order to solve this problem you have to look at the given information. Since we were given the concentration of Hydrochloric acid, it is the same as being given the concentration of hydrogen ions (acids go with hydrogen, bases go with hydronium). Using the chart, we know that in order to get to the pH with the concentration of hydrogen ions we need to do -log [H+].

Other problems require more than one step in order to get the answer. For example, if you are given the concentration of hydronium ions and asked to find the pH, then you need to find the concentration of hydrogen ions, or the pOH first.

This site helped me practice the conversions from this lecture.

Strength of Acids and Bases

Today's lecture was over determining if the salt produced in an acid-base reaction is acidic, basic, or neutral. In order to determine this we have to look at the strength of the parent acid and parent base. When a reaction includes a strong parent acid and strong parent base, the resulting salt is neutral. Likewise, a weak parent acid and weak parent base produces a neutral salt. On the other hand, a strong parent acid and a weak parent base result in an acidic salt, and a weak parent acid and strong parent base produce a basic salt.

In order to determine if the parent acid and parent base are weak or strong you need to know the rule. For acids, the strength is determined by how many oxygens the compound contains. If the oxygen outnumbers the hydrogen present in the compound, then it is a strong acid. Important exceptions to this rule can be found in the acronym BrINCl, pertaining to acids that contain Bromine, Iodine, Nitrogen, and Chlorine. In order to determine if a base is strong or weak, you need to check if the cation is in group 1 or group 2 of the elements. If the cation is in group 1 or group 2 on the periodic table, then it is a strong base and any other cations mean a weak base.

I found this video helpful in remembering how to determine if the acid and base are strong or weak.

Acids vs. Bases

In class today we learned about the differences between acids and bases in physical properties and how they were defined by two different scientists, Arrhenius and Bronsted-Lowery.
Physical properties:
Acids                                    Bases
Taste sour                             Taste bitter
Feel sticky                            Feel slippery
Turns litmus pink                 Turns litmus blue

Arrhenius vs. Bronsted-Lowery

Arrhenius
Acids are those species that produce hydrogen ions in solution (H+)
Bases are those species that produce hydroxide ions in solution (OH-)

Bronsted-Lowery
Acids are those species that donate a proton (H+)
Bases are those species that accept a proton (OH-)

We also learned how to determine the conjugate base and conjugate acid in a reaction. By determining the acid and base using either the Arrhenius or Bronted-Lowery definitions, we can then find its conjugate base and conjugate acid. A conjugate acid is the substance that forms when a proton is added to a base. On the other hand, a conjugate base is the remaining substance when a proton is lost from an acid.

This Link has some helpful practice problems for acids bases.

Molarity in Stoichiometric Calculations

In class today we learned how molarities can be used in stoichiometric calculations. There are two different ways to look at going through these types of problems.

One of these is using a flow chart:




The other is a systemic approach:



We also learned a few more vocabulary words from this unit:
Titrate- acid-base reaction, driven by the production of water
Endpoint- when the moles of acid equals the moles of base

Some links to practice these problems:
Molarity in Stoichiometry
Stoichiometry with Solutions

Murder Investigation Lab

Today in class, my lab partner and I finished conducting the Murder Investigation Lab. On Monday we started by reading about the different suspects for the murder of Miss Scarlet and wrote up a procedure in order to determine who did it. We first did double replacement reactions of potassium iodide and silver nitrate with sodium chloride and sodium carbonate in order to make predictions for the products of mixing the different solutions. Through these double replacement reactions, we were able to predict that the murder weapon was most likely silver nitrate.
By mixing the unknown solution with sodium carbonate, we were able to determine that the murder weapon was in fact silver nitrate based on the reaction forming a solid. We then used the funnel to get the solid that was formed by itself onto the filter paper. After letting the filter paper dry over night, we were able to take the mass of the filter paper with the solid, silver carbonate, that formed. By subtracting the mass of the filter paper from this measurement we can convert the mass of the silver carbonate that formed into moles. We then divide the moles of silver carbonate by the amount of unknown solution used in order to determine the molarity. Doing this, we were able to determine that the murderer was Mr. Green who used silver nitrate as the weapon.







All of the materials we used in this lab.








Taking the mass of the filter paper.

Using a funnel and Urban Meyer flask to separate
the aqueous solution from the solid formed 

The product formed

Links that were helpful in solving the murder:

Calculating Molarity

Molarity

Aqueous Solutions Lab

Today in class we performed a lab in order to see how the concentrations of each solution change in a serial dilution. We started out with a solution containing 10.0 mL of water and 20 drops of blue food coloring. Between each solution we took 1.0 mL from the previous solution and added 9.0 mL of water, doing this until blue coloring was no longer visible. After diluting the color out of the solutions, we calculated the concentration of solution B, solution C, and solution H. In order to make these calculations we started with the formula (2.0)(1.0) = M2(10.0). As we continue through the solutions in the serial dilution, however, the first number in the formula becomes the result of the last calculation. These calculations show the concentration of the solutions in drops.

Links I found helpful with dilutions:

Dilution Calculations

Serial dilutions

Dilutions

Today in class we took notes over dilutions. A dilution is a series starting with a stock solution and adding water to form a new solution. The amount of solution moved from the stock to the next solution is referred to as an aliquot. In order to solve a problem with dilutions you use the formula M1V1=M2V2. The M stands for molarity, the starting and ending, and the V stand for total volume, staring and ending as well.

A dilution


Formula to solve problems with dilutions