Mixing Acids and Bases

Acids and pH: 
A measure of the hydrogen ion concentration [H⁺], pH is calculated using the following formula: 

• pH = -log₁₀[H⁺]   

Hydrogen ion concentration [H⁺] can be calculated using the following formula:  

• [H⁺] = 10^-pH

Example One (pH):

Find the pH of a 0.2 M solution of HCl: 

• Write the balanced equation for the dissociation of the acid:
• HCL ==> H⁺(aq) + Cl^-(aq) 
• Use the equation to find the [H⁺]: 
• 0.2 M x (1/1) = 0.2 M of [H⁺] **HCl is a strong acid which fully dissociates** 
• Calculate pH: 
• pH = -log₁₀[H⁺]   
• pH = -log₁₀[0.2]  
• pH = 0.7 

Example Two (pH):

Find the pH of a 0.2 M solution of H₂SO₄: 

• Write the balanced equation for the dissociation of the acid:
• H₂SO₄ ==> 2H⁺(aq) + SO₄^2- (aq) 
• Use the equation to find the [H⁺]: 
• 0.2 M x (2) = 0.4 M of [H⁺] **H₂SO₄ is another strong acid which fully dissociates** 
• Calculate pH: 
• pH = -log₁₀[H⁺]   
• pH = -log₁₀[0.4]  
• pH = 0.4 

Example Three (pH):

Find the [H⁺]  of a nitric acid solution with a pH of 3.0: 

• Use the formula to find the [H⁺]: 
• pH = 3.0 
• [H⁺]  = 10^-pH 
• [H⁺]  = 10^-3.0 
• [H⁺]  = 0.001 M

• Check it: 
• pH = -log₁₀[H⁺]   
• pH = -log₁₀[0.001]  
• pH = 3.0 

Bases and pOH: 

A measure of the hydroxide ion concentration [OH^-], pOH is calculated using the following formula: 

• pOH = -log₁₀[OH^-]    

Hydroxide ion concentration [OH^-] can be calculated using the following formula:  

• [OH^-] = 10^-pOH 

Example One (pOH):

Find the pOH of a 0.1 M solution of NaOH: 

• Write the balanced equation for the dissociation of the base:
• NaOH ==> Na⁺(aq) + OH^-(aq) 
• Use the equation to find the [OH^-]: 
• 0.1 M x (1) = 0.1 M of [OH^-] **NaOH is a strong base which fully dissociates** 
• Calculate pH: 
• pH = -log₁₀[OH^-]   
• pH = -log₁₀[0.1]  
• pH = 1 

Example Two (pOH):

Find the pOH of a 0.1 M solution of Ba(OH)₂: 

• Write the balanced equation for the dissociation of the base:
• Ba(OH)₂ ==> 2OH^-(aq) + Ba²⁺ (aq) 
• Use the equation to find the [OH^-]: 
• 0.1 M x (2) = 0.2 M of [OH^-] **Ba(OH)₂ is another strong base which fully dissociates** 
• Calculate pH: 
• pH = -log₁₀[OH^-]   
• pH = -log₁₀[0.2]  
• pH = 0.7 

Example Three (pOH):

Find the [OH^-]  of a sodium hydroxide solution with a pOH of 1.0: 

• Use the formula to find the [OH^-]: 
• pH = 3.0 
• [OH^-]  = 10^-pOH 
• [OH^-]  = 10^-1.0 
• [OH^-]  = 0.1 M

• Check it: 
• pH = -log₁₀[OH^-]   
• pH = -log₁₀[0.1]  
• pH = 1.0   

Another formula to know from this unit that greatly simplifies things is:

• **pH and pOH = 14**

- Simon Sierra, George Spencer and Ben Suratos

Copper(II) Chloride Titration Lab

For this lab we had to find out which of the six given test tubes (labelled from A-F) had a concentration of 1.0M. The test tubes all had CuCl2 mixed with water and each one had a different shade of light blue.




Materials :

1 Weighing scale
2 Plastic boats
1 Scoopula
1 Test Tube
1 Graduated Cylinder
1 Erlenmeyer Flask
CuCl2
Water
Flasks

Procedure:

1. We took two plastic boats and put an unknown amount of CuCl2 in one boat and weighed it. It weighed 1.93g without the plastic boat.
2. Then we put 50mL of water into a graduated cylinder and then we transferred the water into an Erlenmeyer flask and mixed the water with the 1.93g of CuCl2.
3. The we took the 1.93g of CuCl2 and converted the units into moles.
4. We took all of the given numbers and put it into the C1V1 = C2V2 dilution formula to find out how much water was needed to add to the solution so that the concentration will be 1.0M.
5. We needed 0.14L of water or 140mL into the first solution.
6. After adding 140mL of water to the original solution we transferred the solution into a test tube and compared it with the test tubes with unknown concentrations. 

Conclusion:

Our test tube colour matched with test tube C.

 -Benedict Suratos

Ion Concentration

Dissociation: 
Ionic compounds are made up of two parts:

• Cation: Positively charged particle 
• Anion: Negatively charged particle 

In a process called dissociation, shown in the video below, the cation and anion in an ionic compound separate from each other.  

Dissociation can be written in a dissociation equation, in which the atoms and charges must balance. The format for such an equation is AB => A + B. Knowing this, we can write the dissociation of the Sodium Chloride from the above video as: 

• NaCl => Na⁺ + Cl^-    

Example One:  

• BaSO4 => Ba ²⁺ + SO4 ^2-  
• (NH4)2CO3 => 2NH4 ⁺ + CO3 ^2-   
• Al2(SO4)3 => 2Al³⁺ + 3SO4 ^2-   

Example Two:  

Determine each of the ion concentrations when a 4.52 M solution of Ammonium Nitrate dissociates.  

• NH4NO3 => NH4 ⁺ + NO3 ^-   

•  4.52 M x (1/1) = 4.52 M 
•                          = [NH4 ⁺] 
•                          = [NO3 ^-] 

Determine each of the ion concentrations when a 0.56 M solution of Iron (III) Hydroxide dissociates.  

• Fe(OH)3 => Fe ³⁺ + 3OH ^-   

•  0.56 M x (1/3) = 0.19 M 
•                          = [Fe ³⁺] 
•  0.56 M x (3/1) = 1.68 M  
•                          = [OH ^- ]  

-Simon Sierra

Dilutions

What is a Dilution?

Dilution is the process of decreasing the concentration of a solution by adding a solvent. The total number of solutes in the solution remains the same after dilution, but the volume of the solution becomes greater, resulting in a lower molarity, ppm, mg/L, or % concentration.  

In the picture above,  the solute is shown as yellow dots and the solvent as solid blue.  The 1 L beaker on the left shows the initial concentration, which represents 13 dots/L.  The beaker on the right is the result of dilution of the left beaker.  After adding more solvent so that the solution's total volume becomes 3 L, the concentration of the diluted beaker is (13 dots)/(3 L), or 4.3 dots/L.

As mentioned, there are two components to any solution and these are:

• Solvent: Does the dissolving (In most cases the solvent is water)
• Solute: Is dissolved  

*An easy way to remember how to distinguish the two and what they do, note that the longer word, SOLVENT takes over the shorter word SOLUTE*  

How Do We Calculate Dilutions:

Recall that concentration (M) is also written as mol/L. From this we can write C=n/V and n=CV. Dilution calculations are simplified by using the following equation:

C₁V₁ = C₂V₂ 

Where: 

• C₁ = concentration of the first solution
• V₁ = volume of the first solution
• C₂ = concentration of the second solution
• V₂ = volume of the second solution

Example One: 

Determine the concentration when 200 mL of 0.60 M HCl is diluted to a final volume of 300 mL:

• C₁ = 0.6 M
• V₁ = 200 mL
• C₂ = ?
• V₂ = 300 mL 

C₁V₁ = C₂V₂  

(C₁V₁) / V₂ = C₂  

((0.60)(200)) / 300 = C₂ 

_{0.4 M = C2}   

Example Two: 

How much water must be added to 5.0mL of 40.0M Na2SO4 to a solution with a concentration of 0.50M?

• C₁ = 40.0 M
• V₁ = 5.0 mL
• C₂ = 0.50 M
• V₂ = ? 

C₁V₁ = C₂V₂  

(C₁V₁) / C₂ = V₂  

((40.0)(5.0)) / 0.5 = V₂ 

_{400 mL = V2}    


Attached is a video with more information on dilutions:

-Simon Sierra

Diluting and Creating Solutions Lab

Titrations

What is Titration? 

Titration is a controlled acid-base neutralization reaction. This experimental procedure is used in chemistry to determine the molarity of an acid or a base. A chemical reaction is set up between a known volume of a solution of unknown concentration and a solution with a known concentration. Essentially, we will use it to determine the concentration of this unknown solution.

What is Used in the Titration Process? 

 

• The BURET contains the known solution, used to dispense known amounts of a liquid reagent in experiments for which such precision is necessary. 
• The STOPCOCK is a valve used to control the flow of solution from the buret.
• The PIPET GLASS TUBE is used to accurately measure the volume of an unknown solution.
• The ERLENMEYER FLASK is the container for the unknown solution.
• The INDICATOR is used to identify the end point of the titration process. In most cases a phenolphthalein indicator will be used.
• The STOCK SOLUTION is the known solution.  

Below is a video which thoroughly explains the steps we will take during next class when we do our titrations lab. 

 

- Simon Sierra

Solution Stoichiometry (Part Two)

Example 1:

250 mL of .300M chromium (II) fluoride reacts with excess zinc. How many grams of chromium are produced?

CrF2  +  Zn  -->  ZnF2  +  Cr

0.250L  x  .300 mol  x 1  x  52.0g  =    3.9 g
                    L            1       mol

- How many moles of ZnF2 are produced?
.250L  x .300 mol  x  1  =    0.075 mol      
                    L            1

Determine [ZnF2]  0.075 mol   x  1      =     0.3 M
                                                  0.250L

Example 2:
A beaker contains 150 mL of 2.5 M HBr. Extra cadmium is added to the beaker.
Determine how many litres of hydrogen gas should be produced at STP.

HBr  +  Cd   -->   H2   +   CdBr2

0.150 L  x  2.5 mol  x  1   x  22.4 L  =   4.2 L
                       L           2        mol
- If 2.50 L of hydrogen gas is actually produced what is the % yield?
                3.5 L  x  100  =  83%
                4.2 L

Example 3:
A 3.00g piece of Iron(II) is added to a beaker containing 100mL of 0.750M of AgNO3. Determine the LR.

0.750mol x 0.100L x 1 x 55.8g = 2.09g of Fe             LR = AgNO3
                                 2     mol

-How many grams of Ag(s) is produced?

0.750mol x 0.100L x 2 x 107.9g = 8.1 g
      L             mol      2      mol

 A video for extra help in Solution Stoichiometry.

 -Ben Suratos