Science 10 ct05d01 resource brown ford ryan ib chem
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Acid/Base Chemistry Chapter 12 (Topic 5) Please note that the chapter does not go in sequence with our lecture material, but you must still understand the concepts. Science 10 CT05D01 Resource: Brown, Ford, Ryan, IB Chem. Topic 05 – Acids/Bases. 5.1 Solutions

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Science 10 CT05D01 Resource: Brown, Ford, Ryan, IB Chem

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Science 10 ct05d01 resource brown ford ryan ib chem

Acid/Base ChemistryChapter 12 (Topic 5)Please note that the chapter does not go in sequence with our lecture material, but you must still understand the concepts

Science 10

CT05D01

Resource:

Brown, Ford, Ryan, IB Chem


Topic 05 acids bases

Topic 05 – Acids/Bases

  • 5.1 Solutions

  • 5.2 Definitions of Acids and Bases

  • 5.3 Properties of Acids and Bases

  • 5.4 Calculating pH, pOH, H+, OH-

  • 5.5 Neutralization equations

  • 5.6 Titrations

  • We will come back to Topic 04 (Redox) following the completion of Acids/Bases


5 1 solutions

5.1 - Solutions

  • 5.1.1 Define solute, solvent, solution, suspension, homogeneous, heterogeneous

  • 5.1.2 Determine the molarity of solutions, given any of the two variables

    • (M = mol / L)

    • L = dm3

  • 5.1.3 Calculate the dilution or concentration of solutions

    • (M1 x V1 = M2 x M2)

  • 5.1.4 Solubility and solubility curves


5 2 definitions of acids and bases

5.2 - Definitions of Acids and Bases

  • 5.2.1 Arrhenius: Acids donate H+, Bases donate OH- ions in solution

  • 5.2.2 Lowry/Bronsted: Acids donate a proton, Bases accept a proton (H+)


5 3 properties of acids and bases

5.3 Properties of Acids and Bases

  • 5.3.1 Acids (donate H+, turn litmus red, corrode active metals, neutralize bases, taste sour)

  • 5.3.2 Bases (donate OH-, turn litmus blue, denature protein, neutralize acids, taste bitter


5 4 calculating ph poh h oh

5.4 Calculating pH, pOH, H+, OH-

  • 5.4.1 Calculate the concentration of ions (H+ and OH-) and acidity (pH and pOH) of strong acids and bases

  • 5.4.2 Calculate the above of a mixture of strong acids and bases


5 5 neutralization equations

5.5 Neutralization equations

  • 5.5.1 Balance simple acid base equations

  • 5.5.2 Conjugate Acid/Base pairs


5 6 titrations

5.6 Titrations

  • 5.6.1 Complete titration calculations

    • (MA x VA= MB x VB)

  • 5.6.2 Identify the parts of a titration curve

  • 5.6.3 Buffer action of weak acids and bases


5 1 solutions1

5.1 - Solutions

  • The solute is the material in the smaller amount that is being dissolved

  • The solvent is the material in the greater quantity in which the solute will dissolve. Water is the universal solvent, but other organic solvents such as ethanol are common as well

  • The Solution is a dissolved homogeneous mixture of two compounds


5 1 solutions2

5.1 - Solutions

  • A suspension is found in a heterogeneous solution, that is due to the presence of insoluble particles added to the solvent material.

  • Turbulent water, and many powders (often protein shakes) can be insoluble


5 1 concentration

5.1 - Concentration

  • The concentration of a solution is a measurement of the amount of solute present in a given quantity of solvent.

  • The solute must be soluble and the solvent a liquid (generally water)

  • The unit common for representing concentration is M or molar concentration

    • The number of moles of solute in 1.0 L of solution


5 1 solutions3

5.1 - Solutions

  • In order to solve for molarity we will often have to make mole conversions

  • Do you remember how to convert between moles and grams, grams and moles, mole and molecules, molecules and moles?

  • How do you calculate the molar mass of a compound?

  • What is avogadros number?

  • How would you convert from mL to L, L to mL?


Science 10 ct05d01 resource brown ford ryan ib chem

Moles to Volume

(Molar Volume of a gas 22.4)

Moles to Molecules

(use Avogadro’s Number)

Moles to Mass

(use Molar Mass)


Simple conversions

Simple Conversions:

- Mole / Mass Conversions -

Use the Molar Mass of a substance to convert from Moles to Mass and Mass to Moles

80. g CuSO4

1 mol CuSO4

Mass to Moles 

= 0.50 mol CuSO4

159.5 g CuSO4

0.50 mol CuSO4

159.5 g CuSO4

Moles to Mass 

= 80. g CuSO4

1 mol CuSO4


Simple conversions1

Simple Conversions:

- Mole / Molecule Conversions -

Use Avogadro’s Number : 6.022 x 1023 molecules (mc) in one mole of the substance

2 mol CuSO4

6.022x1023 (mc) CuSO4

Moles to (mc) 

= 1.2x1024 (mc) CuSO4

1 mol CuSO4

1.2x1024 (mc) CuSO4

1 mol CuSO4

(mc) to Moles 

= 2 mol CuSO4

6.022x1023 (mc) CuSO4


5 1 molarity example

5.1 – Molarity Example

  • A 1.46 M glucose (C6H12O6) solution contains 1.46 moles of the solute in 1L of solution

  • Not all solutions are 1.0 L, what if I only have 500mL of the solution

    • It’s simple, you simply have half the volume so half the moles, molarity stays the same


5 1 molarity example1

5.1 – Molarity Example

  • What is the molarity of a solution containing 1.25 moles sodium chloride in 345 mL of solution?

  • What is the molarity of a solution containing 3.0 g potassium chloride in 500 mL of solution


5 1 dilution of solutions

5.1 – Dilution of Solutions

  • To dilute means to decrease the concentration by increasing the solvent amount

    • Sweet tea is too strong, add water

  • To concentrate means to increase the concentration by decreasing the solvent amount, or adding more solute.

    • Boil salt water

    • Add more Gatorade mix to your drink


5 1 dilution of solutions1

5.1 – Dilution of Solutions

  • In chemical stockrooms, concentrated solutions are often stored and diluted when needed for laboratory purposes. These are called stock solutions.

  • Dilution is a procedure for preparing a less concentrated solution from a more concentrated solution

  • There is a simple dilution equation for this:

    • Where M = molarity, V = volume

    • And 1 = concentrated, 2 = diluted


5 1 dilution of solutions2

5.1 – Dilution of Solutions

  • Suppose your teacher needs to prepare 1.0L of 0.400 M H2SO4 but currently has a large supply of 12.00 M H2SO4. How can this be achieved?

    • Use the dilution equation,

    • M1 = 12.00 M H2SO4

    • V1 = ? (trying to find how much)

    • M2 = 0.400 M H2SO4

    • V2 = 1.0 L

    • When you solve for V1 you will find that you need to dilute 33 mL of the stock solution of 12.00 M H2SO4 to a volume of 1.0 L with water.


5 1 types of solutions

5.1 - Types of solutions

  • Saturated Solution – contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature

  • Unsaturated Solution – contains less than solute than it has capacity to dissolve.

  • Supersaturated Solution – contains more solute than is present in a saturated soln.


5 1 crystallization

5.1 - Crystallization

  • Crystallization is the process in which dissolved solute comes out of solution and forms crystals

  • Usually occurs with a drop in temperature


5 1 changing the temperature of solutions

5.1 - Changing the Temperature of Solutions

  • Remember, a saturated solution is defined at a ‘specific temperature.’

  • If the temperature is changed, the solvent can hold differing amounts of solute.

  • We can demonstrate how this might happen by using a solubility curve.


5 1 solubility curves

5.1 - Solubility Curves

  • Gases decrease in solubility with an increase in temperature

  • Solid and liquid solubility increases


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