Acids and Bases Test - Study Guide Notes

Acids & Bases Test - Study Guide

1. Properties and Definitions

• Properties of Acids and Bases: Students need to know the five properties of acids and bases (not specified in the provided text).
• Arrhenius Definitions: Acids increase the concentration of hydrogen ions (H+) in aqueous solution, while bases increase the concentration of hydroxide ions (OH-).

2. Identifying Acids and Bases

• Chemical Formula Identification: Be able to identify acids and bases from their chemical formulas.
• Examples:
  • Acids: H2SO4, HBr, H3PO4
  • Bases: Pb(OH)2, NaOH, KOH
  • Other: LiCl, HOH (Water), Mg(NO3)2

3. Strength of Acids and Bases

• Factors Determining Strength: Understand what determines the strength or weakness of acids and bases. (This is not elaborated on in the provided text, but could relate to the degree of ionization in solution).

4. Household Examples

• Acids and Bases in Everyday Life: Be familiar with household examples of acids and bases.
• Examples:
  • Milk, eggs, coffee, soda, salt water, ammonia, bleach, oven cleaner, shampoo, rust remover, tap water, dish soap, drain plug remover, jewelry cleaner

5. Neutralization

• Definition: Neutralization is the reaction between an acid and a base.
• Chemical Equation:
  $Acid + Base \rightarrow Salt + Water$
• Examples:
  • $HI + Cu(OH)_2 \rightarrow ?$
  • $HNO_3 + NaOH \rightarrow ?$

6. Lab Understanding

• Part A: Understand the implications of results with red litmus, blue litmus, phenolphthalein, and Universal Indicator.
• Part B & C: Understand neutralization reactions and titration math problems.

7. pH and pOH Calculations

• Formulas:
  • $pH = -log[H_3O^+]$
  • $[H_3O^+] = 10^{-pH}$
  • $pOH = -log[OH^-]$
  • $[OH^-] = 10^{-pOH}$
• Calculations: Be able to calculate pH and pOH from hydronium or hydroxide ion concentrations and vice versa.

8. Ionization Constant for Water

• Kw and its Relation to pH: Know how the ionization constant for water (Kw) is related to acid/base concentration and pH.
• Equation: $Kw = 1.0 \times 10^{-14} = [H_3O^+][OH^-]$ or $pH + pOH = 14$

9. Strength Comparison

• Acidity/Basicity Comparison: Determine how much stronger or weaker an acid or base is compared to another, given pH or pOH values.
• Examples:
  • pH of 3 vs. pH of 5 (both acidic)
  • pH of 6 vs. pH of 12 (both basic)
  • pOH of 1 vs. pOH of 10 (both basic)

10. Solutions, Solvents, and Solutes

• Definitions:
  • Solution: A homogeneous mixture of two or more substances.
  • Solvent: The substance in which the solute is dissolved (typically the substance present in greater amount).
  • Solute: The substance that is dissolved in the solvent.
• Saturation Levels: Know the different levels of saturation.
• Temperature/Pressure Effects: Understand how temperature and pressure affect the solubility of solids, liquids, and gases.

11. Molarity

• Definition: Molarity (M) is a measure of a solution's concentration, defined as moles of solute per liter of solution.
• Formula: $Molarity (M) = \frac{moles <br /> of<br /> solute}{Liters <br /> of<br /> solution}$
• Problem Solving: Solve problems involving molarity, moles of solute, and liters (or mL) of solution (like the Titration Part C of the Acid-Base Lab).

12. Organic Chemistry Basics

• Root Names: Memorize the root names for #1-12 (meth-, eth-, prop-, but-, pent-, hex, hept-, oct-, non-, dec-, undec-, dodec-).
• Formulas: Be able to write chemical, structural, skeletal, and condensed formulas.
• Drawing & Naming: Practice drawing and naming alkanes, alkenes, and alkynes, including those with branches and multiple bonds (double & triple).
• Limitations: No alcohols, halogens, ring structures, or carboxyl groups will be on the test.

Sample Problems

1. pH from [H+]

Find the pH of a hydrogen ion concentration of $1.0 \times 10^{-12} M$?

$pH = -log[1.0 \times 10^{-12}]$
$pH = 12$

2. pOH from [OH-]

Find the pOH of a hydroxide ion concentration of $1.0 \times 10^{4} M$?

$pOH = -log[1.0 \times 10^{4}]$
$pOH = -4$

3. pH from [OH-]

Find the pH of a hydroxide concentration $[OH^-] = 3.29 \times 10^{12} M$?

$pOH = -log[3.29 \times 10^{12}]$
$pOH = -12.517$
$pH = 14-pOH$
$pH = 14-(-12.517) = 26.517$

4. [H+] from pH

Find the $[H^+]$ concentration for a $pH = 6.33$?

$[H^+] = 10^{-pH}$
$[H^+] = 10^{-6.33} = 4.677 \times 10^{-7} M$

5. [OH-] from pOH

Find the $[OH^-]$ concentration for a $pOH = 11.79$?

$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{-11.79} = 1.62 \times 10^{-12} M$

6. [H3O+] from pOH

Find the $[H_3O^+]$ concentration for a $pOH = 4.347$?

$pH = 14 - pOH$
$pH = 14 - 4.347 = 9.653$
$[H<em>3O^+] = 10^{-pH}$$[H</em>3O^+] = 10^{-9.653} = 2.22 \times 10^{-10} M$

7. Molarity Calculation

Find the molarity of 5.23 moles of $KMnO_4$ dissolved into 13.7 Liters of water?

$Molarity = \frac{moles <br /> of<br /> solute}{Liters <br /> of<br /> solution}$
$Molarity = \frac{5.23}{13.7} = 0.382 M$

8. Volume Calculation

How many Liters of water are needed to make a 1.55 M solution of $KMnO_4$ that contains 22.86 moles of potassium permanganate?

$Volume = \frac{moles}{Molarity}$
$Volume = \frac{22.86}{1.55} = 14.75 Liters$

9. Moles Calculation

Find the moles of $KMnO_4$ in a 6.70 M solution that has a volume of 3.88 L?

$Moles = Molarity \times Volume$
$Moles = 6.70 \times 3.88 = 25.996 moles$

10. Molarity with Grams and mL

Find the molarity of a solution with 4500.7 g of $KMnO_4$ , and 32500 mL of water? mm = 158.034 g/mol

Convert grams to moles: $\frac{4500.7 g}{158.034 g/mol} = 28.48 moles$
Convert mL to Liters: $\frac{32500 mL}{1000} = 32.5 L$
$Molarity = \frac{28.48}{32.5} = 0.876 M$

11. Titration Molarity

Find molarity of an unknown HCl solution, if you used 43.5 mL of 2.25 M NaOH that reached the titration end point with 284 mL of the unknown acid.

$Moles <br /> of<br /> NaOH = Molarity \times Volume$
$Moles <br /> of<br /> NaOH = 2.25 \times 0.0435 L = 0.097875 moles$
Since the reaction between HCl and NaOH is 1:1, moles of HCl = moles of NaOH
$Molarity <br /> of<br /> HCl = \frac{Moles <br /> of<br /> HCl}{Volume <br /> of<br /> HCl}$
$Molarity <br /> of<br /> HCl = \frac{0.097875}{0.284} = 0.345 M$