Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Define and identify electrolytes and non-electrolytes.
• Write balanced equations for precipitation reactions.
• Predict precipitate formation using solubility rules.
• Perform stoichiometric calculations with molarity and volume.
• Define acids, bases, and neutralization reactions.

Lesson summary

This week, we delve into the exciting world of reactions happening in water – aqueous solutions. These reactions are fundamental to many processes around us, from the cleaning of our homes with detergents to the industrial production of fertilizers that keep our crops growing. Understanding how these reactions work, and how much of each substance is involved (stoichiometry), is crucial for various fields, including agriculture, environmental science, and medicine.

Lesson notes

2.1 Aqueous Solutions: The Medium of Reactions An aqueous solution is a solution where the solvent is water. Water's polarity makes it an excellent solvent for ionic compounds and polar covalent compounds. When ionic compounds dissolve in water, they dissociate into their constituent ions. 2.2 Electrolytes and Non-electrolytes Electrolytes: Substances that dissociate into ions when dissolved in water, forming a solution that conducts electricity. Examples include NaCl (table salt), KCl, and strong acids like HCl and H2SO

4. Strong electrolytes dissociate completely. Weak electrolytes, like acetic acid (vinegar), only partially dissociate.

Non-electrolytes: Substances that do not dissociate into ions when dissolved in water. Their solutions do not conduct electricity. Examples include sugar (sucrose, C12H22O11) and ethanol (C2H5OH). Why do electrolytes conduct electricity? The presence of mobile ions allows the flow of electric charge through the solution. Non-electrolytes, lacking these mobile ions, cannot conduct electricity. 2.3 Precipitation Reactions and Solubility Rules A precipitation reaction occurs when two aqueous solutions are mixed, and an insoluble solid (the precipitate) forms. To predict whether a precipitate will form, we use solubility rules. These rules are generalizations about the solubility of ionic compounds in water. Here's a simplified version of common solubility rules: Most nitrate (NO3-) salts are soluble. Most alkali metal (Group 1) salts and ammonium (NH4+) salts are soluble. Most chloride (Cl-), bromide (Br-), and iodide (I-) salts are soluble, EXCEPT those of silver (Ag+), lead (Pb2+), and mercury (Hg2+). Most sulfate (SO42-) salts are soluble, EXCEPT those of barium (Ba2+), strontium (Sr2+), lead (Pb2+), and calcium (Ca2+) (Calcium sulfate is only slightly soluble). Most hydroxide (OH-) salts are insoluble, EXCEPT those of alkali metals and barium (Ba2+). Calcium hydroxide (Ca(OH)2) is slightly soluble. Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-) salts are insoluble, EXCEPT those of alkali metals and ammonium (NH4+).

Example: Predict what will happen when aqueous solutions of silver nitrate (AgNO3) and sodium chloride (NaCl) are mixed. AgNO3(aq) + NaCl(aq) → ?

Possible products: AgCl and NaNO3 Solubility rules: Rule 1: NaNO3 is soluble (nitrate salt).

Rule 3: AgCl is insoluble (chloride of silver).

Therefore, a precipitate of AgCl will form.

The balanced equation is: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) 2.4 Stoichiometry of Reactions in Solution Molarity (M) is a measure of concentration, defined as moles of solute per liter of solution: Molarity (M) = moles of solute / liters of solution Stoichiometry allows us to calculate the amounts of reactants and products involved in chemical reactions. When dealing with solutions, we often use molarity and volume to determine the number of moles.

Example: What volume of 0.200 M NaOH solution is needed to completely react with 25.0 mL of 0.150 M H2SO4 solution?

Write the balanced chemical equation: H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

Calculate the moles of H2SO4: moles H2SO4 = Molarity × Volume = 0.150 mol/L × 0.0250 L = 0.00375 mol Use the stoichiometry of the balanced equation to find the moles of NaOH needed: From the equation, 1 mole of H2SO4 reacts with 2 moles of NaOH. moles NaOH = 0.00375 mol H2SO4 × (2 mol NaOH / 1 mol H2SO4) = 0.00750 mol Calculate the volume of 0.200 M NaOH solution needed: Volume NaOH = moles / Molarity = 0.00750 mol / 0.200 mol/L = 0.0375 L = 37.5 mL Therefore, 37.5 mL of 0.200 M NaOH solution is needed. 2.5 Acids, Bases, and Neutralization Reactions Acids: Substances that donate protons (H+) in water or increase the concentration of H+ ions. Examples include hydrochloric acid (HCl) and sulfuric acid (H2SO4).

Bases: Substances that accept protons (H+) in water or increase the concentration of hydroxide ions (OH-). Examples include sodium hydroxide (NaOH) and ammonia (NH3).

Neutralization Reaction: The reaction between an acid and a base, typically producing a salt and water.

Example: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 2.6 Dilution Dilution is the process of decreasing the concentration of a solution by adding more solvent. The number of moles of solute remains constant during dilution.

The dilution equation is: M1V1 = M2V2 Where: M1 = Initial Molarity V1 = Initial Volume M2 = Final Molarity V2 = Final Volume

Example: If 50.0 mL of a 2.0 M solution of NaCl is diluted to a final volume of 250.0 mL, what is the final concentration of the solution? M1 = 2.0 M V1 = 50.0 mL V2 = 250.0 mL M2 = ?

Using the dilution equation: (2.0 M)(50.0 mL) = M2 (250.0 mL) M2 = (2.0 M * 50.0 mL) / 250.0 mL = 0.40 M Guided Practice (With Solutions)

Question 1: Will a precipitate form when aqueous solutions of potassium iodide (KI) and lead(II) nitrate (Pb(NO3)2) are mixed? If so, identify the precipitate and write the balanced chemical equation, including state symbols.