Dr. Fred Omega Garces Chemistry 100 Miramar College 9.01c-PART2 Solutions The Chemistry of Matter in Water Dr. Fred Omega Garces Chemistry 100 Miramar College

Dilution Process Dilution: When the number of Solute:Solvent ratio decreases because the Volume of solvent is increased. 1Can OJ concentrated is diluted to prepare 4-can volume of OJ

Dilute Solutions Dilution Analysis: Suppose 10 ml of the 3.4•10-2 M HNO3 solution is diluted to 50.0mL. What is the new concentration ? 10 ml 3.4•10-2 M HNO3 in 50 ml Volume. What is Conc? 100 mL of 3.4•10-2 M HNO3 Given: C1 = C2 = V1 = V2 = Goal Methodology M1 • V1 = M2 • V2 or C1 • V1 = C2 • V2 1

Dilute Solutions Dilution Analysis: Suppose 10 ml of the 3.4•10-2 M HNO3 solution is diluted to 50.0mL. What is the new concentration ? 10 ml 3.4•10-2 M HNO3 in 50 ml Volume. What is Conc? 100 mL of 3.4•10-2 M HNO3 C1 • V1 = C2 • V2 Before Dilution: 10 mL Aliquot: 10ml of 0.034•10 M Moles in aliquot- Moles = 3.4•10-2 mol • 0.010 L L = 0.00034•10 mol =3.4•10-4 mol After Dilution: 50 mL Volume: 50 mL vol. contains 3.4•10-4 mol Moles of new solution- 3.4•10-4 mol = Mnew • Vol new = Mnew • 0.050 L 6.8•10-3 M = Mnew 1

Dilution Example C1 • V1 = C2 • V2 C1 = ? V1 = ? C2 = ? V2 = ? Suppose you have 0.500 M sucrose stock solution. How do you prepare 250 mL of 0.348 M sucrose solution ? Dilution Example Example (100 RS): There is a bottle of 0.500 M sucrose stock solution in the lab. Give precise directions to your assistant to prepare 250.0 mL of a 0.348 M sucrose solution. Concentration 0.500 M Sucrose 240mL 250.0 mL of 0.348 M sucrose C1 • V1 = C2 • V2 Try to solve this Problem C1 = ? V1 = ? C2 = ? V2 = ?

Dilution Example in Analytical Chemistry 10.0 g MnSO4•H2O is placed in a 1-L volumetric flask. What volume is necessary to prepare 250mL of 0.050% solution? Assume dsoln 1 g/ml 1 Recognize the problem and state it clearly Making the observation i.e., there is a ball of fire in the sky

Solution Stoichiometry How is concentration used in stoichiometry problems?

Solution Stoichiometry: Titration From Previous problem: Suppose 100.0 ml of the 6.8•10-3 M HNO3 is titrated with NaOH. What volume of 0.0010 M NaOH is necessary to neutralize the acid solution. 1 Recognize the problem and state it clearly Making the observation i.e., there is a ball of fire in the sky

Solution Stoichiometry: Titration From Previous problem: Suppose 100.0 ml of the 6.8•10-3 M HNO3 is titrated with NaOH. What volume of 0.0010 M NaOH is necessary to neutralize the acid solution. Reaction: HNO3 (aq) + NaOH (aq)  H2O (l) + NaNO3 (aq) 6.8 e-3 M 1.0 e-3M 100.0 ml V ? F g D Moles HNO3 = 0.100L x (6.8e-3 mol / 1 L) = 6.8 e-4 mol HNO3 D g 4 At neutralization, moles HNO3 = moles NaOH = 6.8 e-4 mol NaOH 4 g 6 Volume NaOH = 6.8 e-4 mol NaOH x (1 L / 1.00e-3 mol NaOH) = 0.680 L Answer: Vol NaOH = 680.0 mL 1 Recognize the problem and state it clearly Making the observation i.e., there is a ball of fire in the sky

Solution Stoichiometry: Titration End of Chapter: Practice problems 8.64 8.111 1 Recognize the problem and state it clearly Making the observation i.e., there is a ball of fire in the sky

Solution Stoichiometry: Titration End of Chapter: Practice problems 8.64 8.111 1 Recognize the problem and state it clearly Making the observation i.e., there is a ball of fire in the sky

Solution Stoichiometry: Titration End of Chapter: Practice problems 8.64 8.111 1 Recognize the problem and state it clearly Making the observation i.e., there is a ball of fire in the sky

Solution Stoichiometry: Titration End of Chapter: Practice problems 8.101 8.118 8.101

Solution Stoichiometry: Titration End of Chapter: Practice problems 8.101 8.118 8.101

Solution Stoichiometry: Titration End of Chapter: Practice problems 8.101 8.118 8.101

Solution at a Glance Solutions can be describe by the following: Solvent The component of a solution present in the greatest quantity Solute The component of solution present in the lesser quantity Solution A homogeneous mixture of two or more substances in which each substance retains its chemical identity Concentration of a Solution The amount of solute in a specific amount of solution. Molarity (M) moles of solute Liters of solution