Solutions Calculations

Transcription

1 Solutions Calculations The Chemistry of Matter in Water Dr. Fred Omega Garces Chemistry 100 Lab Miramar College 1 Solutions

2 Ionic Vs Covalent Compounds: Electrolyte Vs. Nonelectrolyte Substance when dissolve can break-up to ions (NaCl) or stay intact (sugar). Type: % ionization: Solubility Electrolyte: conducts electricity. Strong electrolyte 100 % ionization very soluble weak electrolyte less 100% ionization slightly soluble Nonelectrolyte: no conduction zero ionization insoluble. 2 Solutions

3 9 Types of Solution (derived from 3 phases) Solute Solvent Solution Gas in a gas Air Gas in a liquid carbonated water.. Liquid in a gas fog Liquid in a liquid wine.. Solid in a solid 14 karat gold 3 Solutions

4 Nature of Solute and Solvent Dissolving Process: Why does water not mix with oil? Yet water mixes with alcohol. An oil layer floating on water. For a Solubility Factor: Solute and Solvent characteristic: In aqueous solution, water as the solvent will dissolve only other polar molecules Oil is a nonpolar substance which will only dissolve other nonpolar substances such as organic solvents. The result is the immisciblity of the two liquid. Like Dissolves Like substance to dissolve, the water-water hydrogen bonds must be broken to make a hole for each solute particle. However, the water-water interactions will break only if they are replaced by similar strong interactions with the solute. 4 Solutions

5 Suspended in Solution: Dissolution Solubility - The process in which substances dissolves at the molecular level. Solubility - The maximum amount of that solute that dissolves in a given amount of solvent in a given temperature. Unsaturated Saturated SuperSaturated A solution that has the capacity to dissolve solute A solution that contains all solute it can dissolve (There are no residue) A solution that contains more solute (in dissolved form) than normal Miscible When two liquids are soluble in each other Immiscible When two liquids are not soluble in each other. 5 Solutions

6 Principles of Solubility Homogeneous systems: Solutions Solution - Homogeneous mixture of two or more substances Components of solution Solute - Substance being dissolve Solvent - Substance in which solute is dissolved in. If solvent is water, then solution is considered aqueous. 6 Solutions

7 Solution concentration Concentration- Amount of solute in given amount of solvent. Most common type: Molarity Molarity - moles solute / Liter solution How many moles in 16.0 g CuSO 4 (MW = g/mol)? If this amount of CuSO 4 is dissolved in 0.100L of water what is the Molarity of the solution (mol / L)? To make a 1.00-M solution ofcuso 4, 16.0 g, or mol, of CuSO 4 (the blue crystalline solid) is placed in a 100-mL volumetric flask. Exactly 100 ml of water is measured out and slowly added to the volumetric flask. When enough water had been added so that the solution volume is exactly 100 ml. The emphasizes here is that molar concentrations is defined as moles per liter of solution and not per liter of water or other solvent. 7 Solutions

8 Solution and Concentration 5 ways of expressing concentration- Molarity(M) - moles solute / Liter solution Mass percent (% m)- (grams solute/total grams of solution)*100 Molality * (m) - moles solute / Kg solvent Normality(N) - Number of equivalent / Liter solution mole fraction( χ A) - moles solute / Total moles solution * Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution. 8 Solutions

9 Concentration Relationship! Molecular Weight Moles mass } Solute Molc Wt Density Moles mass Volume } Solvent χ m* %m M * Volume used must be volume of solvent N 9 Solutions

10 % Concentration ppm % Concentration and ppb Concentration w/w = Wt Solute 100 1,000,000 Wt Soln g 100 g % (pph) g g ppm w/v = Wt Solute 100 1,000,000 g 100 g g % ppm (pph) Vol Soln ml v/v = Vol Solute 100 1,000,000 ml 100 g ppm % Vol Soln ml ppm & ppb (For dilute solid in liquid solution) m/v = Wt Solute 1,000,000 g 10 6 g ppm (ppm) Vol Soln ml m/v = Wt Solute 1,000,000,000 g 10 9 g ppb (ppb) Vol Soln ml 10 Solutions

11 Concentration: Molarity Example Suppose g of KMnO 4 is dissolved in enough water to give 250. ml of solution. (i) How many moles of KMnO 4 was used? (ii) What is the molar concentration of KMnO 4? In most stoichiometry problems, the first step is to convert the mass of material to moles. Molar Mass KMnO 4 = g/mol 250 ml 11 Solutions

12 Concentration: Molarity Example Suppose g of KMnO 4 is dissolved in enough water to give 250. ml of solution. (i) How many moles of KMnO 4 was used? (ii) What is the molar concentration of KMnO 4? In most stoichiometry problems, the first step is to convert the mass of material to moles. Molar Mass KMnO 4 = g/mol 12 Solutions

13 Concentration: Molarity Example (1 st ) Suppose g of KMnO 4 is dissolved in enough water to give 250. ml of solution. (i) How many moles of KMnO 4 was used? (ii) What is the molar concentration of KMnO 4? In most stoichiometry problems, the first step is to convert the mass of material to moles. Molar Mass KMnO 4 = g/mol i) g KMnO 4 1 mol KMnO 4 = mole KMnO g KMnO 4 Now that the number of moles of substance is known, this can be combined with the volume of solution which must be in liters to give the molarity ml is equivalent to L. ii) Molarity KMnO 4 = mol KMnO 4 = M L solution 13 Solutions

14 Molarity: second example i) What is the molar concentration of 1.00 g potassium dichromate (K 2 Cr 2 O 7, MW = g/mol) in 250mL solution? ii) What is the concentration as % w/v? Molarity = moles Solute mol g Molar (M) (M) L solution L 14 Solutions

15 Molarity: Second example i) What is the molar concentration of 1.00 g potassium dichromate (K 2 Cr 2 O 7, MW = g/mol) in 250mL solution? ii) What is the concentration as % w/w? Molarity = moles Solute mol g Molar (M) (M) L solution L Calculate the concentration of 1.00 g K 2 Cr 2 O 7 in 250 ml solution? given goal Mass(g) M (g/l) Vol (L) Molar mass K 2 Cr 2 O 7 : g/mol Answer: C g D g F Molarity: 1.00 g mol 1 = mol = M g 0.250L L Assume % w/v = 1.00g K 2 Cr 2 O 7 x 100 = % w/v density = 1.0 g/cc 250 ml solution 15 Solutions

16 Solution Preparation (Molarity as a conversion factor) When a targeted concentration is sought: What mass of K 2 Cr 2 O 7 is needed to prepare 500mL of M solution given goal Conc M mass (g) Vol 0.500L Molar mass K 2 Cr 2 O 7 MW =294.2 g/mol 16 Solutions

17 Solution Preparation (Molarity as a conversion factor) When trying to prepare a specific concentration: What mass of K 2 Cr 2 O 7 is needed to prepare 500.mL of M solution given goal Conc M mass (g) Vol 0.500L Molar mass K 2 Cr 2 O 7 MWt = g/mol Strategy: F g D g C L mol g = g L mol L 0.20 mol g = 29.4 g L mol Weigh 29.4 g of K 2 Cr 2 O 7 placed in 500mL Vol flask and dilute to mark 17 Solutions

18 Solution Preparation (Molarity as a conversion factor) When a targeted concentration is sought: What mass of K 2 Cr 2 O 7 is needed to prepare 500mL of 0.20 M solution given goal Conc M mass (g) Vol 0.500L Molar mass K 2 Cr 2 O g/mol 18 Solutions

19 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 19 Solutions

20 Dilute Solutions HNO 3 dilution: Suppose 10 ml of the M HNO 3 solution is diluted to 50.0mL. What is the new concentration? 10 ml M HNO 3 in 50 ml Volume. What is Conc? 100 ml of M HNO 3 Given: C1 = C2 = V1 = V2 = Goal or Methodology M 1 V 1 = M 2 V 2 C 1 V 1 = C 2 V 2 20 Solutions

21 Dilute Solutions Dilution Analysis: Suppose 10 ml of the M HNO 3 solution is diluted to 50.0mL. What is the new concentration? 10 ml M HNO 3 in 50 ml Volume. What is Conc? 100 ml of M HNO 3 C 1 V 1 = C 2 V 2 Before Dilution: 10 ml Aliquot: 10ml of M Moles in aliquot- C1 * V1 Moles = mol L L = mol After Dilution: 50 ml Volume: 50 ml vol. contains mol Moles of new solution mol = M new Vol new = M new L M = M new 21 Solutions

22 Dilution Example Example (100 RS): There is a bottle of M sucrose stock solution in the lab. Give precise directions to your assistant to prepare ml of a M sucrose solution. 240mL ml of M sucrose Suppose you have M sucrose stock solution. How do you prepare 250 ml of M sucrose solution? Concentration M Sucrose C 1 V 1 = C 2 V 2 C 1 =? V 1 =? C 2 =? V 2 =?.500M? mL 0.500M V 1 = 0.34M8 250mL V 1 = mL = 174mL 0.500M 22 Solutions

23 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? 10.0g MnSO 4 H 2 O 1 L 1 L 1000 ml 100 = 1.00% solution C 1 V 1 = C 2 V 2 V 1 = C 2 V 2 C 1 C 1 = 1.00 % V 1 =? C 2 = % V 2 = 250 ml V 1 = C 2 V 2 C 1 = 0.05 % 250 ml 1.00 % V 1 = 12.5 ml of 1.00 % Solution Aliquot 12.5 ml of 1.00% in to 250mL flask Dilute to the 250mL mark with water. 23 Solutions

24 Solution Stoichiometry How is concentration used in stoichiometry problems? 24 Solutions

25 Solution Stoichiometry: Titration From Previous problem: Suppose ml of the M HNO 3 is titrated with NaOH. What volume of M NaOH is necessary to neutralize the acid solution. 25 Solutions

26 Solution Stoichiometry: Titration From Previous problem: Suppose ml of the M HNO 3 is titrated with NaOH. What volume of M NaOH is necessary to neutralize the acid solution. Reaction: HNO 3 (aq) + NaOH (aq) H 2 O (l) + NaNO 3 (aq) 6.8 e -3 M 1.0 e -3 M ml V? F g D Moles HNO 3 = 0.100L x (6.8e -3 mol / 1 L) = 6.8 e -4 mol HNO 3 D g 4 At neutralization, moles HNO 3 = 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) = L Answer: Vol NaOH = ml 26 Solutions

27 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 27 Solutions