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Ch5_D_Solutions

# Ch5_D_Solutions - REACTIONS IN SOLUTION Section 5.5...

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Unformatted text preview: REACTIONS IN SOLUTION Section 5.5 Terminology Terminology In solution we need to define the • SOLVENT the component whose physical state is preserved when solution forms • SOLUTE the other solution component Concentration of Solute Concentration of Solute The amount of solute in a solution is given by its concentration. concentration Molarity(M) = moles solute liters of solution PROBLEM: Dissolve 5.00 g of NiCl 22•6 H22O in PROBLEM: Dissolve 5.00 g of NiCl •6 H O in enough water to make 250 mL of solution. enough water to make 250 mL of solution. Calculate molarity. Calculate molarity. molarity. Step 1: Calculate moles of NiCl2•6H2O 5.00 g • 1 mol = 0 .0210 mol 237.7 g The Nature of the KMnO 44 Solution The Nature of the KMnO Solution The Nature of a Na 22CO33 Solution The Nature of a Na CO Solution This water-soluble compound is ionic Na2CO3(aq) --> 2 Na+(aq) + CO32-(aq) aq) aq) aq) 3 Na 2CO3 Na If [Na2CO3] = 0.100 M, then [Na+] = 0.200 M [CO32-] = 0.100 M Step 2: Calculate molarity 0.0210 mol = 0 .0841 M 0.250 L [NiCl2•6 H2O ] = 0.0841 M NiCl KMnO4(aq) --> K+(aq) + MnO4-(aq) aq) aq) aq) If [KMnO4 ] = 0.30 M, then [K+] = [MnO 4-] = 0.30 M Page 1 USING MOLARITY USING MOLARITY What mass of oxalic acid, H 2C2O4, is required to make 250. mL of a 0.0500 M solution? Because Conc (M) = moles/volume = mol/V mol/V this means that USING MOLARITY USING MOLARITY What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? Preparing Solutions Preparing Solutions • Weigh out a solid solute and dissolve in a given quantity of solvent. • Dilute a concentrated solution to give one that is less concentrated. moles = M • V Step 1: Calculate moles of acid required. (0.0500 mol/L)(0.250 L) = 0.0125 mol mol/L)(0.250 Step 2: Calculate mass of acid required. (0.0125 mol )(90.00 g/mol) = g/mol) moles = M • V 1.13 g 1.13 PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What NaOH and you want 0.50 M NaOH. What NaOH. do you do? do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What NaOH and you want 0.50 M NaOH. What NaOH. do you do? do you do? H2O PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What NaOH and you want 0.50 M NaOH. What NaOH. do you do? do you do? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution Dilute the solution! But how much water do we add? 3.0 M NaOH Concentrated 0.50 M NaOH Dilute Page 2 PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What NaOH and you want 0.50 M NaOH. What NaOH. do you do? do you do? Moles of NaOH in original solution = PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What NaOH and you want 0.50 M NaOH. What NaOH. do you do? do you do? Conclusion: H2O Preparing Solutions Preparing Solutions by Dilution by Dilution A shortcut shortcut M•V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH mol/L)(0.050 Therefore, moles of NaOH in final solution must also = 0.15 mol NaOH (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L NaOH)(1 mol) or 300 mL = volume of final solution 300 3.0 M NaOH Concentrated add 250 mL of water to to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M 0.50 M NaOH NaOH. NaOH. Dilute Minitial • Vinitial = Mfinal • Vfinal Page 3 ...
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