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Calculations Involving Molarity (Aqueous Solution)

Calculations Involving Molarity (Aqueous Solution). If 100 mL of 1.00 M NaOH and 100 mL of 0.500 M H 2 SO 4 solution are mixed, what will the concentration of the resulting salt solution be? 1 ) Write the balanced equations: 2NaOH + H 2 SO 4  Na 2 SO 4 + 2H 2 O

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Calculations Involving Molarity (Aqueous Solution)

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  1. Calculations Involving Molarity (Aqueous Solution) • If 100 mL of 1.00 M NaOH and 100 mL of 0.500 M H2SO4 solution are mixed, what will the concentration of the resulting salt solution be? • 1) Write the balanced equations: • 2NaOH + H2SO4 Na2SO4 + 2H2O • 2) Determine the starting # of moles of each reactant: • moles NaOH = 1.00 mol/L * 0.100L = 0.100 mol NaOH • moles H2SO4 = 0.500 mol/L * 0.100L = 0.0500 mol H2SO4 • 3) Determine the limiting reagant: • moles of Na2SO4 from NaOH = • 0.100 mol NaOH*(1 mol Na2SO4/2mol NaOH)=0.0500mol Na2SO4 • moles of Na2SO4 from H2SO4 = • 0.0500 mol H2SO4*(1 mol Na2SO4/1mol H2SO4)=0.0500mol Na2SO4 • 4) Determine the final molarity: • M of Na2SO4 = 0.0500mol Na2SO4/(0.100L +0.100L)=0.250 M Na2SO4

  2. If 100 mL of 1.00 M NaOH and 100 mL of 0.500 M H2SO4 solution are mixed, what will the concentration of the resulting salt solution be? 2NaOH + H2SO4 Na2SO4 + 2H2O Rxn Ratio: 2 mol 1 mol 1 mol 2 mol Start: 0.100 mol0.0500 mol 0 mol Change: -0.100 mol -0.0500 mol +0.0500 mol Determine the final molarity: M of Na2SO4 = 0.0500mol Na2SO4/(0.100L +0.100L)=0.250 M Na2SO4

  3. mmoles (millimoles) • mmoles = 10-3 moles & mL = 10-3 L • Molarity = mmol/mL = 10-3 mol/10-3 L = mol/L • If 100 mL of 1.00 M NaOH and 100 mL of 0.500 M H2SO4 solution are mixed, what will the concentration of the resulting salt solution be? • 2NaOH + H2SO4 Na2SO4 + 2H2O • mmoles NaOH = 1.00 mol/L * 100mL = 100 mmol NaOH • mmoles H2SO4 = 0.500 mol/L * 100mL = 50.0 mmol H2SO4 • mmoles of Na2SO4 from NaOH = • 100 mmol NaOH*(1 mmol Na2SO4/2mmol NaOH)=50.0 mmol Na2SO4 • mmoles of Na2SO4 from H2SO4 = • 50.0mmol H2SO4*(1 mmol Na2SO4/1mmol H2SO4)=50.0 mmol Na2SO4 • M of Na2SO4 = 50.0mmol Na2SO4/(100mL + 100mL)=0.250 M Na2SO4

  4. Titrations • Titration - the process by which one determines the volume of a standard solution required to react with a specific amount of another substance. • Standard solutions - solutions of accurately known concentrations. • Primary Standard - a compound that exists in a known high degree of purity, doesn’t react with the atmosphere, is soluble in water, has a high formula weight and reacts according to one invariable reaction. • Secondary Standard - A solution that is standardized by a primary standard. • Standardization - a process by which one determines the concentration of a solution by titration with a standard solution. • Equivalence point - the point at which chemically equivalent amounts of acid and base have reacted (neutralization). • End point - the point at which the indicator changes color. (Slightly past the equivalence point)

  5. Indicator- a highly colored substance that can exist in different forms, with differnet colors that depend on the concentration of protons in soln.

  6. Acid-Base Titration • Common Primary Standard - Potassium Hydrogen Phthalate (KHP) an acidic salt. phthalic acid C6H4(COOH)2 C8H6O4 K C6H4(COO)(COOH) KC8H5O4 KHP + NaOH  NaKP + H2O

  7. Sodium carbonate is another common primary standard • What is the molarity of a sulfuric acid solution if 40.0 mL of the solution neutralizes 0.364 grams of sodium carbonate? • Na2CO3 + H2SO4Na2SO4 + H2O+ CO2 • ? M of sulfuric acid = mol H2SO4/L of soln • We need to find the mol of H2SO4 • 0.364 g NaCO3*(1 mol/106.0g Na2CO3)*(1mol H2SO4/1mol Na2CO3) • = 0.00343 mol H2SO4 • M H2SO4 = 0.00343 mol H2SO4/0.0400L H2SO4 = 0.00858 M H2SO4

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