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TAMU CHEM 101 - pH Scale, Stoichiometry in Aqueous Solutions, & Titrations
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CHEM 101 1st Edition Lecture 12 Outline of Last Lecture I. Quantitative Chemical AnalysisII. CombustionIII. Analysis for Hydrocarbons by CombustionIV. Volumetric Chemical AnalysisOutline of Current Lecture I. Volumetric Chemical AnalysisII. pH ScaleIII. Stoichiometry in Aqueous SolutionsIV. TitrationsV. Acid-Base TitrationsCurrent Lecture- Volumetric Chemical Analysiso Ex: We prepare a 250 mL aqueous solution of 5.00 g of NiCl2 x 6H2O. Calculate themolarity of the solution Step 1: Calculate moles of NiCl2 x 6H2O 5.00 g NiCl2 x 6H2O x 1 mol = 0.0210 mol NiCl2 x 6H2O 237.7g Step 2: Calculate the molarity 0.0210 mol NiCl2 x 6H2O x 1 x 10^(3) mL = 0.0841 M 250.0 mL 1Lo For which concentration is asked for?o Ex: CuCl2(aq)  Cu^(2+)(aq) + 2 Cl^(-)(aq)[CuCl2] = 0.30 M=[Cu^2+)]=0.30 M[Cl^(-)] = 2 x 0.03 M=0.60 Mo Ex: Determine the nitrate ion concentration (in mol/L) of a solution that is made from 50.145 g of iron (III) nitrate diluted to a final volume of 250.0 mLStep 1: Write balanced chemical equation g Fe(NO3)3(aq)  Fe^(3+) + 3NO3^(-)(aq)These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Step 2: g Fe(NO3)3  mol Fe(NO3)3  mol NO3^(-) x 1  mol NO 3 ^(-) Vol (L) Vol (L)Step 3: 50.145g Fe(NO3) 3 x 1 mol Fe(NO 3 ) x 3 mol NO 3 ^(-) x 1 1 1000 mL = 2.488 M 241.88g Fe(NO3) 1 mol Fe(NO3) 3 250 mL 1 L (NO3)3o Ex: How many grams of sodium phosphate are in 35.0 mL of a 1.51 M Na3PO4 solution? mL solution  L  mol Na3PO4  g Na3PO4 35.0 mL x 1L x 1.51 mol Na 3 PO 4 x 163.94 g Na 3 PO 4 = 8.66 g Na3PO4 1000 mL L 1 mol Na3PO4o When more solvent is added to a solution, the concentration of solute decreasesconcentration (M) = moles Solute L Solutionas solvent is added, the denominator increases in magnitudeo Ex: a 50.0 mL of a 0.515 M HCl solution is added to a 500.0 mL volumetric flask. The volumetric flask is then filled to the calibration mark with water. What is the new molarity of the HCl solution? 50.0 Ml HCl x 1 L x 0.515 mol HCl x 1 x 1000 mL = 0.0515 M/L 1000 mL 1 L 500 mL 1L alternate approach to solve the problem: Initial concentration C1 = 0.515 M; initial volume V1= 50.0 mL C2= ? V2= 500 mL C1 x V1 = C2 x V2  C2 =C 1 x V 1 V2o Clicker Question: If you dilute 25 mL of 1.50 M hydrochloric acid to 500 mL, what is the molar concentration of the dilute acid?A. .750 MB. 0.150 MC. 1.50 MD. 0.0750 M (correct)C1 = 1.50 M; initial volume V1= 25.0 mL  1.50 M x 25 mL = 0.0750 MC2= ? V2= 500 mL 500 mL- The pH Scaleo The pH scale allows us to handle concentrations of H3O^(+) ions which describe the acidity of solutions. Acidic solution- has A pH of 7> is considered to have a low pH= high [H^(+)] (lemon juice, acid rain at 4) Basic solution- has A pH of >7 is considered to have a high pH= low [H^(+)] (soap) Neutral solution- has a pH of 7 (pure water)o What we call pH is actually a mathematical function, “p”o p is a shorthand notation for “-log10” (the negative logarithm, based on 10) pH= -log[H3O^(+)] [H3O^(+)] = 10^(-pH)o Ex: A solution has a pH of 4.72, what is the molarity of H3O^(+) in this solution? [H3O^(+)] = 10^(-pH) = 10^(-4.72) (the 4. Is called the characteristic, the .72 called mantissa) = 1.905460718x10^(-5) M 10^(-4.72) = 10^(0.28-5) = 10^(0.28) x 10^(-5) = 1.9 x 10^(-5) M [H3O^(+)]o Clicker question: Which of the solutions listed below has the lowest pH?A. 0.10 M HCl (correct)B. 0.10 M NaOHC. 2.5 x 10^(-5) M NHO3D. Pure H2OAnswer is based on where you can find the highest concentration of H ions. B and C can be ruled out since there are no Hydrogen ions and A’s number was much higher.- Stoichiometry in Aqueous Solutionso Ex: how many grams of calcium carbonate can be consumed by 35.5 mL of 0.125 M H2SO4(aq)? CaCO3(s) + 2H2SO4(aq)  CaSo4(s) CO2(g) +H2O(l) mL H2SO4  mol H2SO4  mol CaCO3  g CaCO3 35.5 mL x L x 0.125 mol H2So4 x 1 mol CaCO3 x 100.1 g CaCO3 = 0.444 g CaCo3 1000 mL L 1 mol H2SO4 1 mol CaCO3- Titrationso Some solutions cannot be accurately made by weight and dilution methods if thesolute is impure or unstable.o If such a solute is dissolved, a solution of approximate concentration of the solute is obtainedo The exact concentration of the solute can then be determined with a standard compound that reacts with the solute in solution.o To determine the exact concentration of the solute, we perform a titration wherea measured volume of the solution of the standard is added to the solutiono An indicator is added to indicate the point where the moles of standard = moles of solute reacted (equivalence point).o Knowing moles and volume, we can calculate the concentration of the solution.- Acid-Base Titrations1. Solution from the buret is added2. Reagent (base) reacts with compound (acid) in solution in the flask3. Indicator shows when exact stoichiometric reaction has occurred4. Net ionic equation: H3O^(+)(aq) + OH^(-)(aq)  2 H2O(l)5. At equivalence point: Moles H3O^(+) = moles


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TAMU CHEM 101 - pH Scale, Stoichiometry in Aqueous Solutions, & Titrations

Type: Lecture Note
Pages: 4
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