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Solution Chemistry

Solution Chemistry. solution. homogeneous mix. of two or more substances. solvent =. substance present in major amount. substance present in minor amount. solute =. gas . air . solvent = N 2. solute = O 2 , Ar, CO 2 , etc. solid . steel . solvent = Fe. solute = C.

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Solution Chemistry

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  1. Solution Chemistry solution homogeneous mix of two or more substances solvent = substance present in major amount substance present in minor amount solute = gas air solvent = N2 solute = O2, Ar, CO2, etc. solid steel solvent = Fe solute = C liquid solvent = H2O solute = salts, covalent compounds

  2. + + H2O - but not equally O and H share electrons separation of charge dipole dipole moment polar solvent hydrogen bonds H-bond need donor H-O acceptor O - + H-N N + - H-F F + -

  3. solvent Aqueous solutions solute H2O NaCl H-bond Ion-dipole Ion-ion Cl- Na+ Cl- H+ O- H+ Na+ O- solvation  Na+ (aq) + Cl- (aq) + H2O (l) NaCl (s)

  4. Non-ionic solutions solvent solute H2O glucose C6H12O6 H-bond H-bond H+ O- H+ O-  C6H12O6 (aq) + H2O (l) C6H12O6 (s) “Likes dissolve likes”

  5. Non-ionic solutions solvent solute H2O octane C8H18 non-polar H-bond H+ O-  no reaction + H2O (l) C8H18 (l)

  6. Properties of aqueous solutions ionic covalent conduct electricity do not conduct electricity NaCl C6H12O6 electrolytes produce ions non-electrolytes mobile, charged salts produce other anions and cations bases produce OH- in aqueous solutions produce H+ in aqueous solutions acids

  7. Electrolytes Rule Exceptions HBr HI 1. Most acids are weak electrolytes HCl HNO3 H2SO4 HClO4 2. Most bases are weak electrolytes LiOH – CsOH Ca(OH)2 – Ba(OH)2 3. Most salts are strong electrolytes HgCl2 Hg(CN)2

  8. Strong Electrolytes dissociate completely form hydrated ions strong acids  H+ (aq) + Cl- (aq) HCl (g) + H2O (l)  strong bases  NaOH (s) + H2O (l) Na+ (aq) + OH- (aq) salts  MgSO4 (s) + H2O (l) Mg2+(aq) + SO42-(aq)

  9. Weak Electrolytes do not dissociate completely weak acids equilibrium all species present  H+ (aq) + F- (aq) HF (g) + H2O (l)  weak bases  NH4+ (aq) + OH- (aq) NH3 (g) + H2O (l) weak electrolytic salts  Hg2+ (aq) + 2Cl- (aq) HgCl2 (s) + H2O (l)

  10. Non- Electrolytes do not dissociate to form ions  CH3CH2OH (aq) + H2O CH3CH2OH (l) 

  11. Solution Composition molarity = mol = M [ ] concentration = amount of solute volume of solution L What is the molarity of a solution prepared by dissolving 23.4 g sodium sulfate in enough water to give 125 mL of solution? 23.4 g Na SO4 1 mol Na2SO4 = 0.165 mol Na2SO4 2 142.0 g Na2SO4 125 mL 1 L = .125 L M = 0.165 mol Na2SO4 = 1.32 M 1000 mL 0.125 L [Na2SO4] = 1.32 M

  12. Solution Composition = mol = M [ ] concentration = amount of solute volume of solution L How many moles of HNO3 are present in 2.0 L of 0.200 M HNO3 solution? 2.0 L = 0.40 mol HNO3 0.200 mol HNO3 L

  13. Solution Composition = mol = M [ ] concentration = amount of solute volume of solution L How many grams of Na2SO4 are required to make 350 mL of 0.500 M Na2SO4? 0.500 mol Na2SO4 = 24.9 g Na2SO4 0.350 L 142.0 g 1 mol Na2SO4 L

  14. Solution Composition = mol = M [ ] concentration = amount of solute volume of solution L stock solution HCl = 12.0 M moles solute before dilution = moles solute after dilution How would you prepare 1.5 L of a 0.10 M HCl solution? 0.10 mol HCl 1.5 L = 0.15 mol HCl moles after dilution L 0.15 mol HCl = 12.0 mol HCl (x) L moles before dilution L (x) = 0.0125 L 12.5 mL of 12.0 M HCl = 1.50 L 0.10 M HCl + 1.4875 L H2O

  15. How would you prepare 1.5 L of a 0.10 M HCl solution, using a 12.0 M stock solution? moles of solute before dilution = moles of solute after dilution Mi x Vi = Mf x Vf (mol/L) (L) x Vi = 12.0 M HCl 0.10 M HCl x 1.5 L Vi = 0.0125 L then add H2O to get to Vf =1.50 L H2O

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