Solution Chemistry
This comprehensive overview delves into the nature of homogeneous mixtures, defining terms such as solute and solvent. We explore common examples such as gas mixtures in air, solutions in water, and the properties of ionic and covalent compounds in aqueous solutions. The text discusses how to calculate molarity, demonstrating through sample problems. It distinguishes between electrolytes and non-electrolytes, emphasizing the concepts of dissociation in solutions. Finally, we outline methods for preparing specific concentrations, guiding through dilution calculations and solute requirements.
Solution Chemistry
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Presentation Transcript
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
+ + 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 + -
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)
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”
Non-ionic solutions solvent solute H2O octane C8H18 non-polar H-bond H+ O- no reaction + H2O (l) C8H18 (l)
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
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
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)
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)
Non- Electrolytes do not dissociate to form ions CH3CH2OH (aq) + H2O CH3CH2OH (l)
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
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
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
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
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 =1.37 L H2O
