Solution Chemistry solution of two or more substances homogeneous mix solvent = solute = substance...
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![Page 1: Solution Chemistry solution of two or more substances homogeneous mix solvent = solute = substance present in major amount substance present in minor amount.](https://reader035.fdocuments.in/reader035/viewer/2022062219/5a4d1b0f7f8b9ab05998e2bf/html5/thumbnails/1.jpg)
Solution Chemistry
solutionof two or more substanceshomogeneous mix
solvent =solute =
substance present in major amountsubstance present in minor amount
gassolute = O2, Ar, CO2, etc.
air solvent = N2
solidsolute = C
steel solvent = Fe
liquid solvent = H2Osolute = salts, covalent compounds
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H2O
O and H share electronsseparation of charge dipole
dipole momentpolar solventhydrogen bonds H-bondneed donor H-O
H-NH-F
acceptor ONF
but not equally
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Aqueous solutionsNaClH2O
solvent solute
H-bondO-H+
Ion-ionNa+ Cl-
Ion-dipoleCl- H+
Na+ O-
solvation
NaCl (s) + H2O (l) Na+ (aq) + Cl- (aq)
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Non-ionic solutions
C6H12O6glucosesolvent soluteH2O
H-bondO-H+
H-bondO-H+
C6H12O6 (s) + H2O (l) C6H12O6 (aq)
“Likes dissolve likes”
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Non-ionic solutions
C8H18octanesolvent soluteH2O
H-bondO-H+
non-polar
C8H18 (l) + H2O (l) no reaction
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Properties of aqueous solutions
ionic covalent
conduct electricity do not conduct electricityNaCl C6H12O6
electrolytes non-electrolytes
acids produce H+ in aqueous solutions
bases produce OH- in aqueous solutions
salts produce other anions and cations
produce ionsmobile, charged
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Electrolytes
Rule Exceptions1. Most acids are weak electrolytes HCl HBr HI
HNO3
H2SO4
HClO4
2. Most bases are weak electrolytesCa(OH)2 – Ba(OH)2
LiOH – CsOH
3. Most salts are strong electrolytes HgCl2
Hg(CN)2
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Strong Electrolytes
dissociate completely form hydrated ions
HCl (g)strong acids
strong basesNaOH (s) + H2O (l) Na+ (aq) + OH- (aq)
saltsMgSO4 (s) + H2O (l) Mg2+(aq) + SO4
2-(aq)
+ H2O (l) H+ (aq) + Cl- (aq)
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Weak Electrolytes
do not dissociate completely
HF (g)weak acids
+ H2O (l) + F- (aq)H+ (aq)
equilibrium all species present
NH3 (g)weak bases
+ H2O (l) + OH- (aq)NH4+ (aq)
HgCl2 (s)weak electrolytic salts
+ H2O (l) + 2Cl- (aq)Hg2+ (aq)
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Non- Electrolytes
do not dissociate to form ions
CH3CH2OH (l)
+ H2O CH3CH2OH (aq)
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Solution Compositionconcentration = amount of solute
volume of solution= mol
L= M
molarity
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 SO421 mol Na2SO4
142.0 g Na2SO4
= 0.165 mol Na2SO4
125 mL 1 L1000 mL
= .125 L M = 0.165 mol Na2SO40.125 L
= 1.32 M
[ ]
[Na2SO4] = 1.32 M
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How many moles of HNO3 are present in 2.0 L of 0.200 M HNO3 solution?
0.200 mol HNO3
L2.0 L = 0.40 mol HNO3
Solution Compositionconcentration = amount of solute
volume of solution= mol
L= M[ ]
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How many grams of Na2SO4 are requiredto make 350 mL of 0.500 M Na2SO4?
0.500 mol Na2SO4
L0.350 L
1 mol Na2SO4
142.0 g = 24.9 g Na2SO4
Solution Compositionconcentration = amount of solute
volume of solution= mol
L= M[ ]
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stock solution HCl = 12.0 Mmoles solute before dilution = moles solute after dilution
How would you prepare 1.5 L of a 0.10 M HCl solution?
0.10 mol HClL
1.5 L = 0.15 mol HCl
0.15 mol HCl =
moles after dilution
moles before dilution12.0 mol HCl L
= 0.0125 L
12.5 mL of 12.0 M HCl + 1.4875 L H2O = 1.50 L 0.10 M HCl
L
(x)
Solution Compositionconcentration = amount of solute
volume of solution= mol
L= M[ ]
(x)
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How would you prepare 1.5 L of a 0.10 M HCl solution, usinga 12.0 M stock solution?
Mi x Vi = Mf x Vf
12.0 M HCl x Vi 0.10 M HCl x 1.5 L
Vi = 0.0125 L
then add H2O to get to Vf
=
moles of solute after dilution moles of solute before dilution =
(mol/L) (L)
=1.37 L H2O