CHEM 130 Chapter 8: Solutions and Aqueous Reactions

Solution Concentration
●Solution: homogeneous mixture of two substances
○Solvent: majority component of a solution
○Solute: minority component of solution
■Solute → what “U” put in
●Aqueous Solution: a solution in which water acts as the solvent
Quantifying Solution Concentration
●Amount of solute in a solution is variable
●Dilute Solution: solution that contains a very small amount of solute relative to the amount of solvent
●Concentrated Solution: solution that contains a large amount of solute relative to the amount of solvent
●Molarity (M): means of expressing solution concentration
○
○Common solutions range from 0 to ~18M
■If concentration is a lot more than 18, double check your calculations
●EX PROBLEM Calculating solution concentration: if you dissolve 25.5g KBr in enough water to make 1.75L solution, what is the molarity?
○GIVEN → 25.5g KBr, 1.75L TOTAL solution, FIND → molarity
○STRAT → Given mass of KBr → use molar mass to find mol KBr → M = mol solute / vol total solution
○SOLVE → 25.5gKBr×1molKBr
119gKBr=0.21429molKBr
■M=0.21429molKBr
1.75Lsolution=0.122M
●Use M as a conversion factor between mols of solute and L of solution
○Ex: 0.5M NaCl solution contains 0.5 mol NaCl for every liter of solution
●EX: how many L of a 0.125M NaOH solution contain 0.255mol NaOH?
○GIVEN → 0.125M NaOH solution, 0.255mol NaOH, FIND → volume NaOH
solution in L
○STRAT → mol NaOH → L solution
○SOLVE → 0.255molNaOH×1Lsolution
0.125molNaOH=2.04Lsolution Solution Dilution
●To save space, labs often store solutions in concentrated forms called stock solutions
○Dilute stock solution to required concentration for lab work
●When we dilute a solution w more solvent, the number of mols of solute do not change; the number of moles are simple dispersed into a greater volume
●Dilution Equation: M1V1 = M2V2
○M1 and V1 are M and V of the initial solution, M2 and V2 are M and V of diluted solution
○Equation works bc MxV gives moles of solute, which is unchanged during dilution
●EX: Need 3.00L of a 0.500M CaCl 2 solution for lab prep. How do we prepare this solution from a 10.0M stock solution?
○M1V1 = M2V2
○V1 = M2V2/M1
○V1 = 0.500mol/Lx3.00L / 10.0mol/L
○V1 = 0.150L
○We dilute 0.150L of stock solution to a total V of 3.00L to make a 0.500M solution
Solution Stoichiometry
●In aqueous reactions, quantities of reactants and products are often specified in terms of
volumes and concentrations which are used to calculate its amount in moles
●We can use stoichiometric coefficients in the chemical equation to convert to the amount
in moles of another reactant or product
●
●EX: What volume (L) of 0.150M KCl solution will completely react with 0.150L of a 0.175M Pb(NO3)2 solution according to the balanced chemical equation: 2KCl + Pb(NO3)2 → PbCl2 + 2KNO3?
○STRAT → L Pb(NO 3)2 → mol Pb(NO3)2 → mol KCl → L KCl solution
○SOLVE → 0.150LPb¿
Types of Aqueous Solutions and Solubility
●When you mix a solid into a liquid solvent, the attractive forces that hold the solid together (solute-solute interactions ) compete with attractive forces between the solvent and the solid ( solvent-solute interactions )
●Ex: Sodium chloride is added to water → competition between attraction of Na+ and Cl– to each other and the attraction of Na+ and Cl– to the water molecules
○H2O’s polar nature attracts Na+ and Cl–
■O in H2O is electron-rich → partial negative charge ( 𝛿–)
■H in H2O is electron-poor → partial positive charge ( 𝛿+)