concentration of solutions
DESCRIPTION
Concentration of solutions. Day 4. Concentration. A measurement of the amount of solute in a given amount of solvent or solution (unit of measurement = molarity ) Kind of like the “strength” of the solution. Molarity. Molarity (M)= mol solute L solution. Molarity. - PowerPoint PPT PresentationTRANSCRIPT
CONCENTRATION OF SOLUTIONSDay 4
CONCENTRATION A measurement of the amount of
solute in a given amount of solvent or solution (unit of measurement = molarity)
Kind of like the “strength” of the solution
MOLARITYMolarity (M)=
mol solute
L solution
MOLARITYExample #1: If we put 0.500 mol of sodium hydroxide into 1.00 L of solution, what is the molarity of the solution?
Molarity= mol solute/L sol’n
M= 1.00 L
0.500 mol NaOH
=0.500 M NaOH
Example #2: What is the molarity if we have 80.0 g NaOH in 1.00 L of solution?
Molarity= mol solute/L sol’n
M=
1.00 L
2.00 mol NaOH=2.00 M NaOH
80.0 g NaOH
40.0 g NaOH
1 mol NaOH =2.00 mol
MOLARITY
Example #3: How many moles of KOH are in 0.500 L of 0.100 M KOH?
Molarity= mol solute cross multiply! 1 L sol’n
mol KOH=
(0.500 L)(0.100 M KOH)
=0.0500 mol KOH
MOLARITY
Ex 4: How many grams of KOH are in 2.75 L of 0.25 M KOH?
MOLARITY - DILUTIONSome chemicals are sold as pre-prepared concentrated solutions (stock solutions).
To be used, stock solutions usually must be diluted.
ALWAYS ADD ACID
2.) MOLARITY - DILUTION
MSVS = MDVD
• MS – Molarity of stock solution
• VS – Volume of stock solution (L or mL)
• MD – Molarity of dilute solution
• VD – Volume of dilute solution (L or mL)
*In dilution calculations, the units for volume must be the same.
2.) MOLARITY - DILUTION
Example #1: How much 12.0 M HCl is required to make 2.50 L
of a 0.500 M solution? Ms * Vs = MD * VD
MS VS = MD VD OR M1 V1 = M2 V2MS = 12.0 M HCl MD = 0.500 M HCl
VS = ?? VD = 2.50 L
VS =(0.500 M HCl)(0.500 M HCl)(2.50 L)(2.50 L)
(12.0 M HCl)(12.0 M HCl)
= 0.104 L HCl0.104 L HCl
2.) MOLARITY - DILUTION
Example #2: What volume of a 1.50 M solution can be made using 0.0250 L of 18.0 M H2SO4
? MS VS = MD VD
MS VS = MD VD
MS = 18.0 M H2SO4 MD = 1.50 M H2SO4
VS = .0250 L VD = ????
VD =(18.0 M (18.0 M H2SO4))(.0250 L)(.0250 L)
(1.50 M (1.50 M H2SO4))= 0.300 L 0.300 L H2SO4