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Suggests increasing number of ions in same volume of water to allow solution to conduct a higher voltage
+ + + + + + + - - - - - - - - - - - - + + + + + Least concentrated Most concentrated Compare the number of ions in the same volume
4.
Graph to compare conductivity and concentration conductivity concentration Increasing concentration — Increasing conductivity Conductivity no longer increases
5.
Eventually Constant Conductivity
Conductivity no longer changes
This suggests that the concentration is no longer changing
Solution must have reached a maximum concentration
Undissolved solid remains at the bottom
The solution is said to be SATURATED.
6.
Lab Procedures
A saturated solutions of several ionic compounds were prepared
Salt was added until solid remained at the bottom
The solution was decanted
The solution was poured down a stirring rod leaving the solid in behind in the beaker
The goal was to isolate the saturated solution
10.0 mL of the saturated solution heated until all the water was removed
This allowed us to measure the amount of salt that was dissolved in the 10.0 mL of solution
7.
Calculating the concentration of the saturated solution
Use the data of # grams in 10.0 mL of solution
Convert grams to moles
Convert mL to L
MOLARITY moles of solute liter of solution
8.
Sample Data
9.
EOS The attractions of water dipoles for ions pulls the ions out of the crystalline lattice and into aqueous solution
10.
Equilibrium in a Saturated Solution
When the solution is saturated, there is no net change- conditions remain constant
The solution and solid are in equilibrium
Both reactions (dissolving and crystallizing) are occurring at equal rates
Example
NaCl(s) Na + (aq) + Cl - (aq)
The ions are dissociating as fast as the ions crystallize
11.
Dynamic Nature of Equilibrium When a system reaches equilibrium , the forward and reverse reactions continue to occur … but at equal rates. We are usually concerned with the situation after equilibrium is reached. After equilibrium the concentrations of reactants and products remain constant.
12.
Dynamic Equilibrium Illustrated NaCl dissolves and recrystallizes continuously. After a time, the solution contains radioactive Na + … NaCl containing radioactive Na + is added to a saturated NaCl solution. … and the added salt now contains some stable Na + .
13.
Calculating an Equilibrium Constant
Since conditions remain constant at equilibrium, an equilibrium constant (Keq) can be calculated
Keq is calculated
as a ratio of concentrations of products compared to reactants at equilibrium
the coefficients in the equation become the exponents in the Keq
Since the concentrations of solids and liquids are constant, they are both left out of the Keq
14.
Ksp- the equilibrium constant for saturated solutions of ionic compounds
NaCl(s) Na + (aq) + Cl - (aq)
Ksp = [Na + ][Cl - ]
If the concentration of dissolved NaCl in a saturated solution is 5 M
Then the concentration of sodium ions must be 5 M and the concentration of chloride ions must be 5 M
Ksp = [Na + ][Cl - ] = (5 M)(5M) = 25
15.
Calculating Concentration given the Ksp
AgCl(s) Ag + (aq) + Cl - (aq)
Ksp = [Ag + ][Cl - ]
The concentration is unknown
Let x = [AgCl], then x also x = [Ag + ] and x = [Cl - ]
Ksp = [Ag + ][Cl - ] = (x)(x) = x 2 = 1.6 x 10 -10
So x = 1.3 x 10 -5 M
Much less silver chloride is present in saturated solution than sodium chloride
The lower Ksp shows the compound is less soluble
16.
Given the Concentration calculate the Ksp
Ag 2 S(s) 2Ag + (aq) + S 2- (aq)
Ksp = [Ag + ] 2 [S 2- ]
The concentration of silver sulfide in a saturated solution is 3.1 x 10 -17 M = [Ag 2 S]
The concentration of sulfide ions must be
3.1 x 10 -17 M = [S 2- ]
The concentration of silver ions must be
2 x 3.1 x 10 -17 M = 6.2 x10 -17 M = [Ag + ]
Ksp = [Ag + ] 2 [S 2- ]
= (6.2 x10 -17 ) 2 (3.1 x 10 -17 )
= 3.2 x 10 -49
A very small Ksp indicates that this compound is barely soluble
17.
Ksp and solubility
Ksp is calculated using the concentrations of dissolved ions in the saturated solution, thus Ksp indicates the relative solubility of compounds
A Ksp greater than 0.1 indicates a compound is soluble
A compound is designated as insoluble is the Ksp is less than 0.1
18.
19.
Calculating the Concentration of Calcium fluoride in a saturated solution
Ksp = 5.3 x 10-11
Reversible reaction at equilibrium
CaF 2 (s) Ca 2+ (aq) + 2F - (aq)
Ksp = [Ca 2+ ][F - ] 2
If x = [CaF 2 ], then x = [Ca 2+ ] and 2x = [F - ]
Ksp = [Ca 2+ ][F - ] 2 = 5.3 x 10 -11 = (x)(2x) 2
(x)(4x 2 ) = 4x 3 = 5.3 x 10 -11
Divide by 4 and then cube root
20.
Preparing a solution of known molarity
Note the final volume of the solution is exactly 250mL.
Calculate the molarity of the is solution
21.
Again, a molar solution The volume is exactly 1.0 L Note that not all of the initial 1.0 L of water was used to prepare the solution
22.
Dilution of a solution
Calculate the molarity of the new solution using M 1 V 1 = M 2 V 2
23.
Other Units of Concentrations
Molality
Moles of solute per kg of solvent
Mole fraction
Moles of solute per total # moles
Percent by Mass
mass of solute/mass of solution x 100
24.
Mole Fraction and Mole Percent The mole fraction ( x i ) of a solution component i is the fraction of all the molecules in the solution that are molecules of i EOS The mole percent of a solution component is its mole fraction multiplied by 100 Sum of x i must equal 1
25.
Calculating other concentrations
1.0L of the 0.100M CuSO 4 solution was prepared by adding 25.0g or 0.100 moles of solid to 992mL of water.
Calculate the molality of the solution
Looking for moles of CuSO 4 per kg of water
Calculate the mole fraction of the solution.
Looking for moles of CuSO 4 per total moles
Total moles = moles of CuSO 4 + moles of water
Calculate the percent by mass of the solution.
Looking for mass of CuSO 4 per total mass
26.
Dilution The desired molarity solutions are often prepared from concentrated stock solutions (routinely used solutions prepared in concentrated form) by adding water. This process is called dilution. Moles of solute before dilution = moles of solute after dilution M 1 V 1 = M 2 V 2
27.
Figure 4.11 (a) A Measuring Pipet (b) A Volumetric (transfer) Pipet
28.
Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of 0.10 M H 2 SO 4 solution? V 1 = Volume before dilution = ? M 1 = Concentration before dilution = 16 M V 2 = Volume after dilution = 1.5 L M 2 = Concentration after dilution = 0.10 M M 1 V 1 = M 2 V 2 V 1 =M 2 V 2 / M 1 = (0.10 M x 1.5 L)/ 16 M = 0.0094 L = 9.4 mL