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 + .
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
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
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