Advance Study Assignment: Determination of the Solubility Product of Pbl 1. Stat
ID: 943111 • Letter: A
Question
Advance Study Assignment: Determination of the Solubility Product of Pbl 1. State in words the meaning of the solubility product equation for Pbl: 2. When 5.00 mL of 0.0120 M Pb(NO)2 are mixed with 5.00 mL of 0.0300 M KI, a yellow precipitate of Pbl2(s) forms a. How many moles of Pb2+ are initially present? (no, moles A =M^ × volume in liters). moles b. How many moles of I- are originally present? moles c. In a colorimeter the equilibrium solution is analyzed for I, and its concentration is found to be 8.0 × 10-3 molc/liter. How many moles of I-are present in the solution (10 mL)? moles d. How many moles of I- precipitated? (You know how many you started with and bow many remain in solution, so?) moles - e. How many moles of Pb2* precipitated? (Note that Pblz is the precipitate, so the amounts of Pb2+ and I-that precipitated must be related.) moles f. How many moles of Pb+ are left in solution? moles moles g. What is the concentration of Pba* in the equilibrium solution? (Still 10 mL) moles/liter h. Find a value for Kp of Pbl, from these data. fconrinuand on folnvins page)Explanation / Answer
Is the product of the concentration at equilibrium of the dissociated ions raised to their stoichiometric coefficient,when the product goes above the Ksp the solid is going to precipitate.
2 Final volume 10mL =0.1L
a) 0.05L*0.012M=0.0006moles
b)0.05L*0.03M=0.0015moles
c) 0.1L*8x10-3=8x10-4moles
d)0.0015moles-8x10-4moles=7x10-4moles
e) from the formula PbI2 it is 2 moles of I for 1 of Pb it means that 7x10-4/2=3.5x10-4moles of Pb
f)0.0006-3.5x10-4=2.5x10-4moles
g)2.5x10-4moles/0.1L=2.5x10-3M
h) Ksp=(2.5x10-3)(8x10-3)2=1.6x10-7
Second page
a) the concentration of I is the double than Pb
b) 5x10-3/2=2.5x10-3M
c) Ksp=(2.5x10-3)(5x10-3)2=6.25x10-8
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