Calculate the solubility of lead(II) iodide, PbI2, in 0.025 M
NaI. Ksp(PbI2) = 7.9×10^-9
A. 4.5 × 10-2 M
B. 2.8 × 10-2 M
C. 8.9 × 10-5 M
D. 5.0 × 10-5 M
E. 1.3 × 10-5 M
The correct answer is E, 1.3*10^-5
Please show me how to get the answer. Please show work, Im having
trouble solving for X
Calculate the solubility of lead(II) iodide, PbI2, in 0.025 M NaI. Ksp(PbI2) = 7.9×10^-9 A. 4.5...
The Ksp of PbI2 is 7.9 x 10-9. What is the molar solubility of lead (II) iodide?
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