Ask a New Question

Question

The Ksp of PbI2 is 8.7x10^-9. Find the solubility of lead(II) iodide on mol/L.

a. 1.3x10^-3
b. 9.3x10^-9
c. 6.6x10^-5
d. 1.6x10^-3
15 years ago

Answers

DrBob222
PbI2 ==>Pb^+2 + 2I^-
Ksp = (Pb^+2)(I^-)^2
Let S = solubility, then (Pb^+2) = S and I^-) = 2S.
Substitute into the Ksp expression and solve for S. Post your work if you get stuck.
15 years ago

Related Questions

What mass of PbI2 can be formed from 75.0 mL of 0.65 M NaI by adding excess Pb(NO3)2? Pb(NO3)2 +... how many milligrams of PbI2 can you dissolve in 300 mL of water at 25 degrees celcius what would happen if PbI2 were to be ingested into your blood stream? A solution of PbI2 has [Pb2+] = 4.5 x 10-5 and [I-] = 6.5 x 10-4. PbI2 has Ksp = 8.7 x 10-9. Write d... What is the solubility of PbI2 in 0.4 M KI given the solubility constant of PbI2 is 7.1 x 10-9 a... The Ksp of PbI2 is 8.73 × 10-9. Find the solubility of PbI2 in M Please guide me , the I C E tabl... In which aqueous system is PbI2 least soluble and why? [A] 0.5 M HI [B] 0.2 M HI [C] 0.8 M KI I... (7x10^5) divided by (7x10^-8)
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use