Saved Bookmarks
| 1. |
The solubility product of PbI2 is 8.0×10−9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x×10−6 mol/L. Given: √2=1.41. The value of x is |
|
Answer» The solubility product of PbI2 is 8.0×10−9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x×10−6 mol/L. Given: √2=1.41. The value of x is |
|