After 250. mL 0.200 M sodium iodide are added to 150. mL 0.100 M lead(II) nitrat

Question

After 250. mL 0.200 M sodium iodide are added to 150. mL 0.100 M lead(II) nitrate, what is the molar solubility of lead(II) iodide (Ksp=7.1 x 10-9)?

A. 1.42 x 10-7 M

B. 3.6 x 10-10 M

C. 2.84 x 10-6 M

D. 8.88 x 10-11 M

After 250. mL 0.200 M sodium iodine are added to 150. mL 0.100 M lead(II) nitrate, how many moles of lead(II) iodine compound are present?

A. 0.0050 mol

B. 1.3 x 10-5 mol

C. 0.0150 mol

D. 0.040 mol

A. 1.42 x 10-7 M

B. 3.6 x 10-10 M

C. 2.84 x 10-6 M

D. 8.88 x 10-11 M

1. 
   

   After 250. mL 0.200 M sodium iodine are added to 150. mL 0.100 M lead(II) nitrate, how many moles of lead(II) iodine compound are present?

   Answer

   A. 0.0050 mol

   B. 1.3 x 10-5 mol

   C. 0.0150 mol

   D. 0.040 mol

Explanation / Answer

1.C. 2.84 x 10-6 M

2.D. 0.040 mol

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