The K_sp values for PbCl_2 and Pbl_2 at room temperature (25 degree C) are given

Question

The K_sp values for PbCl_2 and Pbl_2 at room temperature (25 degree C) are given below. K_sp (PbCl_2) = 1.6 times 10^-5 K_sp (Pbl_2) = 7.1 times 10^9 Calculate the molar solubility of each salt. What is the molar solubility of PbCl_2 if it is dissolved in 0.01 M NaCl? This experiment performed at approximately 25 degree C. If it was conducted at less than 25 degree C, how would your result (i.e. K_sp value) be affected? What if the temperature was warmer than 25 degree C What would happen to the value of the Ksp as a result of the temperature increase? If a negative value of delta H represents an exothermic reaction that gives off energy, and a + value for delta H represents an endothermic reaction that requires energy, what is the sign of the delta H that you determined in Part I of the experiment? Does that make sense in terms of energy required to dissolve a solid into a solution? Explain.

Explanation / Answer

PbCl2 Pb2+ + 2Cl-

Let solubility be s

Ksp = s (2s)2 = 4s3

Ksp = 1.6 x 10-5

4s3= 1.6 x 10-5

s3 = 1.6 x 10-5/ 4

s = 1.59 x 10-2 M

PbI2 Pb2+ + 2I-

Let solubility be s

Ksp = s (2s)2 = 4s3

Ksp = 7.1 x 10-9

4s3= 7.1 x 10-9

s3 = 7.1 x 10-9 / 4

s = 1.21 x 10-3 M

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.