A saturated solution of lead(II) iodide, Pblz has an iodide concentration of 3.0

Question

A saturated solution of lead(II) iodide, Pblz has an iodide concentration of 3.0 x103 mol/L (see photo). 1. a. What is the molar solubility of Pblh? b. Determine the solubility constant, Ksp, for lead(lI) iodide. Does the molar solubility of lead(II) iodide increase, decrease, or remain unchanged with the addition of potassium iodide to the solution? Explain. c. 2. What is the indicator used to detect the endpoint in the titration for this experiment? What is the expected color change at the endpoint in this experiment? (Hint: Textbook Table 17.3) 3.

Explanation / Answer

Dear student, I will answer question 1 with all subparts (a, b and c), please for the other answers post another question and we will gladly answer them

1.a) PbI2(s) Pb2+(aq) + 2I-(aq)
From this equation 1.0 mol of PbI2 gives 2.0 mol of I- in solution:

1.0 mol of PbI2 ---------------------- 2.0 mol of I-

X mol of PbI2 --------------------- 3.0x10-3 mol of I-   

X= 1.5×10-3 M

so the solubility of PbI2 is 1.5×10-3 M

1.b) Ksp = [Pb2+][I-]^2

[I-] = 3.0×10-3 M [Pb2+] = 1.5×10-3 M
Ksp = [1.5×10-3]×[3.0×10-3]2 = 1.35×10-8

1.c) From Le Chatelier's Principle if you add I- to the equilibrium the equation will shift to lower the added I-. The PbI2 will increase in amount so is less soluble (it is called the common ion effect).

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