If 400 mL of some Pb(NO3)2 solution is mixed with 150 mL of 1.80 x 10−2 M NaCl solution, what is the maximum concentration of the Pb(NO3)2 solution added if no solid PbCl2 forms? (Assume Ksp = 2.00 x 10−5 M at this temperature.) Enter the concentration in M.

The maximum concentration of Pb(NO3)2 is 2.68 M

Explanation:

Step 1: Data given

Volume of the Pb(NO3)2 solution = 400 mL = 0.400 L

Volume of the 0.01 M NaCl solution = 150 mL = 0.150 L

Ksp = 2.00 * 10^-5

Step 2: The equations

NaCl → Na+ + Cl -

PbCl2 → Pb^2+ + 2Cl-

Step 3: Calculate moles

Moles = volume * molarity

Moles NaCl = 0.150 L * 0.01 M

Moles NaCl = 0.0015 moles

Moles Cl- = 0.0015 moles

Step 4: Calculate total volume

Total volume = 400 + 150 = 550 mL

Step 5: Calculate [Cl-]

Molarity = moles / volume

Molarity Cl- = 0.0015 moles / 0.550 L

Molarity = 0.00273 M

Step 6: Calculate Ksp

Ksp = [Pb^2+]*[Cl-]²

2.00 * 10^-5 = [Pb^2+]* 0.00273²

[Pb^2+] = 2.68 M

[Pb^2+] = [Pb(NO3)2] = 2.68 M

The maximum concentration of Pb(NO3)2 is 2.68 M

the correct answer is:

a. formation of precipitation