At 25 oC the solubility of lead(II) chloride is 1.59 x 10-2 mol/L. Calculate the value of Ksp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect). [a]

The Ksp at 25°C is 1.6 * 10^-5

Explanation:

Step 1: Data given

Temperature = 25°C

Solubility of lead(II) chloride = 1.59 * 10^-2 mol/L

Step 2: The balanced equation

PbCl2(s) <===> Pb2+(aq) + 2Cl-(aq)

Step 3: Calculate the Ksp

Ksp = [Pb2+][Cl-]²

Ksp = [Pb2+][Cl-]²

[Pb2+] = 1.59 *10-2  = 0.0159 M

[Br-] = 2 x 1.59*10-2 = 3.18 *10-2 M

Ksp = (1.59*10-2)(3.18*10-2)²

Ksp =1.6 * 10^-5

The Ksp at 25°C is 1.6 * 10^-5

the answer is 1 i believe