Answer:
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 value of solubility constant (Ksp) at 25°C in scientific notation is 1.6 × 10-⁵.
According to this question, the following parameters are given:
The balanced equation is as follows:
PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq)
The solubility constant can be calculated as follows:
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
Therefore, the solubility constant (Ksp) at 25°C is 1.6 × 10-⁵.
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