what [I-]should be maintained in KI(aq) to produce a solubility of 1.3*10^-5 mol PbI2/L when PbI2 is added?
PbI2==> Pb^+2 + 2I^-
Ksp = (Pb^+2)(I^-)^2
Look up Ksp, plug in the solubility you want to maintain for (Pb^+2) and solve for (I^-)
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What should be maintained in to produce a solubility of 1.3×10−5 when is added?
Express your answer using two significant figures.
=2.3×10−2 M
Correct
FeedbackYour answer was either rounded differently or used a different number of significant figures than required for this item. If you need the result for a later calculation in this item, please use the MasteringChemistry value 2.3×10−2 to avoid rounding errors.
To determine the [I-] concentration required to achieve a solubility of 1.3*10^-5 mol PbI2/L, we can use the solubility product constant (Ksp) expression for PbI2.
The Ksp expression for PbI2 is: Ksp = [Pb2+][I-]^2
Given that PbI2 is a sparingly soluble salt, we can assume that the majority of PbI2 dissociates into Pb2+ and I- ions:
PbI2 (s) ⇌ Pb2+ (aq) + 2I- (aq)
Since the stoichiometric ratio between Pb2+ and I- ions is 1:2, we can express the concentration of Pb2+ as half the concentration of I-. Let's denote the concentration of [I-] as x.
Therefore, [Pb2+] = 0.5x and [I-] = x.
Substituting these values into the Ksp expression:
Ksp = (0.5x)(x)^2
1.3*10^-5 = (0.5x)(x)^2
Now, we can solve this quadratic equation to find the value of x, which represents the concentration of [I-]:
1.3*10^-5 = (0.5x^3)
x^3 = (2 * 1.3 * 10^-5) / 0.5
x^3 = 5.2 * 10^-5
Taking the cube root of both sides:
x = (5.2 * 10^-5)^(1/3)
Now calculate the value of x using a calculator or computing software, which yields:
x ≈ 0.037 mol/L
Therefore, to achieve a solubility of 1.3*10^-5 mol PbI2/L, the concentration of [I-] in the solution should be approximately 0.037 mol/L.