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how many moles of Ca(NO3)2 must be added to 1.oL of a 0.203MKF solution to begin precipitation of CaF2? For CaF2 ,ksp = 4.0x10 e-11

User Jamex
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Final answer:

To find the number of moles of Ca(NO3)2 to be added to start precipitation of CaF2, use the Ksp formula. Let the number of moles of Ca(NO3)2 be x, and set up and solve the equation 4.0x10e-11 = x(0.203M + 2x)^2.

Step-by-step explanation:

In order to determine the moles of Ca(NO3)2 that should be added to the 0.203MKF solution to start precipitation of CaF2, we need to apply the solubility product constant formula, which is Ksp=[Ca2+][F-]^2.

Assuming [F-] derives from KF’s concentration, it is 0.203M. However, as CaF2 produces 2 moles of F- ions per mole, [F-] will become twice as the number of moles of Ca(NO3)2.

So, let's assume x moles of Ca(NO3)2 are added, then [F-] = 0.203M + 2x. As the Ksp expression for CaF2 is 4.0x10e-11 = [Ca2+][F-]^2 = x(0.203M + 2x)^2. Now, solve this equation for x to find the moles of Ca(NO3)2.

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