Optimistic Post: Discovering the Minimum Concentration of Cr3+ Needed for Precipitation

What is the minimum concentration of Cr3+ required to initiate a precipitate of chromium(III) fluoride in a solution of 0.095 M NaF?

A) 6.6 x 10^-9 M
B) 6.6 x 10^-4 M
C) 6.6 x 10^-6 M
D) 6.6 x 10^-11 M

Final Answer:

The minimum concentration of Cr3+ that must be added to 0.095 M NaF to initiate a precipitate of chromium(III) fluoride is 6.6 x 10^-9 M.

To calculate the minimum concentration of Cr3+ required to initiate a precipitate of chromium(III) fluoride in a solution of 0.095 M NaF, we need to consider the solubility product (Ksp) of chromium(III) fluoride. The ion product (Q) is calculated by multiplying the concentrations of the ions involved in the equilibrium equation.

If the value of Q is greater than the Ksp value, a precipitate will form. By substituting the given concentrations of NaF and the fluoride ions into the equation, we can solve for the minimum concentration of Cr3+ needed, which is 6.6 x 10^-9 M.

This result indicates that even with a relatively low concentration of Cr3+, a precipitate of chromium(III) fluoride can be initiated in the solution. Understanding the minimum concentration required for precipitation provides valuable insights into the chemical behavior of the system.