How many moles of Ca(NO3)2 must be added to 1.0 L of a 0.255 M HF solution to begin precipitation of CaF2(s)

6.1×10^(-10) mol

Explanation:

CaF2(s) > Ca2+(aq) +2F-(aq)

          Ksp = [Ca2+][F-]^2

  Given data

   [F-] = 0.255 M

∴ The concentration of Ca2+required to start the precipitation

                 [Ca2+] = Ksp / [F-]2

                            = (4.0 X 10^-11) /(0.255)^2

                            = 6.1×10^(-10) M

Here the volume is 1.0 L  

∴ The number of molesof  Ca(NO3)2 must be added to 1.0 L of a1.0 L of a 0.255 M HF solution to begin precipitation ofCaF2 (s)=   6.1×10^(-10) mol

h let me think

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l