Question

what mass of Ca(NO₃)₂ must be added to 1.0 L of a 1.0 M HF solution to begin precipitation of CaF₂. You may assume no volume change on the addition of Ca(NO₃)₂. K_sp for CaF2 = 4.0 x 10^-11 and k_a for HF=7.2 x 10^-4.

Answer 1

Given that 1.0 L of 1.0 M HF (1.0 L * 1.0 M = 1 mol) istaken. The concentration of F^- from the given solutionis governed by the equilibrium:

               HF(aq) ↔ H⁺ +F^-            Ka = 7.2 * 10^-4

    Initial:     1M       0      0

    Change:    -x       +x      + x

Equilibrium: (1-x)     x        x

                   Ka = [H⁺][F^-] / [HF]

           7.2 * 10^-4 = (x)(x) / (1-x)

                   x = 2.65 * 10^-2 M

                  [F^-] = 2.65 * 10^-2 M

Since the concentration of  F^- inthe solution is known to be 2.65 * 10^-2 M, the greatestconcentration of Ca⁺² that can be present withoutcausing the precipitation of CaF₂ can be calculated fromthe K_sp expression.

                 CaF₂ ↔ Ca⁺² + 2F^-                 K_sp = 4.0 * 10^-11    [ F^-] = 2( 2.65 *10^-2)M

           =  5.3 * 10^-2M                      

     K_sp =[Ca⁺²] [ F^-]²

4.0 * 10^-11 =  [Ca⁺²](5.3 * 10^-2 )²

        [Ca⁺²] = 1.42 * 10^-8 M

A concentration of Ca⁺² in excess of 1.42 *10^-8 M, will cause the precipitaion of CaF₂

  

Calculation of amount of Ca(NO₃)₂in grams present in 1.42 * 10^-8 M

1.42 * 10^-8 M corresponds to 1.42 *10^-8 M moles Ca(NO₃)₂ as thevolume is 1.0 L

    Molarity =  1.42 *10^-8 M

                =  1.42 * 10^-8 mol /1.0 L

Mass of Ca(NO₃)₂ = moles * molarmass(g/mol)

                                =1.42 * 10^-8 mol * 164.09g/mol

                                =2.33 * 10^-6 g

Amount of Ca(NO₃)₂ greaterthan  2.33 * 10^-6 g will cause theprecipitation of CaF₂