Question 19.10: Predicting Whether a Precipitate Will Form Problem A common ......

Plan First, we decide which slightly soluble salt could form, look up its K_{sp} value in Appendix C, and write a dissolution equation and ion-product expression. To see whether mixing these solutions forms the precipitate, we find the initial ion concentrations by calculating the amount (mol) of each ion from its concentration and volume, and dividing by the total volume because each solution dilutes the other. Finally, we substitute these concentrations to calculate Q_{sp}, and compare Q_{sp} with K_{sp}.

Solution The ions present are Ca^{2+},  Na^+,  F^−, and NO_3^−. All sodium and all nitrate salts are soluble (Table 4.1), so the only possible compound that could precipitate is CaF_2  (K_{sp}  =  3.2×10^{−11}). Writing the equation and ion-product expression:

Calculating the ion concentrations:

            \text{Amount (mol) of }Ca^{2+}  =  0.30  M  Ca^{2+ } ×  0.100  L  =  0.030\text{ mol }Ca^{2+}
            [Ca^{2+}]_{\text{init}}  =  \frac{0.030\text{ mol }Ca^{2+} }{0.100  L  +  0.200  L}  =  0.10  M  Ca^{2+}
            \text{Amount (mol) of }F^−  =  0.060  M  F^−  ×  0.200  L  =  0.012\text{ mol }F^−
            [F^−]_{\text{init}}  =  \frac{0.012\text{ mol }F^−}{0.100  L  +  0.200  L}  =  0.040  M  F^−

Substituting into the ion-product expression and comparing Q_{sp} with K_{sp}:
            Q_{sp}  =  [Ca^{2+}]_{\text{init}}[F^−]^2_{\text{init}}  =  (0.10)(0.040)^2  =  1.6×10^{−4}
Because Q_{sp}  >  K_{sp},  CaF_2 will precipitate until Q_{sp}  =  3.2×10^{−11}.

Check Make sure you round off and quickly check the math. For example, Q_{sp}  =  (1×10^{−1})(4×10^{−2})^2  =  1.6×10^{−4}. With K_{sp} so low, CaF_2 must have a low solubility, and given the sizable concentrations being mixed, we would expect CaF_2 to precipitate.