Question

You are a CEM 262 TA and are preparing a 1.55 M ZnCl2 solution as an unknown for the EDTA titration experiment. The pH must be below what value to avoid precipitation of zinc hydroxide in the solution? The solubility product of Zn(OH)2 is 3×10-16. For this calculation, assume that the solution is ideal.

Answer 1

To avoid the precipitation of zinc hydroxide in the 1.55 M ZnCl2 solution, the pH must be below a certain value determined by the solubility product. Keeping the concentration of the common ion, OH-, low is essential. By calculating the pOH and pH values, we can determine the maximum pH value to avoid precipitation.

Step-by-step explanation:

To avoid the precipitation of zinc hydroxide in the 1.55 M ZnCl2 solution, the pH must be below a certain value. The solubility product of Zn(OH)2 is 3×10-16. The relatively high concentration of the common ion, OH-, results in a very low concentration of zinc ion. Therefore, to prevent precipitation, we need to keep the concentration of OH- low, which can be achieved by maintaining a pH below the pOH value calculated from the solubility product.

Using the equation pOH = -log(OH-) and the given solubility product (Ksp) of 3×10-16, we can calculate the pOH value as follows:

pOH = -log(3×10-16).

Solving for pOH gives pOH = 15.52. To find the pH, we can use the equation pH + pOH = 14. Substituting the pOH value into the equation, we get:

pH + 15.52 = 14.

Solving for pH gives pH = -1.52.

Answer 2

To avoid the precipitation of zinc hydroxide in a ZnCl2 solution, the pH must be maintained below a certain value which can be calculated using the solubility product (Ksp) for Zn(OH)2 and the concentration of Zn2+ in the solution.

Step-by-step explanation:

The question asks about the pH below which a ZnCl2 solution must be maintained to prevent the precipitation of zinc hydroxide, given a solubility product (Ksp) for Zn(OH)2. The Ksp is an indication of the product of the molar concentrations of zinc ions and hydroxide ions in a saturated solution of zinc hydroxide. As the concentration of Zn2+ increases in the presence of a common ion like Cl-, the solubility of Zn(OH)2 decreases, and thus, the solution can tolerate fewer OH- ions before precipitation occurs. This can be managed by adding a strong acid such as HCl to increase the H+ concentration, which decreases the OH- concentration by the reaction H+ + OH- → H2O. To calculate the specific pH required, the concentration of Zn2+ and the Ksp value would need to be put into the solubility product expression, and the resulting OH- concentration converted to pH.