You are investigating the effect of strong base on the solubility of zinc hydroxide, Zn(OH)2 (s).
(Ksp= 3 x 10- 15 for Zn(OH)2 (s))
You prepare 1.0 L solution of 0.10M NaOH and add excess zinc hydroxide, Zn(OH)2 (s).
Instead of suppresing the solubility, you observe extra solubility.
Your research advisor tells you that zinc ions react with four hydroxide ions in a solution to form the complex ion Zn(OH)42- (aq).
(Kf = 2 x 10 15 for Zn(OH)42- (aq))
Calculate the pH of your solution.
I would do this.
Zn(OH)2 ==> Zn^2+ 2OH Ksp = ?
Zn^2+ + 4OH^- ==> [Zn(OH)4]^2- Kf=?
Add these two equations.
Zn(OH)2 + 2OH^- ==> [Zn(OH)4]^2-
.solid....+2x.........-x
Krxn = Ksp*Kf = approx 6
Then [(x)/(2x)^2] = 6
Solve for x and convert to pH.
To calculate the pH of the solution, we need to consider the reactions happening in the solution and determine the concentration of the relevant species. Here's how you can approach the problem:
1. Start by writing the balanced equation for the reaction between zinc hydroxide (Zn(OH)2) and sodium hydroxide (NaOH):
Zn(OH)2 (s) + 2 NaOH (aq) -> Na2Zn(OH)4 (aq)
2. Since you have added excess zinc hydroxide, we can assume that the reaction goes to completion, and all the zinc hydroxide reacts to form the complex ion Na2Zn(OH)4.
3. Calculate the concentration of Na2Zn(OH)4 in the solution. Since you prepared a 1.0 L solution of 0.10 M NaOH, the concentration of NaOH is 0.10 M. From the balanced equation, we can see that the stoichiometry between NaOH and Na2Zn(OH)4 is 2:1. So, the concentration of Na2Zn(OH)4 will also be 0.10 M.
4. Now, we need to determine the concentration of the hydroxide ion, OH-, in the solution. Since each molecule of Na2Zn(OH)4 releases four hydroxide ions, the concentration of OH- will be four times the concentration of Na2Zn(OH)4. Therefore, the concentration of OH- is 4 * 0.10 M = 0.40 M.
5. Since OH- is a strong base, it will react with water to produce hydroxide ions and raise the pH of the solution. We can ignore the self-ionization of water in this calculation since the concentration of OH- is relatively high compared to the concentration of H+.
6. To calculate the pH, we need to find the pOH first. The pOH is calculated by taking the negative logarithm (base 10) of the concentration of OH-. In this case, pOH = -log(0.40) = 0.4.
7. Finally, to calculate the pH, we subtract the pOH from 14 (the negative logarithm of the concentration of H+ in pure water). pH = 14 - pOH = 14 - 0.4 = 13.6.
Therefore, the pH of the solution is 13.6.