Consider the following reaction that describes the solubility equilibria of solid Zn(OH)2 in aqueous solution (Ksp of Zn(OH)2 4.0x10^-17. Calculate molar solubility of Zn(OH)s in an acidic solution of pH5.0.
Zn(OH)2 ==> Zn^2+ + 2OH^-
Ksp = 4.0E-17 = (Zn^+)(OH^-)^2
pH = 5.0; use
pH + pOH = pKw = 14 to solve for pOH, then -log(OH^-) to solve for (OH^-). Substitute OH into Ksp expression and solve for Zn^2+. That is the solubility of Zn(OH)2.
I followed your direction and I calculated the solubility to be 40. Am I correct?
To calculate the molar solubility of Zn(OH)2 in an acidic solution of pH 5.0, we need to determine the concentration of Zn2+ ions in the solution.
First, let's write down the balanced equation for the dissociation of Zn(OH)2:
Zn(OH)2(s) ⇌ Zn2+(aq) + 2OH-(aq)
According to the solubility product constant (Ksp), the equilibrium expression for the solubility of Zn(OH)2 can be written as:
Ksp = [Zn2+][OH-]^2 = 4.0x10^-17
The concentration of OH- ions can be determined using the pH of the solution. In an acidic solution, the concentration of H+ ions is greater than the concentration of OH- ions.
Since water is neutral, the concentration of H+ ions is equal to the concentration of OH- ions in pure water, which is 1.0x10^-7 M at 25°C. However, in an acidic solution, the concentration of H+ ions is higher than 1.0x10^-7 M.
To calculate the concentration of OH- ions in an acidic solution, we can use the autoionization constant of water (Kw), which is equal to [H+][OH-] = 1.0x10^-14 at 25°C.
Since the solution is acidic with a pH of 5.0, the concentration of H+ ions can be determined using the equation:
[H+] = 10^(-pH) = 10^(-5.0) M = 1.0x10^-5 M
Using the equilibrium expression, we can rearrange it to solve for [Zn2+]:
[Zn2+] = Ksp / [OH-]^2
Substituting the known values:
[Zn2+] = (4.0x10^-17) / (1.0x10^-5)^2 = 4.0x10^-7 M
Therefore, the molar solubility of Zn(OH)2 in an acidic solution of pH 5.0 is approximately 4.0x10^-7 M.