SOMEONE PLEASE EXPLAIN AND ANSWER THIS QUESTION :
Will Co(OH)2 precipitate from solution if the pH of a 0.034 M solution of Co(NO3)2 is adjusted to 9.45? (Ksp of Co(OH)2 is 1.3 10-15)
Support your answer by calculating the ion product (Q) of the Co(OH)2 solution.
Co(OH)2 will ppt if Q > Ksp =1.3E-15
Q = (Co^2+)(OH^-)^2
(Co^2+) = 0.034 from the problem.
(OH^-) = from pH 9.45
pH = 9.45, pOH = 4.55
4.55 = -log(OH^-)
(OH^-) = 2.82E-5
Q = (0.034)(2.82E-5)^2 = 2.7E-11
If Q > Ksp, Co(OH)2 will ppt.
If Q < Ksp, Co(OH)2 will not ppt.
THANX!!!
To determine if Co(OH)2 will precipitate from a solution when the pH is adjusted to 9.45, we can calculate the ion product (Q) and compare it to the solubility product constant (Ksp) of Co(OH)2.
The balanced equation for the dissociation of Co(OH)2 in water is:
Co(OH)2(s) ⇌ Co2+(aq) + 2OH-(aq)
The solubility product constant expression for Co(OH)2 is:
Ksp = [Co2+][OH-]^2
Given:
Ksp = 1.3 × 10^-15
pH = 9.45
[M] = 0.034 M (concentration of Co(NO3)2)
First, we need to determine the concentration of Co2+ and OH- in the solution at pH 9.45.
Since the pH is above 7, we know that OH- will be present in the solution due to the dissociation of water:
2H2O ⇌ OH- + H3O+
Using the pH, we can determine the concentration of OH-:
pOH = 14 - pH
pOH = 14 - 9.45
pOH = 4.55
Now, we can convert pOH to the concentration of OH-:
OH- = 10^(-pOH)
OH- = 10^(-4.55)
Next, we need to determine the concentration of Co2+, which can be calculated using the initial concentration of Co(NO3)2:
Co(NO3)2 → Co2+ + 2NO3-
Since the concentration of Co(NO3)2 is 0.034 M, the concentration of Co2+ is also 0.034 M.
Finally, we can calculate the ion product (Q):
Q = [Co2+][OH-]^2
Q = (0.034)(10^(-4.55))^2
By calculating the value of Q, we can compare it to the Ksp value of Co(OH)2.
Let's calculate the value of Q:
To determine whether Co(OH)2 will precipitate from the solution, we need to compare the ion product (Q) of Co(OH)2 with its solubility product constant (Ksp). If Q is greater than Ksp, the compound will precipitate.
The balanced equation for the dissociation of Co(OH)2 in water is:
Co(OH)2 ⇌ Co2+ + 2OH-
First, we'll calculate the concentrations of Co2+ and OH- ions in the solution:
Given that the initial concentration of Co(NO3)2 is 0.034 M, the concentration of Co2+ is also 0.034 M (since it dissociates completely). Since Co(NO3)2 is a strong electrolyte, it will dissociate into Co2+ and 2NO3- ions.
Next, we need to consider the OH- concentration to calculate Q. To do this, we need to determine the OH- concentration from the pH value of the solution. The pH value of a solution is defined as the negative logarithm of the hydrogen ion concentration (pH = -log[H+]). Since we're given the pH, we can calculate the hydroxide ion concentration (OH-) using the relationship:
[H+][OH-] = 1.0 x 10^-14 (at 25°C)
By taking the negative logarithm of both sides, we get:
-log([H+]) - log([OH-]) = -log(1.0 x 10^-14)
-pH - log([OH-]) = -log(1.0 x 10^-14)
-log([OH-]) = -log(1.0 x 10^-14) + pH
-log([OH-]) = 14 + pH
Therefore, the hydroxide concentration can be calculated as:
[OH-] = 10^(-14 - pH)
Substituting the given pH value (9.45) into the equation, we find:
[OH-] = 10^(-14 - 9.45) = 3.162 x 10^-6 M
Now that we have the concentrations of Co2+ and OH-, we can calculate Q by multiplying their concentrations:
Q = [Co2+][OH-]^2
Q = (0.034 M)(3.162 x 10^-6 M)^2
Calculating Q, we find:
Q ≈ 3.496 x 10^-14
Comparing Q with the Ksp of Co(OH)2 (1.3 x 10^-15), we see that Q is significantly larger than Ksp. Therefore, Co(OH)2 will precipitate from the solution because the ion product exceeds the solubility product constant.