Calculate the molar solubility of calcium hydroxide in a solution buffered at each pH.
a. pH=4
b. pH= 7
c. pH=9
To calculate the molar solubility of calcium hydroxide (Ca(OH)2) in a solution buffered at a specific pH, we need to consider the equilibrium constant expression for its solubility. Calcium hydroxide dissociates in water to form calcium ions (Ca2+) and hydroxide ions (OH-). The dissociation reaction is as follows:
Ca(OH)2 ⇌ Ca2+ + 2OH-
The solubility equilibrium constant, Ksp, is the ratio of the concentrations of the ions at equilibrium. It can be defined as:
Ksp = [Ca2+][OH-]^2
The pH of a solution is a measure of its acidity or basicity. It is defined as the negative logarithm of the concentration of hydrogen ions (H+). In a buffered solution, the pH is maintained at a specific value by the presence of a weak acid and its conjugate base (such as a weak acid and its salt).
Now, let's calculate the molar solubility at each pH value:
a. pH = 4:
At pH 4, the solution is acidic. This means that there is a relatively high concentration of hydrogen ions (H+). The solubility of calcium hydroxide is generally low in acidic solutions due to the increased concentration of H+ ions.
b. pH = 7:
At pH 7, the solution is considered neutral. The concentration of hydrogen ions (H+) is equal to the concentration of hydroxide ions (OH-). In this case, the solubility of calcium hydroxide is moderate.
c. pH = 9:
At pH 9, the solution is alkaline or basic. The concentration of hydroxide ions (OH-) is relatively high, and this leads to the solubility of calcium hydroxide being high in basic solutions.
So, the molar solubility of calcium hydroxide in a solution buffered at each pH will depend on the specific pH value and the concentration of hydrogen ions (H+) or hydroxide ions (OH-).
To calculate the molar solubility of calcium hydroxide (Ca(OH)2) in a solution buffered at a specific pH, you need to consider the equilibrium reaction of Ca(OH)2 dissociating into Ca2+ and OH- ions:
Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq)
The solubility of Ca(OH)2 is directly related to the concentration of OH- ions in solution, so you need to calculate the concentration of OH- ions for each pH.
a. pH = 4:
At pH 4, you need to consider that the solution is acidic, and the concentration of H+ ions will be relatively high. To calculate the concentration of OH- ions, you can use the following formula:
pOH = 14 - pH
pOH = 14 - 4 = 10
Now, convert pOH to concentration:
[OH-] = 10^(-pOH) = 10^(-10) = 1 x 10^(-10)
Since the concentration of OH- ions is equal to the concentration of Ca2+ ions in this reaction, the molar solubility of Ca(OH)2 is 1 x 10^(-10) M.
b. pH = 7:
At pH 7, the solution is neutral, which means the concentration of H+ ions is equal to the concentration of OH- ions. Therefore, the concentration of OH- ions can be calculated using the formula:
pOH = 14 - pH
pOH = 14 - 7 = 7
[OH-] = 10^(-pOH) = 10^(-7) = 1 x 10^(-7)
Again, since the concentration of OH- ions is equal to the concentration of Ca2+ ions, the molar solubility of Ca(OH)2 is 1 x 10^(-7) M.
c. pH = 9:
At pH 9, the solution is basic, and the concentration of OH- ions will be relatively high. Using the same formula as before:
pOH = 14 - pH
pOH = 14 - 9 = 5
[OH-] = 10^(-pOH) = 10^(-5) = 1 x 10^(-5)
Once again, the concentration of OH- ions is equal to the concentration of Ca2+ ions, so the molar solubility of Ca(OH)2 is 1 x 10^(-5) M.
.......Ca(OH)2 ==> Ca^2+ + 2OH^-
I.......solid.......0.......0
C.......solid.......x.......2x
E.......solid......x.......2x
(H^+) for pH 4 is 1E-4
Ksp = (Ca^2+)(OH^-)
From the above, (Ca^2+) = x
(OH^-) = x + 1E-4. Substitute and solve for x = solubility.
Note: for the total OH^-, the 2x is that from Ca(OH)2 and the 10^-4 is contributed by the solution buffered at pH of 4. So total is 2x+1E-4. Sometimes the 2x can be ignored; sometimes it can't be ignored (which avoids a quadratic equation) and the quadratic equation must be solved.