In this experiment the only source of Ca2+ ions is Ca(OH)2. Using the experimental Ksp value, calculate the slubility of Ca(OH)2 (in mol/L) in .30 M NaOH.
Experiment Ksp = 1.69x10^-5
To calculate the solubility of Ca(OH)2 in NaOH, we need to use the concept of the common ion effect. The common ion effect states that the solubility of a salt decreases when another salt containing a common ion is added to the solution. In this case, both Ca(OH)2 and NaOH produce OH- ions, so the OH- ion is the common ion.
The equation for the dissociation of Ca(OH)2 is:
Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq)
The equation for the dissociation of NaOH is:
NaOH(s) ⇌ Na+(aq) + OH-(aq)
Since we are given that the only source of Ca2+ ions is Ca(OH)2, we can consider the Ca2+ concentration as the limiting factor in the solubility of Ca(OH)2.
Let's assume that the solubility of Ca(OH)2 in NaOH is x mol/L. This means that [Ca2+] = x mol/L and [OH-] = 0.30 M + x mol/L (since we are adding 0.30 M NaOH).
According to the equation for the dissociation of Ca(OH)2, the solubility product expression (Ksp) is:
Ksp = [Ca2+][OH-]^2
Substituting the concentrations into the Ksp expression:
Ksp = (x)(0.30 + x)^2
Given that the Ksp value is 1.69x10^-5, we can set up the equation:
1.69x10^-5 = (x)(0.30 + x)^2
Now, we can solve this quadratic equation for x using algebraic methods or by plugging it into an online calculator or using software like MATLAB. By solving the quadratic equation, we will get the value of x, which represents the solubility of Ca(OH)2 in mol/L in the presence of 0.30 M NaOH.
Ca(OH)2 ==> Ca^2+ + 2OH^-
NaOH ==> Na^+ + OH^-
Ksp = 1.69E-5 = (Ca^2+)(OH^-)^2
Set up an ICE chart.
(Ca^2+) = x
(OH^-) from Ca(OH)2 = x
(OH^-) from NaOH = 0.3
Substitute into Ksp expression to obtain
Ksp = (x)(x+0.3)^2
Solve for x.