Using the systematic method and information in the Appendices F and J (note that log value is given in Appendix J), determine [SCN-], [Hg2+], and [HgSCN+] at equilibrium when Hg(SCN)2(s) is dissolved in water.
If you will look up those values and post them here we may be able to help. Include infor on HgSCN+. Why? My table values may not agree (probably will not) so we can't get the same answer. Show what you know to do also.
I posted the whole question again. Hg2+and SCN-=HgSCN+ =logarithim for the reaction is 9.64 = 10^9.64= 4.3*10^9
Ksp HG2(SCN)2= 3.0*10^-20
I have reactions Hgscn2> hg2+2SCN
Hg2+ + SCN->Hgscn
cations= anions
2[hg2+]+[Hgscn]=[Scn-]
Mass balance
[hgscn+]+[Hg2+]=[HGscn+]+[Scn-]
To solve this problem, we will use the systematic method and the information provided in Appendices F and J. Appendix F contains the values of solubility product constants (Ksp) for various compounds, while Appendix J provides logarithmic values (log K) for these Ksp values.
First, let's write the balanced chemical equation for the dissociation of Hg(SCN)2(s) in water:
Hg(SCN)2(s) ⇌ Hg2+(aq) + 2SCN-(aq)
Now, we need to find the values of the solubility product constants (Ksp) for the substances involved. In this case, the Ksp for Hg(SCN)2 can be found in Appendix F. Let's assume the value is Ksp = 1.5 × 10^-10.
Next, we need to consider the initial concentrations of the species involved. When a solid compound is dissolved in water, it dissociates to some extent, so initially, all of the Hg(SCN)2 will be in the solid form.
At equilibrium, let's assume the concentration of Hg(SCN)2(s) that dissolves is "x" mol/L. The concentration of Hg2+ will be "x" mol/L, and the concentration of SCN- will be "2x" mol/L since the coefficient in front of SCN- is 2 in the balanced equation.
Now, we construct an ICE (Initial, Change, Equilibrium) table:
Hg(SCN)2(s) ⇌ Hg2+(aq) + 2SCN-(aq)
Initial: x 0 0
Change: -x +x +2x
Equilibrium: 0 x 2x
Using the Ksp expression for Hg(SCN)2, we can express it as:
Ksp = [Hg2+] * [SCN-]^2
Substituting the equilibrium concentrations into the expression, we get:
1.5 × 10^-10 = x * (2x)^2
1.5 × 10^-10 = 4x^3
Now, let's solve this equation for "x". We can take the cube root of both sides:
x = (1.5 × 10^-10 / 4)^(1/3)
x ≈ 6.31 × 10^-4
Now that we have the value of "x", we can determine the concentrations at equilibrium.
[SCN-] = 2x ≈ 2 * (6.31 × 10^-4) ≈ 1.26 × 10^-3 mol/L
[Hg2+] = x ≈ 6.31 × 10^-4 mol/L
[HgSCN+] = 0 mol/L (since it is not mentioned in the equation)
So, at equilibrium, the concentrations are approximately:
[SCN-] ≈ 1.26 × 10^-3 mol/L
[Hg2+] ≈ 6.31 × 10^-4 mol/L
[HgSCN+] ≈ 0 mol/L
Please note that the actual values may vary depending on the specific solubility product constant in Appendix F. Make sure to check the appendix for the most accurate and up-to-date values.