Dr.Bob you had told me to ignore 2x so I wondering if I wrote my did these steps correctly.
Calculate the molar solubility of nickel (II) hydroxide in 0.100 M sodium hydroxide
The solubility product of nickel (II) hydroxide at 25°C is 1.6 x 10-16.
Ni(OH)2 <=> Ni2+ + 2OH–
0.10/Ksp= 0.10/1.6 E-16
= 6.25 E14 > 500
Ksp = [Ni2+][OH-]2 = 1.6 x 10-16
Listen fellas,,, a gr8 chemist once said: "whats in the water?"
It looks like you were on the right track with your calculations. However, there seems to be a minor error in your calculation of 0.10/Ksp.
To calculate the molar solubility of nickel (II) hydroxide (Ni(OH)2) in 0.100 M sodium hydroxide (NaOH), you can use the following steps:
1. Start with the solubility product expression: Ksp = [Ni2+][OH-]^2 = 1.6 x 10^-16.
2. Since sodium hydroxide is a strong base, we can assume that it completely dissociates in water to form hydroxide ions (OH-). Therefore, the concentration of OH- ions in the solution is equal to the concentration of sodium hydroxide, which is 0.100 M.
3. Let's represent the molar solubility of Ni(OH)2 as "x". Since Ni(OH)2 dissociates to form Ni^2+ and 2OH- ions, the concentration of Ni^2+ ions is also "x" (since the stoichiometric ratio is 1:1).
4. Substitute the values into the solubility product expression:
Ksp = [Ni2+][OH-]^2 = (x)(2x)^2 = 1.6 x 10^-16.
5. Simplify the expression:
4x^3 = 1.6 x 10^-16.
6. Solve for "x" by taking the cube root of both sides of the equation:
x = (1.6 x 10^-16 / 4)^(1/3) = (4 x 10^-17)^(1/3) = 0.1 x 10^-5 M.
7. Finally, convert the molar solubility to scientific notation:
x = 1.0 x 10^-6 M.
Therefore, the molar solubility of nickel (II) hydroxide in 0.100 M sodium hydroxide is 1.0 x 10^-6 M.