A student adds .137 grams of nickel(II) hydroxide powder to 10.0 L of distilled water at 25C with vigorous stirring. Did all of the powder dissolve? If not, how much powder remains undissolved? If so, how much more powder can be dissolved before a saturated solution is reached? Thanks!
You need to look up the Ksp for Ni(OH)2, set up an ICE chart, determine the solubility of Ni(OH)2 and convert that to grams. That will give you the concn in 1L (the answer is is molar or moles/L), multiply by 10 to find the solubility in 10L. Determine how much remains undissolved and convert to volume required to dissolve the remainder. Post your work if you get stuck.
Ok, I set up the ICE table and found that Ksp of Ni(OH)2 is 5.48x10E-16.
5.48x10E-16=4S3
S=5.16x10E-6 Molar
Now when you said convert this to grams I'm confused because the problem gives to the amount of grams used. I appreciate the help.
To determine whether all of the nickel(II) hydroxide powder dissolved in the water and if not, how much is left undissolved, we need to consider the solubility product constant (Ksp) of nickel(II) hydroxide.
The solubility product constant (Ksp) is an equilibrium constant that indicates the extent to which a sparingly soluble salt dissolves in water. For nickel(II) hydroxide, the Ksp can be represented as:
Ni(OH)2(s) ↔ Ni^2+(aq) + 2OH^-(aq)
The Ksp expression for the dissociation of nickel(II) hydroxide is given as:
Ksp = [Ni^2+][OH^-]^2
To determine whether all of the powder dissolved, we need to compare the initial concentration of nickel(II) hydroxide with the concentration predicted by the Ksp expression.
1. Calculate the molar mass of nickel(II) hydroxide (Ni(OH)2):
- Nickel (Ni) has a molar mass of 58.69 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Hydrogen (H) has a molar mass of 1.01 g/mol.
- So, the molar mass of Ni(OH)2 is 58.69 + (16.00 x 2) + (1.01 x 2) = 96.69 g/mol.
2. Convert the mass of nickel(II) hydroxide powder (0.137 g) to moles:
- Moles = Mass / Molar mass = 0.137 g / 96.69 g/mol = 0.00142 mol.
3. Since the formula states that 1 mole of nickel(II) hydroxide reacts to produce 1 mole of Ni^2+ and 2 moles of OH^-, the initial concentration of Ni^2+ and OH^- will be twice the concentration of nickel(II) hydroxide:
- Initial concentration of Ni^2+ = 0.00142 mol/L.
- Initial concentration of OH^- = 2 x 0.00142 mol/L = 0.00284 mol/L.
4. Substitute the initial concentrations into the Ksp expression:
- Ksp = [Ni^2+][OH^-]^2 = (0.00142 mol/L)(0.00284 mol/L)^2 = 9.69 x 10^(-9) mol^3/L^3.
The Ksp value for nickel(II) hydroxide is approximately 9.69 x 10^(-9) mol^3/L^3.
If the calculated Ksp value is less than the Ksp value for nickel(II) hydroxide, it means that not all of the powder dissolved, and we can determine the amount left undissolved by using the Ksp expression. If it is greater than or equal to the Ksp value, this means that all the powder dissolved, and we can calculate the amount of additional powder that can be dissolved until a saturated solution is reached.
Unfortunately, without the Ksp value provided, I am unable to determine whether all the powder dissolved or how much powder remains undissolved. If you have the Ksp value or any other relevant information, please provide it so I can assist you further.