The solubility of Chromium (II) Hydroxide is .0035g/100mL
- Calculate the solubility product of Chromium (II) Hydroxide.
I know that we have to find Ksp. but i do not understand how to make the Ksp equation to calculate the solubility product, if it is already given to us...
To calculate the solubility product (Ksp), you can use the given solubility value. The solubility product equation, based on the balanced chemical equation for the dissociation of Chromium (II) Hydroxide, is:
Cr(OH)2(s) ↔ Cr2+ (aq) + 2OH- (aq)
The dissociation equation shows that one mole of Cr(OH)2 produces one mole of Cr2+ ions and two moles of OH- ions. Therefore, the expression for the solubility product is:
Ksp = [Cr2+][OH-]^2
Since the concentration of OH- ions is twice that of Cr2+ ions, you can substitute [OH-]^2 as 4x^2, where x is the concentration of Cr2+ ions.
Now, let's calculate the solubility product (Ksp):
Given solubility of Chromium (II) Hydroxide: 0.0035 g/100 mL
First, convert the solubility to the molar concentration (mol/L) of Cr2+ ions using the molar mass of Cr(OH)2, which is 110.0 g/mol.
Mass of Cr(OH)2 in 100 mL = 0.0035 g
Molar concentration of Cr2+ ions = (0.0035 g / 110.0 g/mol) / (0.100 L)
Now, substitute the concentration of Cr2+ ions into the Ksp expression:
Ksp = [Cr2+][OH-]^2 = (concentration of Cr2+ ions) * (4x^2)
Substituting the concentration of Cr2+ ions, the expression becomes:
Ksp = ((0.0035 g / 110.0 g/mol) / (0.100 L)) * 4x^2
Simplify the expression by calculating the numerical value of (0.0035 g / 110.0 g/mol) / (0.100 L) and substituting it with its decimal value, and you will have the Ksp value for Chromium (II) Hydroxide.
To calculate the solubility product (Ksp) of Chromium (II) Hydroxide (Cr(OH)2), you can use the given solubility of .0035g/100mL.
The solubility of Cr(OH)2 can be expressed as the molar solubility (S) by converting grams to moles.
First, you need to calculate the molar mass of Cr(OH)2.
The molar mass of Cr = 52 g/mol
The molar mass of O = 16 g/mol
The molar mass of H = 1 g/mol
Molar mass of Cr(OH)2 = (52 g/mol) + 2*(1 g/mol) + 2*(16 g/mol) = 84 g/mol
Now, calculate the molar solubility (S) using the given solubility:
0.0035 g/100 mL = (0.0035 g/100 mL) * (1 mol/84 g) * (1000 mL/1 L) = 0.00417 mol/L
Next, write the solubility equilibrium expression for Cr(OH)2 and calculate the Ksp.
The balanced chemical equation for the dissociation of Cr(OH)2 is:
Cr(OH)2(s) ⇌ Cr2+(aq) + 2OH-(aq)
The solubility equilibrium expression is:
Ksp = [Cr2+][OH-]2
Since the stoichiometric coefficients for Cr2+ and OH- are 1 and 2, respectively, the Ksp expression can be written as:
Ksp = S*(2S)^2 = 4S^3
Substitute the molar solubility (S) value into the expression to calculate the Ksp:
Ksp = 4*(0.00417 mol/L)^3 ≈ 2.63 x 10^-8
Therefore, the solubility product (Ksp) of Chromium (II) Hydroxide is approximately 2.63 x 10^-8.
You are not given Ksp, That's what you are to calculate.
You have g/100 Cr(OH)3. Multiply by 10 gives you g/L. Divide by molar mass to get mols/L.
Then plug that into Ksp expression fopr (Cr^3+). That times 3 gives you the (OH^-). Then solve for Ksp. Post your work if you get stuck.