the solubility product constant (Ksp) of ca(oh)2 at 25 C is 4.42*10^-5. a 500 ml of saturated solution of ca(oh)2 is mixed with equal volume of .4M naoh. what mass of ca(oh)2 is precipitated?
You really need to find the caps key on your keyboard and use it. First, calculate the solubility, S, in the saturated solution of Ca(OH)2. That is
......Ca(OH)2==>Ca^2+ + 2OH^-
I.....solid......0.......0
C.....solid......S.......2S
E.....solid......S.......2S
Plug the E line into Ksp expression and solve for S. This gives you the solubility of Ca(OH)2 in a saturated solution in mols/L. Convert to grams/L with grams = mols x molar mass = ?g
Then we add 500 mL 0.4M NaOH.
(NaOH) = 0.4M x (500/1000) = 0.2M.
We solve the expression again for S but this time with the added NaOH.
.....Ca(OH)2 ==> Ca^2+ + 2OH^-
I....solid.......0......0.2M
C....solid.......S........2S
E....solid.......S......2S+0.2
Plug the E line into Ksp expression and solve for S.
Convert to grams.
How much ppts? You know the solubility with no NaOH and you know the solubility with NaOH (it will be less). Subtract to find grams that ppt. Post your work if you get stuck.
To find the mass of Ca(OH)2 precipitated, we need to determine the concentration of Ca2+ ions in the saturated solution and the excess concentration of OH- ions after the neutralization reaction between Ca(OH)2 and NaOH.
1. Firstly, let's find the initial concentration of Ca2+ ions in the saturated solution by using the Ksp expression for Ca(OH)2:
Ksp = [Ca2+][OH-]^2
Since the concentration of OH- ions is twice the concentration of Ca2+ ions in a saturated solution, we can write:
Ksp = [Ca2+][(2[Ca2+])^2]
Substituting the given value for Ksp:
4.42 x 10^-5 = [Ca2+][4Ca2+]^2
Simplifying:
4.42 x 10^-5 = [Ca2+][16[Ca2+]^2]
Setting [Ca2+] as x:
4.42 x 10^-5 = x(16x^2)
Rearranging:
16x^3 = 4.42 x 10^-5
Solving for x:
x = (4.42 x 10^-5 / 16)^(1/3)
Calculating:
x ≈ 0.0244 M (concentration of Ca2+ ions in the saturated solution)
2. Now that we know the concentration of Ca2+ ions, let's calculate the excess concentration of OH- ions after mixing the saturated solution with NaOH.
When equal volumes of the saturated solution and 0.4 M NaOH are mixed, the concentration of OH- ions from NaOH will be halved. Thus, the concentration of OH- ions would be 0.2 M.
3. To determine the amount of Ca(OH)2 precipitated, we need to calculate the number of moles of OH- ions reacting with Ca2+ ions in the saturated solution.
Since the stoichiometry of the reaction between Ca(OH)2 and OH- is 1:2, the number of moles of OH- ions reacting is twice the number of moles of Ca2+ ions in the saturated solution.
Moles of Ca2+ ions = concentration x volume
Moles of Ca2+ ions = 0.0244 M x 0.5 L
= 0.0122 moles
Moles of OH- ions reacting = 2 x 0.0122 moles
= 0.0244 moles
4. The molar mass of Ca(OH)2 is:
Ca: 40.08 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Molar mass of Ca(OH)2 = (40.08) + (2 x 16.00) + (2 x 1.01)
= 74.12 g/mol
5. Finally, we can calculate the mass of Ca(OH)2 precipitated using the moles of OH- ions reacting and the molar mass of Ca(OH)2.
Mass of Ca(OH)2 = moles of OH- ions reacting x molar mass of Ca(OH)2
= 0.0244 moles x 74.12 g/mol
= 1.789 g
Therefore, approximately 1.789 grams of Ca(OH)2 will precipitate.
To calculate the mass of Ca(OH)2 precipitated, we need to determine the concentration of Ca2+ ions in the saturated solution of Ca(OH)2 before and after mixing with NaOH. Then, we can calculate the moles of Ca2+ ions that reacted with NaOH and finally convert it to the mass of Ca(OH)2.
Step 1: Calculate the concentration of Ca2+ ions in the saturated solution.
The solubility product constant (Ksp) of Ca(OH)2 is given as 4.42 * 10^-5.
Since Ca(OH)2 dissociates into one Ca2+ ion and two OH- ions, we can write the equation for the dissociation reaction as follows:
Ca(OH)2 ⇌ Ca2+ + 2OH-
Let's assume 'x' as the molar solubility of Ca(OH)2.
The concentration of Ca2+ ions in the saturated solution will be 'x' M.
Step 2: Calculate the initial moles of Ca2+ ions in the saturated solution.
We have a 500 mL solution of Ca(OH)2, which is equal to 0.5 L.
The initial moles of Ca2+ ions can be calculated using:
Moles of Ca2+ ions = Concentration × Volume
Moles of Ca2+ ions = x M × 0.5 L = 0.5x moles
Step 3: After mixing with NaOH, Ca2+ reacts with OH- to form precipitate.
Since we add an equal volume of NaOH, the final volume of the solution will be doubled, i.e., 1 L.
Step 4: Calculate the moles of OH- ions added.
The concentration of NaOH is given as 0.4 M.
The moles of OH- ions added can be calculated as:
Moles of OH- ions = Concentration × Volume
Moles of OH- ions = 0.4 M × 0.5 L = 0.2 moles
Step 5: Determine the moles of Ca2+ ions that react with OH- ions.
From the balanced equation, we know that 1 mole of Ca(OH)2 produces 1 mole of Ca2+ ions.
Therefore, the moles of Ca2+ ions that react with OH- ions will also be 0.2 moles.
Step 6: Convert moles of Ca2+ ions to mass of Ca(OH)2.
The molar mass of Ca(OH)2 can be calculated as follows:
Molar Mass of Ca(OH)2 = (40.08 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol) = 74.10 g/mol
Mass of Ca(OH)2 = Moles of Ca(OH)2 × Molar Mass of Ca(OH)2
Mass of Ca(OH)2 = 0.2 moles × 74.10 g/mol
Finally, calculate the mass of Ca(OH)2 precipitated using the given values.