The solubility product constant of Calcium hydroxide is 6.5 * 10^-6 . If 0.10 mo of sodium hydroxide is added to 1 L of 0.0001 M Ca(OH)2, what is the final concentration of the calcium ion? SHOW your work.

Again, I have no idea where to start

This is a common ion problem where 0.10M NaOH is added into a solution containing 0.0001M Ca(OH)2.

Ca(OH)2 <=> Ca^+2 + 2OH^-
Ci: --- 0M 0.10M
∆Ci: --- +x +2x
Ceq: --- x 0.10+2x~0.10M

Ksp = [Ca^+2][OH^-]^2
6.5x10^-6 = [Ca^+2](0.10)^2
Solve for [Ca^+2] ...
= (6.5x10^-6)/(0.10)^2
= 6.5x10-4M

Qsp = (Ca^2+)(OH^-)^2 = (1E-4)(0.1)^2 = 1E-6 < Ksp = 6.5E-6 so no ppt of Ca(OH)2 occurs and (Ca^2+) must be 1E-4M

To find the final concentration of calcium ions, we need to consider the reaction between calcium hydroxide (Ca(OH)2) and sodium hydroxide (NaOH). The balanced chemical equation for this reaction is:

Ca(OH)2 + 2NaOH --> Ca(OH)4 + 2NaOH

From the equation, we can see that 1 mole of calcium hydroxide reacts with 2 moles of sodium hydroxide, resulting in the formation of 1 mole of calcium hydroxide and 2 moles of sodium hydroxide.

Given:
- Solubility product constant (Ksp) for calcium hydroxide = 6.5 * 10^-6
- Volume of solution = 1 L
- Initial concentration of calcium hydroxide (Ca(OH)2) = 0.0001 M
- Amount of sodium hydroxide (NaOH) added = 0.10 mol

First, let's calculate the number of moles of calcium hydroxide initially present in 1 L of 0.0001 M solution:
Initial moles of Ca(OH)2 = Initial concentration (M) * Volume (L)
= 0.0001 mol/L * 1 L
= 0.0001 mol

Next, since 1 mole of Ca(OH)2 reacts with 2 moles of NaOH, we know that the amount of Ca(OH)2 reacted will be half of the amount of NaOH added:
Moles of Ca(OH)2 reacted = 0.10 mol / 2
= 0.05 mol

The final moles of Ca(OH)2 will be the difference between the initial moles and the moles reacted:
Final moles of Ca(OH)2 = Initial moles - Moles reacted
= 0.0001 mol - 0.05 mol
= -0.0499 mol

Since the amount of Ca(OH)2 cannot be negative, the reaction is limited by the Ca(OH)2, and all the Ca(OH)2 will react.

Now, let's find the final concentration of calcium ions (Ca2+) in the solution:
Final concentration of Ca2+ = Final moles of Ca2+ / Final volume (L)

The final volume is the same as the initial volume since no water has been added:
Final volume = Initial volume = 1 L

Final concentration of Ca2+ = -0.0499 mol / 1 L
= -0.0499 M

However, the concentration of calcium ions cannot be negative. Therefore, the final concentration of calcium ions is 0 M.

In conclusion, the final concentration of calcium ions in the solution after adding 0.10 moles of sodium hydroxide is 0 M.

To determine the final concentration of the calcium ion in the solution, we need to consider the reaction between calcium hydroxide (Ca(OH)2) and sodium hydroxide (NaOH).

The balanced equation for the reaction is as follows:
Ca(OH)2 + 2NaOH -> Ca(OH)4 + 2NaOH

Now let's break down the given information:

Initial concentration of Ca(OH)2 = 0.0001 M
Volume of solution = 1 L
Moles of NaOH added = 0.10 mol

First, we need to find the moles of Ca(OH)2 present initially:
Moles = Concentration * Volume
Moles of Ca(OH)2 = 0.0001 M * 1 L = 0.0001 mol

Next, since Ca(OH)2 reacts with NaOH in a 1:2 ratio, we need to determine the limiting reagent. The limiting reagent is the one that is completely consumed first. Here, the moles of NaOH added are higher than the initial moles of Ca(OH)2. So, Ca(OH)2 is the limiting reagent.

As per the reaction, 1 mole of Ca(OH)2 produces 1 mole of calcium ions (Ca^2+), so the number of moles of Ca^2+ produced is equal to the moles of Ca(OH)2 added.
Hence, the moles of Ca^2+ produced = 0.0001 mol

Now, to find the final concentration of Ca^2+, we need to consider the final volume of the solution. Since no new solution is added, the volume remains 1 L.

Finally, the final concentration of Ca^2+ is:
Concentration = Moles / Volume
Concentration of Ca^2+ = 0.0001 mol / 1 L = 0.0001 M

Therefore, the final concentration of the calcium ion is 0.0001 M.