Background information: The concentration of various cations in seawater in moles per liter are Na+ (0.46M); Mg2+ (0.056M); Ca2+ (0.01M); Al3+ (4x10^-7M); Fe3+ (2x10^-7M)

Question:
If enough OH- ion is added to precipitate 50% of the Mg2+, what percentage of other the other ions will be precipitated. ( I know the answer is virtually all- but how do I get to that answer?)

Mg = 0.056 M

(Mg^+2)(OH^-)^2 = Ksp
Take a 1 L sample.
You know Mg^+2, divide that by 2, you know Ksp, calculate (OH^-).
Using that OH^-, set up Ksp expressions for the other ions. For example,
(M^+)(OH^-) = Ksp
Plug in Ksp and OH^- and calculate (M^+), then (M^+)/(concn in sea water) = fraction remaining in solution, that subtracted from 1 gives the fraction unpptd. None of the Na^+ will ppt.

thanks again

To find out the percentage of other ions that will be precipitated when enough OH- ions are added to precipitate 50% of Mg2+, we need to understand the chemistry of precipitation reactions and the concept of solubility product.

When a compound is dissolved in water, it dissociates into its constituent ions. For example, Mg2+ ions are released when Mg(OH)2 is dissolved in water:

Mg(OH)2(s) ⇌ Mg^2+(aq) + 2OH^-(aq)

The solubility product (Ksp) is the equilibrium constant for the dissolution of a solid compound into its constituent ions. For Mg(OH)2, the Ksp is given as:

Ksp = [Mg^2+][OH^-]^2

Now, if enough OH- ions are added to precipitate 50% of the Mg2+, it means that half of the Mg2+ ions will combine with OH- ions to form Mg(OH)2(s). The remaining Mg2+ ions will remain in solution. In terms of moles, this means that:

0.028 moles of Mg2+ (50% of 0.056M) will be precipitated as Mg(OH)2
0.028 moles of Mg2+ will be left in solution

Since Mg(OH)2 is a 1:2 electrolyte (1 mole of Mg2+ reacts with 2 moles of OH-), we need 0.056 moles of OH- ions to precipitate 0.028 moles of Mg2+. The remaining OH- ions will react with other cations present in the solution.

Now let's consider each ion present in the seawater and calculate the amount that will be precipitated:

1. Na+ (0.46M): Na+ ions will not react with OH- ions to form any precipitate because NaOH is highly soluble in water. Therefore, none of the Na+ ions will be precipitated.

2. Ca2+ (0.01M): Ca2+ ions can react with OH- ions to form Ca(OH)2, which is sparingly soluble in water. To calculate the amount of Ca2+ precipitated, we need to compare the solubility product of Ca(OH)2 (Ksp = [Ca^2+][OH^-]^2) with the concentration of Ca2+ ions:

Ksp = [Ca^2+][OH^-]^2
For Ca^2+, [Ca^2+] = 0.01M
For OH^-, [OH^-] = 2 × [Ca^2+] (due to the stoichiometry of the reaction)

Now you can substitute these values into the Ksp expression, solve for [OH^-], and determine the percentage of Ca2+ precipitated.

3. Al3+ (4x10^-7M): Similar to Ca2+, Al3+ ions can react with OH- ions to form Al(OH)3, which is also sparingly soluble in water. Calculate the solubility product of Al(OH)3 (Ksp = [Al^3+][OH^-]^3) and compare it with the concentration of Al3+ ions to find the percentage of Al3+ precipitated.

4. Fe3+ (2x10^-7M): Similarly, calculate the solubility product of Fe(OH)3 (Ksp = [Fe^3+][OH^-]^3) and compare it with the concentration of Fe3+ ions to find the percentage of Fe3+ precipitated.

By performing these calculations, you will find that a significant percentage of Ca2+, Al3+, and Fe3+ ions will be precipitated, while the Na+ ions will remain in solution. Remember to account for the stoichiometry of the reactions and the solubility product constants for each compound.