Question

Question #b98fe

Answer 1

1. `1 mol K^+, 1 molCl^-`
2. `1 mol Mg^(2+), 2 mol NO_3^-`

Explanation:

In one liter of a one molar solution, the number of moles of the solute is

`(M)(V) = (1 (mol)/L)(1L) = 1 mol`

For `KCl`, there are one atom of each per formula unit, so there are one mole of each `K^+` and `Cl^-` in one liter of a `1M` solution of potassium chloride.

For `Mg(NO_3)_2`, there are two moles of `NO_3^-` per one mole of `Mg^(2+)`, so there is one mole of `Mg^(2+)` and two moles of `NO_3^-` in one liter of a `1M` solution of magnesium nitrate.

Answer 2

`KCl`: `2 mol`
`Mg(NO_(3))_(2)`: (Approximately) `3 mol`.

Explanation:

Both `KCl` and `Mg(NO_(3))2` are salts. Thus they completely ionize when dissolved in water (to yield `K^(+)`, `Cl^(-)`, `Mg^(+2)` and NO_(3)^(2-), respectively.)
The number of moles of ions produced from `KCl` dissolving:
`1 mol K^+ +1 mol Cl^(-)=2 mol (ions)`
The number of moles of ions generated from `Mg(NO_(3))2` dissolving:
`1 mol Mg^(2+) +2 mol(NO_(3)^(-))=3 mol (ions)`

However, if you take the reversible reaction into account: `Mg^(2+)+2(OH^-) rightleftharpoonsMg(OH)2`
`K_(sp)=5.61×10−12` for `Mg(OH)2`.
Let the number of `Mg^(2+)` converted in this reaction equals to `x`, using the RICE table,
`(1-x)*(10^(-7)-2x)=5.61×10^(−12)`, `x=4.99972*10^-8`.
This value is too small to be considered.
Therefore the moles of ions in `1L` of `1M` `Mg(NO_(3))_(2)` solution is `3 mol`.

(reference: [https://en.wikipedia.org/wiki/Magnesium_hydroxide](https://en.wikipedia.org/wiki/Magnesium_hydroxide) )
[What's the meaning of the ICE table.. as seen in this equilibrium solution?],(https://socratic.org/questions/what-s-the-meaning-if-ice-table-as-seen-in-this-equilibrium-solution).