Question

Anyone have an idea about this chemistry problem?

Answer 1

`"Na"_2"S"`, `color(white)(x)"pH" gt 7`

Explanation:

Let's Start with The simplest one, `"KCl"`.

It does ionize, and fully ionize(strong electrolyte) when dissolved in water, but doesn't undergo hydrolysis.

`"KCl" + "H"_2"O" rightleftharpoons "K"^+ + "Cl"^-) + "H"^+ + "OH"^-`

`"pH"` will stay `7` if the water used for dissolving `"KCl"` was distilled. Basically, `"KCl"` doesn't produce any ions to cause appearance of acidic or basic properties in the solution.

Now, the Barium Acetate, `"(CH"_3"COO)"_2"Ba"`.

It also ionizes, but not fully(weak electrolyte) when dissolved in water, and also doesn't undergo hydrolysis.

`"(CH"_3"COO)"_2"Ba" + "H"_2"O" rightleftharpoons "H"^+ + "OH"^-) + "Ba"^(2+) + 2"CH"_3"COO"^-`

Here, acetate anion with hydrogen ion is a weak acid, whereas `"Ba(OH)"_2` is rather a strong base. So, both interact to change the solution `"pH"` to be around `8` to `9`, and it depends upon the concentration.

But, In case of `"Na"_2"S"`, it's not the same.

It ionizes and also hydrolyses to form `"H"_2"S"` and `"NaOH"`.

`"Na"_2"S" + 2"H"_2"O" -> 2"Na"^+ + "S"^(2-) + 2"H"^+ + 2"OH"^-`

And then, `"H"^+` and `"S"^(2-)` combine to form `"H"_2"S"`, and that escapes as gas, leaving behind `"NaOH"`.

`"NaOH"` is a base, so the solution must have a `"pH"` greater than 7.

That's why, you should not keep aqueous sodium sulphide solution for a long time. For use, use freshly prepared solution instead.

Hope this helps.

Answer 2

See Below

Explanation:

`Na_2S` in water comes apart into two `Na^+` ions and a `S^-2` ion. The `S^-2` ion will interact with water to form a basic solution.

`S^-2 + 2H_2O = H_2S + 2OH^-` There will be two sodium ions on both sides of the equal sign. This will make a basic solution, pH>7

Barium Acetate, `Ba(CH_3COO)` will come apart to give `Ba^+2` ions that don't do anything...and acetate ions, `CH_3COO^-` ions that do interact with water to make basic solutions.
`2CH_3COO^-1 + 2H_2O = 2CH_3COOH + 2OH^-1` There will be barium ions on both sides of this equation.
This will make a basic solution, with pH>7

`CH_3COO^-1` has a `K_B` values (acetate ion is a weak base), so you can figure out how basic they'll be, but suffice it to say, they'll both be pH>7. Sulfide will be very high pH since it is a strong base, while acetate solution will be about 8.5-9.0 pH.

KCl comes apart to give K+ and Cl- ions, and they don't do anything to the water with respect to pH.