Apr 9, 2018

$\text{Na"_2"S}$, $\textcolor{w h i t e}{x} \text{pH} > 7$

Explanation:

Let's Start with The simplest one, $\text{KCl}$.

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

${\text{KCl" + "H"_2"O" rightleftharpoons "K"^+ + "Cl"^-) + "H"^+ + "OH}}^{-}$

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

Now, the Barium Acetate, $\text{(CH"_3"COO)"_2"Ba}$.

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

${\text{(CH"_3"COO)"_2"Ba" + "H"_2"O" rightleftharpoons "H"^+ + "OH"^-) + "Ba"^(2+) + 2"CH"_3"COO}}^{-}$

Here, acetate anion with hydrogen ion is a weak acid, whereas ${\text{Ba(OH)}}_{2}$ is rather a strong base. So, both interact to change the solution $\text{pH}$ to be around $8$ to $9$, and it depends upon the concentration.

But, In case of $\text{Na"_2"S}$, it's not the same.

It ionizes and also hydrolyses to form $\text{H"_2"S}$ and $\text{NaOH}$.

${\text{Na"_2"S" + 2"H"_2"O" -> 2"Na"^+ + "S"^(2-) + 2"H"^+ + 2"OH}}^{-}$

And then, ${\text{H}}^{+}$ and ${\text{S}}^{2 -}$ combine to form $\text{H"_2"S}$, and that escapes as gas, leaving behind $\text{NaOH}$.

$\text{NaOH}$ is a base, so the solution must have a $\text{pH}$ 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.

Apr 9, 2018

See Below

Explanation:

$N {a}_{2} S$ in water comes apart into two $N {a}^{+}$ ions and a ${S}^{-} 2$ ion. The ${S}^{-} 2$ ion will interact with water to form a basic solution.

${S}^{-} 2 + 2 {H}_{2} O = {H}_{2} S + 2 O {H}^{-}$ There will be two sodium ions on both sides of the equal sign. This will make a basic solution, pH>7

Barium Acetate, $B a \left(C {H}_{3} C O O\right)$ will come apart to give $B {a}^{+} 2$ ions that don't do anything...and acetate ions, $C {H}_{3} C O {O}^{-}$ ions that do interact with water to make basic solutions.
$2 C {H}_{3} C O {O}^{-} 1 + 2 {H}_{2} O = 2 C {H}_{3} C O O H + 2 O {H}^{-} 1$ There will be barium ions on both sides of this equation.
This will make a basic solution, with pH>7

$C {H}_{3} C O {O}^{-} 1$ 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.