Anyone have an idea about this chemistry problem?

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2 Answers

#"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.

Apr 9, 2018

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.