Question #b41c8

You know that a weak monoprotic acid will only partially ionize to produce hydronium cations and the conjugate base of the acid in #1:1# mole ratios, so if you take #"HA"# to be your generic monoprotic weak acid, you can say that you have

#"HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "A"_ ((aq))^(-)#

Now, you know that you start with #1.75# moles of weak acid and that, at equilibrium, the solution contains #1.15# moles of weak acid.

This implies that

#"175 moles " - " 1.15 moles" = "0.60 moles"#

of weak acid have ionized to produce hydronium cations and the conjugate base of the weak acid.

This implies that, at equilibrium, the solution contains #0.60# moles of hydronium cations and #0.60# moles of #"A"^(-)# #-># this is the case because for every #1# mole of weak acid that dissociates, you get #1# mole of each product.

Use the volume of the solution to calculate the equilibrium concentrations of the species involved in the reaction.

#["HA"] = "1.15 moles"/"1.5 L" = "0.767 mol L"^(-1)#

#["H"_3"O"^(+)] = "0.60 moles"/"1.5 L" = "0.40 moles L"^(-1)#

#["A"^(-)] = "0.60 moles"/"1.5 L" = "0.40 moles L"^(-1)#

By definition, the acid dissociation constant, #K_a#, which uses equilibrium concentrations, is equal to

#K_a = (["H"_3"O"^(+)] * ["A"^(-)])/(["HA"])#

Plug in your values to find--I'll leave the answer without added units

#K_a = (0.40 * 0.40)/(0.767) = color(darkgreen)(ul(color(black)(0.21)))#

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the solution.

Finally, the #"pH"# of the solution is equal to

#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#

In your case, the solution will have a #"pH"# of

#"pH" = - log(0.40) = color(darkgreen)(ul(color(black)(0.40)))#

The answer is rounded to two decimal places because you have two sig figs for the volume of the solution.

Now, notice that the #"pH"# of the solution is quite low. However, you can say that you're dealing with a weak acid because strong acids are characterized by the fact that they ionize completely, i.e. #1# mole of acid produces #1# mole of hydronium cations, in aqueous solution to produce hydronium cations.

In your case, the acid does not ionize completely, i.e. #1# mole of acid does not produce #1# mole of hydronium cations, so you can classify it as a weak acid despite the very low #"pH"# value of its solution.