Question #eae43

#(2g)/L * (mol)/(138g) approx 1.45*10^-2M#

#C_7H_6O_3(aq) rightleftharpoons H^(+)(aq)+C_7H_5O_3^(-)(aq)#

#K_a = ([H^+][C_7H_5O_3^(-)])/([C_7H_6O_3])#

#1.06*10^-3 = x^2/(0.0145 - x)#
#therefore x = [H^+] approx 0.00343M#

#%I = ([H^+])/([C_7H_6O_3]) = (0.00343)/(0.0145M)*100% approx 23.7%#

This is basically an equilibrium problem where you know the initial concentration (a usual one). But your cited #"pH"# is not accurate. I used your #K_a# to get a percent dissociation of #ul(17.7%)#.

[Your cited #"pH"# suggests a percent dissociation that is far too low.]

Write the reaction first, so you have a way to organize:

#"HC"_7"H"_5"O"_3(aq) rightleftharpoons "C"_7"H"_5"O"_3^(-)(aq) + "H"^(+)(aq)#

#"I"" "["HC"_7"H"_5"O"_3]_i" "" "" ""0 M"" "" "" "" ""0 M"#
#"C"" "" "-x" "" "" "" "" "+x" "" "" "+x#
#"E"" "["HC"_7"H"_5"O"_3]_(i) - x" "x" M"" "" "" "" "x" M"#

METHOD 1: USING THE #bb("K"_a)# (this works)

Next, apparently we know #["HC"_7"H"_5"O"_3]_(i)# already (not equilibrium but initial). However, in water at #25^@ "C"#, the cited solubility is #"2.48 g/L"#, so that's what I'll use.

If you want to use a solubility, you must cite its temperature.

#color(green)(["HC"_7"H"_5"O"_3]_(i) = (2.48 cancel"g")/"L" xx "1 mol"/(138.12 cancel("g H"_7"H"_5"O"_3))#

#=# #color(green)("0.0179"_6 "M")#

Using your cited #K_a#, which is actually correct, we can do this in one of two ways. We know #["HC"_7"H"_5"O"_3]_(i)#, so...

#K_a = 1.06 xx 10^(-3) = x^2/("0.01796 M" - x)#

Solving for the quadratic form,

#x^2 + 1.06 xx 10^(-3) x - 1.06 xx 10^(-3) cdot 0.01796 = 0#,

and the result is #x = ul(["H"^(+)]_(eq) = "0.00387 M")# from the quadratic formula.

The percent dissociation is

#color(blue)(%"dissoc") = x/(["HA"]_i) xx 100%#

#= ("0.00387 M")/("0.0179"_6 "M" + "0.00387 M") xx 100%#

#= color(blue)ul(17.7%)#

And that's reasonable. Salicylic acid has a non-negligible percent dissociation with a #K_a# larger than #10^(-5)#.

METHOD 2: TRYING THE pH (this won't work)

With the #"pH"#,

#10^(-"pH") = ul(["H"^(+)]_(eq)) = x = 10^(-4.96) "M" = ul(1.10 xx 10^(-5) "M")#

As before, the solubility was #"0.01796 M"#.

This is pretty big in comparison to #x#. As a result, the percent dissociation using the cited #"pH"# will be far too low.

#color(red)(%"dissoc") = x/(["HA"]_i) xx 100%#

#= (10^(-4.96) "M")/("0.0179"_6 "M") xx 100% = color(red)(0.0611%)#

which doesn't make sense how low this is, given the size of #K_a#.

In fact, the #K_a# obtained from this percent dissociation is

#color(red)(K_a) = (10^(-4.96) "M")^2/("0.0179"_6 "M" - 10^(-4.96)"M") = color(red)(6.70 xx 10^(-9))#,

which is far from the actual.