Calculate the limiting value of the fraction of dissociation (α) of a weak acid (pKa = 5.00) as the concentration of HA approaches 0. Repeat the same calculation for pKa = 9.00.?

Let's first calculate the expression for α for a weak acid #"HA"#.

#color(white)(mmmmmmm)"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"#
#"I/mol·L"^"-1": color(white)(mml)c color(white)(mmmmmmml)0color(white)(mmm)0#
#"C/mol·L"^"-1": color(white)(ml)"-"αc color(white)(mmmmmm)"+"αc color(white)(ml)"+"αc#
#"E/mol·L"^"-1": color(white)(m)c("1-α") color(white)(mmmmmll)αc color(white)(mml)αc#

#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = (α^2 stackrelcolor(blue)(c)(color(red)(cancel(color(black)(c^2)))))/(color(red)(cancel(color(black)(c)))("1-α")) = (α^2c)/("1-α")#

#α^2c + K_text(a)α -K_text(a) = 0#

#color(blue)(bar(ul(|color(white)(a/a)α =("-"K_text(a) + sqrt(K_text(a)^2 + 4K_text(a)c))/(2c)color(white)(a/a)|)))" "#

Fortunately, "they" didn't specify how we were to calculate the limit.

I chose to calculate the limit by inserting progressively smaller values for the molar concentration #c#.

I used Microsoft Excel to calculate the values of α for concentrations ranging from #"100 mol/L"# to #10^"-7" "mol/L"# for #"p"K_text(a) = 5.00# and for #"p"K_text(a) = 9.00#

I got the following table:

Normally, we will not encounter solutions as concentrated as #"100 mol/L"# nor as dilute as #10^"-7" color(white)(l)"mol/L"#, so these represent the extremes of concentrations.

Next, I plotted #α# vs. #log(c)# for the two values of #K_text(a)#.

The graph shows that a weak acid like acetic acid is almost completely dissociated when #c = 10^"-7"color(white)(l)"mol/L"#.

That is, #lim_(c→0)α = 1#.

However, an even weaker acid like #"HCN"# is only about 10 % dissociated even at extreme dilution.

The fraction of dissociation #alpha -> 1.00# as #["HA"]_i -> K_a# for a monoprotic acid #"HA"#, i.e. as concentration decreases, fraction of dissociation increases until it gets to the maximum of #1.00#, or #100%# dissociation.

The only thing that changes with a smaller #K_a# is that #["HA"]_i# can be smaller before it becomes nonphysically small, and that #alpha# approaches #1# more slowly.

All that matters for convergence of #x_n# is that #K_a/(["HA"]_i) < 1#.

First, let's consider what fraction of dissociation means.

#alpha = (["HA"]_"lost")/(["HA"]_i)#

The #"HA"# lost was dissociated away into #"A"^(-)# and #"H"^(+)#. Since that is the case, #x = ["H"^(+)]_(eq) = ["A"^(-)]_(eq) = ["HA"]_"lost"#, and

#alpha = (["H"^(+)])/(["HA"]_i) -= x/(["HA"]_i)#

#x# is considered small if #K_a# is small (usually if #K_a# is #10^(-5)# or less), and when #K_a# is small, the fraction of dissociation is small. As #["HA"]_i# decreases, a greater fraction of the particles should be able to dissociate, so logically,

#bb(lim_(["A"]_i -> 0) alpha = 1)#.

The condition, as shown in this video, is that

#K_a/(["HA"]_i) < 1#

for the small #x# approximation to work at all (not necessarily well), and #"<<"# #1# to work well.

What I shall do is decrease #["HA"]_i# and calculate #x# at each iteration for a given #K_a = 10^(-5)# and decreasing concentration #["HA"]_i#.

At each iteration up until the final one, we have:

#x_1 ~~ sqrt(["HA"]_i cdot K_a)#

#x_2 ~~ sqrt((["HA"]_i - x_1)cdot K_a)#

#x_3 ~~ sqrt((["HA"]_i - x_2)cdot K_a)#

#vdots#

#x_N ~~ sqrt((["HA"]_i - x_(N-1))cdot K_a)#

With that in mind, I did this in Excel just now, supposing that #K_a = 10^(-5)#. The graph for this plotted against the iteration number becomes:

The data graphed was:

Things to scan for on the data:

• Decreasing concentrations #["HA"]_i# were used, starting from #["HA"]_i = K_a# and doubling from right to left.

• As #["HA"]_i# decreases, more iterations are required for convergence. When #["HA"]_i# gets too small (smaller than #K_a#), convergence is impossible (see the sine wave?).

• There was an if conditional, i.e.

#"If K"_a//["HA"]_i < 1#, then say #"YES"#.
#"Else"#, say #"NO!"#

I was showing here then, that if #K_a//["HA"]_i >= 1#, the small #x# approximation fails catastrophically.

• I calculated the #%"dissoc"#, which was #0.22, 0.30, 0.39, 0.50, 1.00#, i.e. it approaches #bb1# as #["HA"]_i# decreases. It is #1# when #K_a = ["HA"]_i#.

We can now see the issue: since #alpha <= 1#, and #alpha = x_n/(["HA"]_i)#, it means #x_n <= ["HA"]_i#.

• If we have the #(n-1)#th value, #x_(n-1)#, less than #["HA"]_i#, then #x_n# converges monotonically.
• If we have the #(n-1)#th value, #x_(n-1)#, equal to #["HA"]_i#, then #x_n# oscillates between #0# and #K_a#.
• If we have the #(n-1)#th value, #x_(n-1)#, greater than #["HA"]_i#, then #x_n# is imaginary.

I did the same thing for #K_a = 10^(-9)#, the only thing that changes is how low #["HA"]_i# can get before it is nonphysical. Otherwise it looks identical.

So now, let's consider #alpha# for different #K_a# values but the same chosen values of #["HA"]_i# (on the order of #10^(-5) "M"#).

This graph is then:

And here we do see that #alpha -> 1# as #["HA"]_i -> 0#. However, it is stopped at the value of #K_a#, showing that #["HA"]_i# must be less than or equal to #K_a#.