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.?

2 Answers
Jan 30, 2018

lim_(c→0)α = 1

Explanation:

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:

TableTable

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).

GraphGraph

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.

Jan 30, 2018

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.