Question

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.?

Answer 1

`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:

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.

Answer 2

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`.