Ka, pKa and Acid Strength: HSC Chemistry Guide

Two acid solutions can have exactly the same concentration but very different pH values. One might produce plenty of \(\ce{H3O+}\), while the other barely dissociates. So concentration alone cannot tell you how strongly an acid gives up a proton.

Suppose two acids have \(K_a\) values of \(1.0 \times 10^{-3}\) and \(1.0 \times 10^{-5}\). Before doing any maths, predict which acid produces more \(\ce{H3O+}\) at the same concentration.

It is the acid with \(K_a = 1.0 \times 10^{-3}\). A larger \(K_a\) means the dissociation equilibrium lies further towards the products. More acid molecules have transferred their proton to water.

That idea is the heart of \(K_a\). \(pK_a\) is simply a more convenient way of expressing the same information.

01What \(K_a\) is actually measuring

Consider a generic monoprotic acid, \(\ce{HA}\). When it dissolves in water, some molecules transfer a proton to water:

\[ \ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)} \]

You can picture \(\ce{HA}\) as someone refusing to let go of a borrowed hoodie. The hoodie is the proton. Some acids are extremely clingy; others hand it over to water much more readily.

The analogy stops there. Molecules aren’t making decisions, and acid strength comes from the energetics of the particles and their interactions with the solvent. But the useful picture is simple: stronger acids let go of their proton more readily.

For a weak acid, both the forward and reverse reactions occur. Eventually the system reaches dynamic equilibrium.

The acid dissociation constant is

\[ K_a = \frac{[\ce{H3O+}][\mathrm{A}^{-}]}{[\ce{HA}]} \]

where:

• \(K_a\) is the acid dissociation constant,
• \([\ce{H3O+}]\) is the equilibrium hydronium concentration,
• \([\mathrm{A}^{-}]\) is the equilibrium concentration of the conjugate base, and
• \([\ce{HA}]\) is the equilibrium concentration of undissociated acid.

The square brackets mean equilibrium concentration, usually measured in \(\mathrm{mol\,L^{-1}}\) in HSC calculations.

Liquid water does not appear in the expression because its concentration is effectively constant and is incorporated into \(K_a\).

Larger \(K_a\) means a stronger acid

Look again at

\[ K_a = \frac{[\ce{H3O+}][\mathrm{A}^{-}]}{[\ce{HA}]} \]

If the equilibrium lies further to the right, there is more \(\ce{H3O+}\) and \(\mathrm{A}^{-}\), and less \(\ce{HA}\). The value of \(K_a\) therefore increases.

So, at the same temperature:

The qualification about temperature matters. \(K_a\) is an equilibrium constant, so changing temperature can change its value.

Changing the acid concentration does not give the acid a new \(K_a\) at the same temperature. The equilibrium concentrations adjust, but the equilibrium constant remains the same.

02Why chemists use \(pK_a\)

Acid dissociation constants often involve awkward powers of ten. For example,

\[ K_a = 0.000032 \]

is much harder to compare quickly than a small number such as 4.49.

So we define

\[ pK_a = -\log_{10}(K_a) \]

This creates one important reversal:

larger \(K_a\) means stronger acid, but smaller \(pK_a\) means stronger acid.

For example:

\[ K_a = 1.0 \times 10^{-3} \]

gives

\[ pK_a = -\log_{10}(1.0 \times 10^{-3}) = 3.00 \]

while

\[ K_a = 1.0 \times 10^{-5} \]

gives

\[ pK_a = 5.00 \]

The first acid is stronger.

The \(pK_a\) scale is logarithmic

A difference of one \(pK_a\) unit represents a factor of 10 in \(K_a\).

A difference of two \(pK_a\) units represents a factor of 100.

For example, if:

• acid X has \(pK_a = 3.0\), and
• acid Y has \(pK_a = 5.0\),

then

\[ K_{a,\mathrm{X}} = 100K_{a,\mathrm{Y}} \]

That does not mean X’s pH must be 100 times lower. \(K_a\) describes acid strength; pH describes a particular solution.

03Calculating \(K_a\) from the pH of a weak acid

Suppose you know the initial concentration of a weak acid and its measured pH.

The pH gives you \([\ce{H3O+}]\):

\[ [\ce{H3O+}] = 10^{-\mathrm{pH}} \]

If the acid is the only important source of \(\ce{H3O+}\), every mole of \(\ce{HA}\) that dissociates produces one mole of \(\ce{H3O+}\) and one mole of \(\mathrm{A}^{-}\).

If \(x\) mol \(\mathrm{L^{-1}}\) dissociates:

Here \(C\) is the initial acid concentration.

Therefore,

\[ K_a = \frac{x^2}{C-x} \]

The \(C-x\) term matters because \(K_a\) uses the equilibrium concentration of \(\ce{HA}\), not its starting concentration.

04Going the other way: using \(K_a\) to find pH

Sometimes the question gives you \(K_a\) and the acid concentration, then asks for pH.

If a weak monoprotic acid begins at concentration \(C\), let \(x\) be the concentration that dissociates.

Then

\[ K_a = \frac{x^2}{C-x} \]

You may sometimes see the approximation

\[ K_a \approx \frac{x^2}{C} \]

because \(C-x \approx C\) when only a very small fraction of the acid dissociates.

That is an approximation, not the definition of \(K_a\).

A useful check is the percentage dissociation:

\[ \%\text{ dissociation} = \frac{x}{C}\times100 \]

If the percentage is only a few per cent, treating \(C-x\) as approximately \(C\) may be reasonable. If a substantial fraction dissociates, use the full expression.

05Acid strength is not the same thing as pH

This is the most tempting mistake in the topic.

A student sees two bottles:

• solution X has pH 2,
• solution Y has pH 3.

It feels natural to say, “X contains the stronger acid.”

You cannot conclude that from pH alone.

Acid strength describes the tendency of an acid to dissociate. It is represented by \(K_a\) or \(pK_a\).

pH describes the hydronium concentration in a particular solution:

\[ \mathrm{pH}=-\log_{10}[\ce{H3O+}] \]

That hydronium concentration depends on several things, including how strong the acid is and how much acid is present.

A concentrated weak acid can therefore have a lower pH than a very dilute stronger acid.

Keep these ideas separate:

This distinction becomes particularly useful when interpreting titration curves and conductivity graphs.

06What dilution does to a weak acid

Imagine diluting a weak acid with water. The concentration of acid falls.

What happens to the percentage of acid molecules that dissociate: decrease, stay constant, or increase?

It increases.

Dilution shifts the equilibrium towards the side containing more dissolved particles:

\[ \ce{HA + H2O <=> H3O+ + A-} \]

So a larger fraction of the remaining acid dissociates.

This does not mean the actual \([\ce{H3O+}]\) increases. The solution still normally becomes less acidic overall, so its pH rises. The point is subtler: the acid is more dilute, but a greater percentage of it is dissociated.

And \(K_a\) has not changed.

At the same temperature, the new equilibrium concentrations still satisfy

\[ K_a = \frac{[\ce{H3O+}][\mathrm{A}^{-}]}{[\ce{HA}]} \]

This is why treating a weak acid like a completely dissociated strong acid can give the wrong effect of dilution.

07A compact decision rule

When an HSC question gives you \(K_a\), \(pK_a\), pH, or concentration, first ask what kind of quantity you actually have.

• To compare intrinsic acid strength, compare \(K_a\) or \(pK_a\).
• Larger \(K_a\) means stronger acid.
• Smaller \(pK_a\) means stronger acid.
• To convert \(K_a\) into \(pK_a\), use \(pK_a=-\log_{10}(K_a)\).
• To convert \(pK_a\) into \(K_a\), use \(K_a=10^{-pK_a}\).
• To calculate \(K_a\) from pH, find \([\ce{H3O+}]\), construct the equilibrium concentrations, then substitute.
• To calculate pH from \(K_a\), solve for the equilibrium \([\ce{H3O+}]\), then take \(-\log_{10}\).
• Never use pH alone as proof that one acid is intrinsically stronger than another.

08Questions and solutions

Question 1

At the same temperature, acid P has \(K_a = 4.0 \times 10^{-5}\), while acid Q has \(K_a = 8.0 \times 10^{-6}\).

Calculate the \(pK_a\) of each acid and identify the stronger acid.

Solution 1

Acid P is stronger; its \(pK_a\) is 4.40, compared with 5.10 for acid Q.

For acid P:

\[ pK_a = -\log_{10}(4.0 \times 10^{-5}) = 4.40 \]

For acid Q:

\[ pK_a = -\log_{10}(8.0 \times 10^{-6}) = 5.10 \]

Acid P has the larger \(K_a\) and smaller \(pK_a\), so its equilibrium lies further towards \(\ce{H3O+}\) and \(\mathrm{A}^{-}\).

A common mistake is to choose the acid with the larger \(pK_a\). The logarithm reverses the direction: stronger acid means larger \(K_a\), but smaller \(pK_a\).

Question 2

A \(0.200\ \mathrm{mol\,L^{-1}}\) solution of a monoprotic weak acid \(\ce{HA}\) has a pH of 2.54.

Calculate:

a. the equilibrium \([\ce{H3O+}]\),

b. \(K_a\),

c. \(pK_a\), and

d. the percentage of the acid that has dissociated.

Assume the acid is the only significant source of \(\ce{H3O+}\).

Solution 2

The acid has \(K_a \approx 4.22 \times 10^{-5}\), \(pK_a \approx 4.37\), and about \(1.44\%\) of it is dissociated.

First find the hydronium concentration:

\[ [\ce{H3O+}] =10^{-2.54} =2.884\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Because each dissociation produces one \(\ce{H3O+}\) and one \(\mathrm{A}^{-}\),

\[ [\mathrm{A}^{-}] =2.884\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

The equilibrium concentration of \(\ce{HA}\) is

\[ [\ce{HA}] =0.200-0.002884 =0.197116\ \mathrm{mol\,L^{-1}} \]

Therefore,

\[ K_a =\frac{(2.884\times10^{-3})^2}{0.197116} =4.22\times10^{-5} \]

Then

\[ pK_a =-\log_{10}(4.22\times10^{-5}) =4.37 \]

The percentage dissociation is

\[ \frac{2.884\times10^{-3}}{0.200}\times100 =1.44\% \]

The tempting shortcut is to put the initial concentration \(0.200\ \mathrm{mol\,L^{-1}}\) directly into the denominator. That would be close numerically here because dissociation is small, but \(K_a\) is defined using equilibrium concentrations. The proper denominator is \(0.197116\ \mathrm{mol\,L^{-1}}\).

Question 3

A monoprotic weak acid has \(K_a = 1.6 \times 10^{-5}\).

a. Calculate the pH of a \(0.0500\ \mathrm{mol\,L^{-1}}\) solution.

b. The solution is diluted until its acid concentration is \(0.00500\ \mathrm{mol\,L^{-1}}\). Calculate the new pH.

c. Explain why decreasing the acid concentration by a factor of 10 does not increase the pH by exactly 1.00.

Solution 3

The pH changes from approximately 3.05 to 3.56, an increase of only about 0.51 because dilution causes a greater fraction of the weak acid to dissociate.

For the original solution, let \(x=[\ce{H3O+}]\).

\[ 1.6\times10^{-5} =\frac{x^2}{0.0500-x} \]

Rearranging,

\[ x^2+(1.6\times10^{-5})x-(8.0\times10^{-7})=0 \]

The positive solution is

\[ x=8.86\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

so

\[ \mathrm{pH} =-\log_{10}(8.86\times10^{-4}) =3.05 \]

After dilution,

\[ 1.6\times10^{-5} =\frac{x^2}{0.00500-x} \]

which gives

\[ x=2.75\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Therefore,

\[ \mathrm{pH} =-\log_{10}(2.75\times10^{-4}) =3.56 \]

The original percentage dissociation is

\[ \frac{8.86\times10^{-4}}{0.0500}\times100 =1.77\% \]

After dilution it becomes

\[ \frac{2.75\times10^{-4}}{0.00500}\times100 =5.50\% \]

A tempting approach is to treat the weak acid as if \([\ce{H3O+}]\) were always equal to its formal concentration. For a completely dissociated monoprotic acid, a tenfold dilution would reduce \([\ce{H3O+}]\) tenfold and raise pH by 1.

This acid does not behave that way. Dilution encourages further dissociation, partially replacing the \(\ce{H3O+}\) lost through dilution. The concentration of \(\ce{H3O+}\) therefore falls by less than a factor of 10.

Question 4

A \(0.100\ \mathrm{mol\,L^{-1}}\) solution of a weak acid \(\ce{HA}\) has \(pK_a=4.00\).

Hydrochloric acid is added, without significant volume change, so that immediately before \(\ce{HA}\) re-establishes equilibrium the hydrochloric acid has supplied \(0.0100\ \mathrm{mol\,L^{-1}}\) of \(\ce{H3O+}\).

Assume hydrochloric acid is completely dissociated.

a. Does the \(K_a\) of \(\ce{HA}\) change?

b. Calculate the additional concentration of \(\ce{HA}\) that dissociates after equilibrium is established.

c. Compare the percentage dissociation of \(\ce{HA}\) with the percentage dissociation it would have in water without the added hydrochloric acid.

Solution 4

The \(K_a\) remains \(1.00\times10^{-4}\), and the added \(\ce{H3O+}\) suppresses the dissociation of \(\ce{HA}\) from about \(3.11\%\) to about \(0.908\%\).

First convert \(pK_a\) into \(K_a\):

\[ K_a=10^{-pK_a}=10^{-4.00}=1.00\times10^{-4} \]

Adding hydrochloric acid does not change \(K_a\) at the same temperature. It changes the equilibrium position.

Let \(x\) be the additional concentration of \(\ce{HA}\) that dissociates.

At equilibrium,

\[ [\ce{HA}]=0.100-x \]

\[ [\mathrm{A}^{-}]=x \]

and

\[ [\ce{H3O+}]=0.0100+x \]

Therefore,

\[ 1.00\times10^{-4} =\frac{(0.0100+x)x}{0.100-x} \]

Rearranging gives

\[ x^2+(0.0101)x-1.00\times10^{-5}=0 \]

The positive solution is

\[ x=9.08\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

So the percentage dissociation is

\[ \frac{9.08\times10^{-4}}{0.100}\times100 =0.908\% \]

Now consider the same \(0.100\ \mathrm{mol\,L^{-1}}\) acid without hydrochloric acid:

\[ 1.00\times10^{-4} =\frac{x^2}{0.100-x} \]

Solving gives

\[ x=3.11\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

so

\[ \%\text{ dissociation} =\frac{3.11\times10^{-3}}{0.100}\times100 =3.11\% \]

It is tempting to think that adding another acid should somehow make \(\ce{HA}\) “more acidic” and therefore make more of it dissociate. The equilibrium does the opposite. The extra \(\ce{H3O+}\) is already a product of

\[ \ce{HA + H2O <=> H3O+ + A-} \]

so increasing \([\ce{H3O+}]\) favours the reverse reaction. Less \(\ce{HA}\) dissociates.

The equilibrium position changes. The acid’s \(K_a\) does not.

Question 5

Two monoprotic weak-acid solutions A and B both have pH 3.00 at the same temperature.

Solution A has an acid concentration of \(0.100\ \mathrm{mol\,L^{-1}}\).

Solution B has an acid concentration of \(0.0100\ \mathrm{mol\,L^{-1}}\).

Assume water’s contribution to \([\ce{H3O+}]\) is negligible.

a. Calculate \(K_a\) and \(pK_a\) for each acid.

b. Identify the stronger acid.

c. Each solution is then diluted by the same factor of 10. Calculate the new pH of each solution.

d. Which diluted solution has the lower pH? Explain why this result does not contradict your answer to part b.

Solution 5

Acid B is intrinsically stronger, but after both solutions are diluted tenfold, solution A has the slightly lower pH: approximately 3.505 compared with 3.549 for B.

Initially, both solutions have pH 3.00, so

\[ [\ce{H3O+}] =10^{-3.00} =1.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

For solution A,

\[ [\mathrm{A}^{-}]=1.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

and

\[ [\ce{HA}] =0.100-0.00100 =0.0990\ \mathrm{mol\,L^{-1}} \]

Therefore,

\[ K_{a,\mathrm A} =\frac{(1.00\times10^{-3})^2}{0.0990} =1.01\times10^{-5} \]

and

\[ pK_{a,\mathrm A} =-\log_{10}(1.01\times10^{-5}) \approx5.00 \]

For solution B,

\[ [\ce{HB}] =0.0100-0.00100 =0.00900\ \mathrm{mol\,L^{-1}} \]

so

\[ K_{a,\mathrm B} =\frac{(1.00\times10^{-3})^2}{0.00900} =1.11\times10^{-4} \]

and

\[ pK_{a,\mathrm B} =-\log_{10}(1.11\times10^{-4}) =3.95 \]

Acid B is therefore stronger because it has the larger \(K_a\) and smaller \(pK_a\).

There is also a useful way to see this before calculating. Both solutions somehow produce the same \([\ce{H3O+}]\), even though solution B contains only one-tenth as much acid. A much larger fraction of B must therefore be dissociating.

Now dilute both solutions tenfold.

For A, the new formal concentration is

\[ 0.0100\ \mathrm{mol\,L^{-1}} \]

and its \(K_a\) remains \(1.01\times10^{-5}\):

\[ 1.01\times10^{-5} =\frac{x^2}{0.0100-x} \]

Solving gives

\[ x=3.13\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

so

\[ \mathrm{pH}_\mathrm A =-\log_{10}(3.13\times10^{-4}) =3.505 \]

For B, the new concentration is

\[ 0.00100\ \mathrm{mol\,L^{-1}} \]

and

\[ 1.11\times10^{-4} =\frac{x^2}{0.00100-x} \]

Solving gives

\[ x=2.82\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

so

\[ \mathrm{pH}_\mathrm B =-\log_{10}(2.82\times10^{-4}) =3.549 \]

Therefore, after dilution, solution A has the lower pH even though acid B is stronger.

That sounds contradictory only if “stronger acid” and “lower-pH solution” are treated as the same statement.

They are not.

After dilution, solution A still contains ten times the formal acid concentration of solution B. Acid B compensates by dissociating much more strongly: about \(28\%\) of B is dissociated, compared with only about \(3.1\%\) of A. But B begins with so much less acid that its absolute \([\ce{H3O+}]\) ends up slightly lower.

The tempting conclusion from the original equal pH values is that the acids must have equal strength. The concentration data destroy that conclusion. The second tempting conclusion is that the stronger acid must always produce the lower pH. That also fails unless concentration and the rest of the solution conditions are controlled.

This separation between acid strength and the acidity of a particular solution is exactly what \(K_a\) and \(pK_a\) are designed to make clear.

The next useful step is to apply the same equilibrium idea when both an acid and its conjugate base are present. That leads directly to buffer behaviour, where the relationship between \(pH\) and \(pK_a\) becomes especially useful. See Buffers in Natural Systems: HSC Chemistry Explained for that next connection.