Ka vs pKa: Which Number Means the Stronger Acid?

Two beakers each contain a \(0.0800\ \mathrm{mol\,L^{-1}}\) solution of a different weak acid. Both are kept at \(25^\circ\mathrm{C}\). A calibrated pH probe gives a reading of 2.75 for acid A and 3.75 for acid B.

Which acid is stronger?

Acid A is stronger. Because the solutions have the same initial concentration and temperature, the lower pH indicates that acid A has produced more hydronium ions through dissociation.

The general rule is straightforward:

• Larger \(K_a\) = stronger acid.
• Smaller \(pK_a\) = stronger acid.

These numbers describe the same acid property, but their numerical directions are opposite. Understanding why will help you compare acids even when a question gives you one \(K_a\) value and one \(pK_a\) value.

01What are we actually measuring?

Before comparing numbers, consider what happens when a weak acid dissolves in water.

A monoprotic acid, represented by \(\ce{HA}\), transfers a proton to water:

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

Some acid molecules dissociate, producing hydronium ions, \(\ce{H3O+}\), and conjugate base ions, \(\mathrm{A}^{-}\). Other molecules remain as undissociated \(\ce{HA}\).

Eventually, the forward and reverse reactions occur at equal rates. The solution has reached dynamic equilibrium.

Imagine comparing two acids at the same starting concentration.

• Acid A dissociates to a greater extent.
• Acid B retains a greater proportion of its original HA molecules.

Acid A produces more hydronium ions under these matched conditions. It is the stronger acid.

But how do we describe that difference mathematically?

The acid dissociation constant, \(K_a\)

The acid dissociation constant measures the position of the equilibrium.

For the reaction above:

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

The square brackets represent equilibrium concentrations.

Water is omitted because it is the solvent, and its concentration is effectively constant.

Notice what happens when more acid dissociates:

• \([\ce{H3O+}]\) increases.
• \([\mathrm{A}^{-}]\) increases.
• \([\ce{HA}]\) decreases.

The resulting \(K_a\) is larger.

Therefore, a larger \(K_a\) indicates that the acid dissociation equilibrium favours the products more strongly.

This gives us our first comparison rule:

When two acids are compared at the same temperature and in the same solvent, the acid with the larger \(K_a\) is stronger.

All concentrations in the calculations below are expressed in \(\mathrm{mol\,L^{-1}}\). We use the concentration-based equilibrium expression expected in HSC calculations. More advanced thermodynamic treatments use activities instead of concentrations.

02What does pKa mean?

Acid dissociation constants are often small numbers.

For example:

\[
K_a=0.000001
\]

Writing and comparing several numbers like this can become inconvenient.

Chemists solve this problem by using a logarithmic scale called \(pK_a\).

It is defined as:

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

The negative sign is important.

It reverses the direction of the scale.

For example:

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

gives:

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

Whereas:

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

gives:

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

The first acid has the larger \(K_a\), so it is stronger. It also has the smaller \(pK_a\).

The relationship is:

\[
\boxed{\text{Larger }K_a
\Longleftrightarrow
\text{Smaller }pK_a}
\]

How much stronger is an acid with a lower pKa?

This is where the logarithmic scale becomes particularly useful.

A difference of one pKa unit represents a tenfold difference in \(K_a\).

A difference of two pKa units represents a hundredfold difference.

Consider the following illustrative data.

These are illustrative values, not measurements of named acids.

Notice that acid A has a \(K_a\) value 100 times greater than acid C.

This means its acid dissociation constant is 100 times larger, even though their pKa values differ by only two.

You can calculate the ratio directly:

\[
\frac{K_{a,A}}{K_{a,B}}
=10^{pK_{a,B}-pK_{a,A}}
\]

For example, if acid A has \(pK_a=3\) and acid B has \(pK_a=5\):

\[
\frac{K_{a,A}}{K_{a,B}}
=10^{5-3}=100
\]

Acid A has a \(K_a\) value 100 times larger.

Don’t interpret this as saying its pH must be 100 times lower. pKa measures acid strength, while pH measures the hydronium concentration in a particular solution.

03How to compare Ka and pKa in an HSC question

Questions don’t always give you the same type of value.

You might receive two \(K_a\) values, two pKa values, or a mixture.

The safest approach is to put both acids on the same scale before comparing them.

Use these rules:

Remember the conversion equations:

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

\[
K_a=10^{-pK_a}
\]

04Connecting Ka to actual acid dissociation

So far, we’ve used \(K_a\) and pKa to compare acid strength.

But what do these values mean for the actual concentrations of particles in a solution?

Consider a weak monoprotic acid with initial concentration \(C\).

Let \(x\) represent the concentration of acid that dissociates.

The equilibrium concentrations can be organised as follows:

We are assuming that the acid is the only significant source of hydronium ions.

Substituting these equilibrium concentrations into the \(K_a\) expression gives:

\[
K_a=\frac{x\times x}{C-x}
\]

Therefore:

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

This equation connects the equilibrium constant to the actual extent of dissociation.

When can we use the small-x approximation?

Sometimes, only a very small fraction of the original acid dissociates.

In this situation:

\[
x\ll C
\]

This means:

\[
C-x\approx C
\]

We can simplify the equilibrium expression:

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

Rearranging:

\[
\boxed{x\approx\sqrt{K_aC}}
\]

This approximation is useful because it allows us to estimate the hydronium concentration without solving a quadratic equation.

However, there is a catch.

You cannot automatically use it just because the acid is weak. A weak acid can still dissociate substantially when its solution is sufficiently dilute.

A common HSC check is the 5% rule:

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

If the percentage is below approximately 5%, the small-x approximation is generally reasonable.

If the percentage is much larger, use the full equilibrium expression.

06The important distinction: Acid strength is not pH

Here is a situation that catches students out.

Suppose you have two solutions:

• Solution S contains a stronger acid with \(pK_a=5.00\).
• Solution W contains a weaker acid with \(pK_a=6.00\).

You might expect solution S to have the lower pH.

But what if solution S is much more dilute?

Let’s compare the following illustrative solutions.

Both calculations use the small-x approximation.

For solution S:

\[
x_S\approx\sqrt{(1.0\times10^{-5})(0.0100)}
\]

\[
x_S=3.16\times10^{-4}\ \mathrm{mol\,L^{-1}}
\]

Its percentage dissociation is approximately 3.16%.

For solution W:

\[
x_W\approx\sqrt{(1.0\times10^{-6})(0.200)}
\]

\[
x_W=4.47\times10^{-4}\ \mathrm{mol\,L^{-1}}
\]

Its percentage dissociation is approximately 0.224%.

Both approximations are reasonable.

The result is surprising.

Solution W contains the weaker acid, but it has the lower pH.

Why?

Because pH measures the hydronium concentration in a particular solution. It does not directly measure the intrinsic strength of the acid.

The more concentrated weak-acid solution produces more hydronium overall, even though a smaller fraction of its acid molecules dissociates.

Keep these quantities separate:

Exam rule: Use \(K_a\) or pKa to compare acid strength. Use pH to compare the hydronium concentrations of solutions.

For a more detailed explanation of the equilibrium calculations and their applications, see Ka, pKa and Acid Strength: HSC Chemistry Guide.

07Questions and solutions

The following questions move from straightforward comparisons to situations where concentration, equilibrium, and approximation limits interact.

Question 1

Two weak monoprotic acids have the following dissociation constants at the same temperature:

\[
K_{a,A}=6.0\times10^{-5}
\]

\[
K_{a,B}=2.0\times10^{-5}
\]

Which acid is stronger, and how many times larger is its \(K_a\)?

Solution 1

Acid A is stronger, and its \(K_a\) is three times larger than that of acid B.

A larger acid dissociation constant indicates that the equilibrium favours dissociation more strongly.

Calculate the ratio:

\[
\frac{K_{a,A}}{K_{a,B}}
=
\frac{6.0\times10^{-5}}
{2.0\times10^{-5}}
\]

\[
=3
\]

Therefore, acid A has a \(K_a\) three times larger than acid B.

The important point is that we compare the numerical magnitudes of the \(K_a\) values, not the exponents in isolation.

Question 2

Acid P has a pKa of 3.65, while acid Q has a dissociation constant of \(1.1\times10^{-5}\).

Determine which acid is stronger and calculate approximately how many times larger its \(K_a\) is.

Solution 2

Acid P is stronger, with a \(K_a\) approximately 20 times larger than that of acid Q.

The information is given in two different forms, so we must convert one value before comparing.

First, calculate the dissociation constant of acid P:

\[
K_{a,P}=10^{-3.65}
\]

\[
K_{a,P}=2.24\times10^{-4}
\]

Acid Q already has its \(K_a\) value:

\[
K_{a,Q}=1.1\times10^{-5}
\]

Now compare:

\[
\frac{K_{a,P}}{K_{a,Q}}
=
\frac{2.24\times10^{-4}}
{1.1\times10^{-5}}
\]

\[
=20.4
\]

Therefore, acid P has a \(K_a\) approximately 20.4 times larger.

We can also check using pKa.

\[
pK_{a,Q}=-\log_{10}(1.1\times10^{-5})
\]

\[
pK_{a,Q}=4.96
\]

Acid P has pKa 3.65, while acid Q has pKa 4.96.

The smaller pKa confirms that acid P is stronger.

Question 3

A \(0.00200\ \mathrm{mol\,L^{-1}}\) solution of a weak monoprotic acid has:

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

A student estimates the hydronium concentration using:

\[
[\ce{H3O+}]\approx\sqrt{K_aC}
\]

Determine whether this approximation is valid. Calculate the actual equilibrium hydronium concentration and pH using the full equilibrium expression.

Solution 3

The small-x approximation is not sufficiently accurate here. The correct hydronium concentration is \(4.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\), giving a pH of approximately 3.40.

Start with the full equilibrium expression:

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

Substitute the supplied values:

\[
1.00\times10^{-4}
=
\frac{x^2}{0.00200-x}
\]

Rearrange:

\[
x^2=2.00\times10^{-7}
-1.00\times10^{-4}x
\]

Therefore:

\[
x^2+1.00\times10^{-4}x
-2.00\times10^{-7}=0
\]

Using the quadratic formula and taking the positive root:

\[
x=4.00\times10^{-4}\ \mathrm{mol\,L^{-1}}
\]

This is the equilibrium hydronium concentration.

Now calculate the percentage dissociation:

\[
\%\text{ dissociation}
=
\frac{4.00\times10^{-4}}{0.00200}
\times100
\]

\[
=20.0\%
\]

The approximation is not justified because 20% of the acid has dissociated.

Finally, calculate the pH:

\[
\mathrm{pH}
=-\log_{10}(4.00\times10^{-4})
\]

\[
\boxed{\mathrm{pH}=3.40}
\]

The tempting shortcut would give:

\[
x\approx\sqrt{2.00\times10^{-7}}
\]

\[
x\approx4.47\times10^{-4}
\]

This is noticeably higher than the correct answer.

The mistake is assuming \(C-x\approx C\) when \(x\) is actually a substantial fraction of the original concentration.

Question 4

Two monoprotic weak acids are tested at the same temperature.

The following measurements are obtained:

Assume both solutions have negligible water autoionisation and satisfy the small-x approximation.

(a) Which acid is stronger?

(b) Which solution has the greater hydronium concentration?

(c) Estimate the ratio of the initial concentrations of the two acid solutions, \(C_X/C_Y\).

Solution 4

Acid Y is stronger, but solution X has the greater hydronium concentration. Solution X is approximately 25 times more concentrated than solution Y.

(a) Compare acid strength.

Acid X has:

\[
pK_{a,X}=6.00
\]

Acid Y has:

\[
pK_{a,Y}=5.00
\]

Since a smaller pKa indicates a stronger acid, acid Y is stronger.

Its \(K_a\) is ten times larger.

(b) Compare hydronium concentrations.

Use:

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

For solution X:

\[
[\ce{H3O+}]_X=10^{-3.40}
\]

\[
=3.98\times10^{-4}\ \mathrm{mol\,L^{-1}}
\]

For solution Y:

\[
[\ce{H3O+}]_Y=10^{-3.60}
\]

\[
=2.51\times10^{-4}\ \mathrm{mol\,L^{-1}}
\]

Solution X has the greater hydronium concentration.

This is not a contradiction. Acid strength and solution pH describe different things.

(c) Calculate the initial concentration ratio.

From the small-x approximation:

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

Rearranging:

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

First, convert the pKa values into \(K_a\).

For X:

\[
K_{a,X}=10^{-6}
\]

For Y:

\[
K_{a,Y}=10^{-5}
\]

Now calculate the initial concentration of each solution.

For X:

\[
C_X\approx
\frac{(3.98\times10^{-4})^2}
{1.00\times10^{-6}}
\]

\[
C_X\approx0.158\ \mathrm{mol\,L^{-1}}
\]

For Y:

\[
C_Y\approx
\frac{(2.51\times10^{-4})^2}
{1.00\times10^{-5}}
\]

\[
C_Y\approx0.00631\ \mathrm{mol\,L^{-1}}
\]

Calculate the ratio:

\[
\frac{C_X}{C_Y}
=
\frac{0.158}{0.00631}
\]

\[
\boxed{\frac{C_X}{C_Y}\approx25.1}
\]

The initial concentration of solution X is approximately 25 times greater than that of solution Y.

We can check the small-x assumption:

• X: approximately 0.251% dissociation.
• Y: approximately 3.98% dissociation.

Both are below 5%, so the approximation is reasonable.

The key lesson is that solution X has a higher hydronium concentration because it is much more concentrated, not because its acid is stronger.

Question 5

Two separate 100.0 mL samples contain different monoprotic weak acids, A and B.

Both samples are at \(25^\circ\mathrm{C}\), and both initially have a measured pH of 3.00.

Their initial analytical concentrations are:

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

\[
C_B=0.100\ \mathrm{mol\,L^{-1}}
\]

Assume that each sample initially contains only its respective weak acid, with no added conjugate base.

Both samples are then diluted separately to ten times their original volumes.

Ignore water autoionisation and activity effects.

(a) Calculate \(K_a\) and pKa for each acid. Determine which acid is stronger and calculate the ratio of their dissociation constants.

(b) After the tenfold dilution, calculate the new hydronium concentration and pH of each solution. Use the full equilibrium expression where necessary.

(c) A student uses the small-x approximation for both diluted solutions and concludes that acid A will produce more hydronium ions than acid B. Explain why this prediction is incorrect.

Solution 5

Acid A is approximately 11 times stronger than acid B. However, after dilution, solution B has a slightly greater hydronium concentration and a slightly lower pH. The difference arises because the small-x approximation is no longer valid for diluted acid A.

This question combines acid strength, pH, equilibrium calculations, dilution, and the limitations of approximations.

(a) Calculate the dissociation constants.

Both solutions initially have:

\[
\mathrm{pH}=3.00
\]

Therefore:

\[
[\ce{H3O+}]=10^{-3}
\]

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

Because the acid is monoprotic, the amount of conjugate base formed is also \(x\).

Use:

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

For acid A:

\[
K_{a,A}
=
\frac{(1.00\times10^{-3})^2}
{0.0100-0.00100}
\]

\[
K_{a,A}
=
\frac{1.00\times10^{-6}}{0.00900}
\]

\[
\boxed{K_{a,A}=1.11\times10^{-4}}
\]

For acid B:

\[
K_{a,B}
=
\frac{(1.00\times10^{-3})^2}
{0.100-0.00100}
\]

\[
K_{a,B}
=
\frac{1.00\times10^{-6}}{0.0990}
\]

\[
\boxed{K_{a,B}=1.01\times10^{-5}}
\]

Now calculate their pKa values.

For acid A:

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

\[
pK_{a,A}=3.95
\]

For acid B:

\[
pK_{a,B}
=-\log_{10}(1.01\times10^{-5})
\]

\[
pK_{a,B}=5.00
\]

Acid A has the larger \(K_a\) and smaller pKa.

Calculate the ratio:

\[
\frac{K_{a,A}}{K_{a,B}}
=
\frac{1.11\times10^{-4}}
{1.01\times10^{-5}}
\]

\[
\boxed{\frac{K_{a,A}}{K_{a,B}}=11.0}
\]

Acid A is therefore approximately 11 times stronger according to its dissociation constant.

Notice that both solutions initially have exactly the same pH, despite having different acid strengths.

The reason is that acid B starts with ten times the analytical acid concentration.

(b) Calculate the pH after dilution.

Each solution is diluted tenfold.

The new analytical concentrations are:

For acid A:

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

For acid B:

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

An important point: dilution does not change \(K_a\) at constant temperature.

The equilibrium concentrations change, but the dissociation constant remains the same.

We therefore use the original \(K_a\) values.

For acid A:

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

Rearranging:

\[
x^2+1.11\times10^{-4}x
-1.11\times10^{-7}=0
\]

Using the quadratic formula and taking the positive root:

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

Therefore:

\[
[\ce{H3O+}]_A
=2.82\times10^{-4}
\ \mathrm{mol\,L^{-1}}
\]

Calculate pH:

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

\[
\boxed{\mathrm{pH}_A=3.55}
\]

Now calculate acid B.

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

Rearranging:

\[
x^2+1.01\times10^{-5}x
-1.01\times10^{-7}=0
\]

Solving using the quadratic formula:

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

Therefore:

\[
[\ce{H3O+}]_B
=3.13\times10^{-4}
\ \mathrm{mol\,L^{-1}}
\]

Calculate pH:

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

\[
\boxed{\mathrm{pH}_B=3.50}
\]

Compare the results:

Solution B now has the slightly lower pH, even though acid A is stronger.

(c) Explain why the small-x prediction is incorrect.

The tempting approach is to use:

\[
x\approx\sqrt{K_aC}
\]

For acid A, this gives:

\[
x_A\approx
\sqrt{(1.11\times10^{-4})(0.00100)}
\]

\[
x_A\approx3.33\times10^{-4}
\]

For acid B:

\[
x_B\approx
\sqrt{(1.01\times10^{-5})(0.0100)}
\]

\[
x_B\approx3.18\times10^{-4}
\]

This approximation predicts that acid A produces slightly more hydronium.

However, check the percentage dissociation.

For acid A, the approximate percentage is:

\[
\frac{3.33\times10^{-4}}{0.00100}
\times100
=33.3\%
\]

That is far too large for the small-x approximation.

For acid B:

\[
\frac{3.18\times10^{-4}}{0.0100}
\times100
=3.18\%
\]

The approximation is reasonable for B, but not A.

The full quadratic calculation gives actual dissociation percentages of approximately:

• Acid A: 28.2%.
• Acid B: 3.13%.

The original acid concentration in solution A is now sufficiently low that a substantial fraction dissociates. We cannot treat \(C-x\) as approximately equal to \(C\).

The denominator in the full equilibrium expression matters:

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

Ignoring the decrease in undissociated acid gives the wrong comparison of hydronium concentrations.

The final conclusion: Acid A remains the stronger acid because its \(K_a\) is larger and its pKa is smaller. However, solution B has the lower pH after dilution because its final hydronium concentration is slightly greater.

This is why acid strength and solution pH must always be treated as separate quantities.

08What to remember in the HSC exam

When comparing \(K_a\) and pKa, keep these relationships in mind:

There is one more useful connection.

If you are given a weak-acid titration curve rather than a table of \(K_a\) or pKa values, you can determine pKa from the half-equivalence point, where the concentrations of the weak acid and its conjugate base are approximately equal.

You can learn that method in How to Find pKa from a Weak-Acid Titration Curve.

The next time an HSC question gives you a pKa value, don’t just memorise that smaller means stronger. Connect it to the equilibrium: a lower pKa corresponds to a larger \(K_a\), which means the acid dissociation equilibrium favours the formation of ions more strongly.