pH, pOH and Ion Concentrations for HSC Chemistry

A solution has pH 3. Another has pH 4. The first one has a lower pH, but how much more \(\mathrm{H}^{+}\) does it actually contain? Twice as much? Three times?

It contains ten times the hydrogen ion concentration.

That is the main idea behind pH: the scale compresses enormous changes in ion concentration into small changes in numbers. Once you understand that logarithmic relationship, pH, pOH, \([\mathrm{H}^{+}]\), and \([\ce{OH-}]\) stop looking like four separate things. They become four ways of describing the same aqueous solution.

01Start with the ions already present in water

Even pure water contains a tiny amount of ions. Water molecules can transfer a proton between themselves:

\[
\ce{2H2O(l) <=> H3O+(aq) + OH-(aq)}
\]

The \(\ce{H3O+}\) ion is the hydronium ion. In HSC calculations, it is common to use \(\mathrm{H}^{+}\) as shorthand, so we will write \([\mathrm{H}^{+}]\) for the hydrogen ion concentration.

At \(25^\circ\text{C}\), the concentrations are related by the ionic product of water:

\[
K_w=[\mathrm{H}^{+}][\ce{OH-}]=1.0\times10^{-14}
\]

Here, \([\mathrm{H}^{+}]\) and \([\ce{OH-}]\) are measured in \(\text{mol L}^{-1}\).

This gives us an important constraint. If one ion concentration rises, the other must fall so that their product remains \(1.0\times10^{-14}\) at \(25^\circ\text{C}\).

For example, suppose

\[
[\mathrm{H}^{+}]=1.0\times10^{-3}\text{ mol L}^{-1}.
\]

Predict \([\ce{OH-}]\) before calculating it. Should it also be \(10^{-3}\)?

It cannot be. Their product would then be \(10^{-6}\), nowhere near \(10^{-14}\).

Instead,

\[
[\ce{OH-}]
=\frac{K_w}{[\mathrm{H}^{+}]}
=\frac{1.0\times10^{-14}}{1.0\times10^{-3}}
=1.0\times10^{-11}\text{ mol L}^{-1}.
\]

A high \([\mathrm{H}^{+}]\) therefore comes with a low \([\ce{OH-}]\), and vice versa.

02What pH actually measures

Writing concentrations such as \(0.0000010\text{ mol L}^{-1}\) becomes awkward very quickly. pH is a way of compressing those powers of ten.

The definition is

\[
\text{pH}=-\log_{10}[\mathrm{H}^{+}].
\]

If

\[
[\mathrm{H}^{+}]=1.0\times10^{-6}\text{ mol L}^{-1},
\]

then

\[
\text{pH}
=-\log_{10}(1.0\times10^{-6})
=6.00.
\]

The minus sign is important. Without it, common hydrogen ion concentrations would produce negative numbers.

You can think of pH as a volume knob marked in powers of ten rather than equal-sized concentration steps. Moving the knob from pH 5 to pH 4 increases \([\mathrm{H}^{+}]\) by a factor of 10. Moving from pH 5 to pH 3 increases it by a factor of 100.

The analogy has a limit: pH is not simply a general “amount of acid” knob. It specifically describes hydrogen ion concentration. Two acids with the same formal acid concentration can have different pH values because they may ionise to different extents.

03pOH works the same way

pOH measures hydroxide ion concentration:

\[
\text{pOH}=-\log_{10}[\ce{OH-}].
\]

The reverse equations are just as important:

\[
[\mathrm{H}^{+}]=10^{-\text{pH}}
\]

and

\[
[\ce{OH-}]=10^{-\text{pOH}}.
\]

These are worth understanding rather than memorising as unrelated formulas.

If pH is the negative logarithm of \([\mathrm{H}^{+}]\), reversing the logarithm must return the original concentration.

So:

• concentration \(\rightarrow\) pH or pOH: use \(-\log_{10}\)
• pH or pOH \(\rightarrow\) concentration: use \(10^{-x}\)

That decision rule prevents a surprising number of calculator errors.

04Why pH and pOH add to 14

At \(25^\circ\text{C}\),

\[
[\mathrm{H}^{+}][\ce{OH-}]=1.0\times10^{-14}.
\]

Take the negative logarithm of both sides:

\[
-\log_{10}\left([\mathrm{H}^{+}][\ce{OH-}]\right)
=-\log_{10}(1.0\times10^{-14}).
\]

Using the logarithm rule \(\log(ab)=\log a+\log b\),

\[
-\log_{10}[\mathrm{H}^{+}]
-\log_{10}[\ce{OH-}]
=14.00.
\]

Those two terms are pH and pOH, so

\[
\text{pH}+\text{pOH}=14.00
\]

at \(25^\circ\text{C}\).

This gives you two possible routes between the ions.

If you know pH, you could calculate \([\mathrm{H}^{+}]\), use \(K_w\) to find \([\ce{OH-}]\), and then calculate pOH.

Usually, that is unnecessarily long. At \(25^\circ\text{C}\), simply use

\[
\text{pOH}=14.00-\text{pH}.
\]

Then find \([\ce{OH-}]\) if the question actually asks for it.

05A compact calculation map

A useful habit is to identify what kind of quantity you have before touching the calculator. A concentration and a pH are not interchangeable numbers.

06Acidic, basic, and neutral solutions

The most fundamental comparison is not actually pH versus 7. It is \([\mathrm{H}^{+}]\) versus \([\ce{OH-}]\).

A solution is:

• acidic when \([\mathrm{H}^{+}]>[\ce{OH-}]\)
• neutral when \([\mathrm{H}^{+}]=[\ce{OH-}]\)
• basic when \([\mathrm{H}^{+}]<[\ce{OH-}]\)

At \(25^\circ\text{C}\), neutral water has

\[
[\mathrm{H}^{+}]=[\ce{OH-}].
\]

Because their product is \(1.0\times10^{-14}\),

\[
[\mathrm{H}^{+}]=[\ce{OH-}]=1.0\times10^{-7}\text{ mol L}^{-1}.
\]

Therefore,

\[
\text{pH}=7.00.
\]

That is why, at \(25^\circ\text{C}\), we normally use:

There is an important condition attached to this table: \(25^\circ\text{C}\). We will return to that shortly.

07A one-unit pH change is a tenfold concentration change

Suppose solution A has pH \(2.0\), while solution B has pH \(5.0\).

The difference is only three pH units. It can look small on a scale marked from roughly 0 to 14.

But compare their hydrogen ion concentrations:

\[
[\mathrm{H}^{+}]_{\text{A}}=10^{-2.0}=1.0\times10^{-2}\text{ mol L}^{-1}
\]

and

\[
[\mathrm{H}^{+}]_{\text{B}}=10^{-5.0}=1.0\times10^{-5}\text{ mol L}^{-1}.
\]

The ratio is

\[
\frac{1.0\times10^{-2}}{1.0\times10^{-5}}
=10^3
=1000.
\]

Solution A has 1000 times the hydrogen ion concentration of solution B.

More generally, if two solutions differ by \(\Delta\text{pH}\),

\[
\frac{[\mathrm{H}^{+}]_{\text{lower pH}}}
{[\mathrm{H}^{+}]_{\text{higher pH}}}
=10^{\Delta\text{pH}}.
\]

So:

• a difference of \(1\) pH unit means a factor of \(10\)
• a difference of \(2\) means a factor of \(100\)
• a difference of \(0.30\) means a factor of about \(2\)

That last result is useful because pH differences do not need to be whole numbers.

08Do not average pH values

Imagine equal volumes of two solutions with pH \(2.00\) and pH \(4.00\).

A tempting prediction is that their mixture should have pH

\[
\frac{2.00+4.00}{2}=3.00.
\]

That feels reasonable because we often average measurements directly. But pH is already a logarithm. What physically mixes are the ions and their amounts, not the pH numbers printed on the bottles.

Ignoring any additional acid-base chemistry, the original hydrogen ion concentrations are

\[
10^{-2.00}=1.00\times10^{-2}\text{ mol L}^{-1}
\]

and

\[
10^{-4.00}=1.00\times10^{-4}\text{ mol L}^{-1}.
\]

For equal volumes, the concentration after mixing is their concentration average:

\[
[\mathrm{H}^{+}]
=\frac{1.00\times10^{-2}+1.00\times10^{-4}}{2}
=5.05\times10^{-3}\text{ mol L}^{-1}.
\]

Therefore,

\[
\text{pH}
=-\log_{10}(5.05\times10^{-3})
=2.30.
\]

Not \(3.00\).

The lower-pH solution dominates because its hydrogen ion concentration is 100 times larger before mixing.

09pH does not directly tell you the original acid concentration

Suppose a solution has pH \(3.00\). You know immediately that

\[
[\mathrm{H}^{+}]=1.0\times10^{-3}\text{ mol L}^{-1}.
\]

Can you now say the acid concentration must also be \(1.0\times10^{-3}\text{ mol L}^{-1}\)?

Not without more information.

That would work for a strong monoprotic acid under the usual dilute-solution assumptions because each mole of acid supplies approximately one mole of \(\mathrm{H}^{+}\).

A weak acid only partially ionises. Its equilibrium acid concentration can therefore be much larger than its \([\mathrm{H}^{+}]\).

Likewise, acids capable of donating more than one proton need extra care.

So keep these two ideas separate:

• pH tells you \([\mathrm{H}^{+}]\)
• finding the original acid concentration may require acid ionisation or stoichiometry as well

That distinction becomes important when you move into weak-acid equilibria.

10Neutral does not fundamentally mean pH 7

At \(25^\circ\text{C}\),

\[
K_w=1.0\times10^{-14}
\]

and neutral water has pH \(7.00\).

But water ionisation changes with temperature. At another temperature, \(K_w\) can have a different value.

Neutrality still means

\[
[\mathrm{H}^{+}]=[\ce{OH-}],
\]

but the equal concentrations no longer have to be \(1.0\times10^{-7}\text{ mol L}^{-1}\).

For neutral water,

\[
[\mathrm{H}^{+}]^2=K_w.
\]

Therefore,

\[
[\mathrm{H}^{+}]=\sqrt{K_w}.
\]

If, at some temperature,

\[
K_w=4.0\times10^{-13},
\]

then neutral water has

\[
\begin{aligned}
[\mathrm{H}^{+}]
&=\sqrt{4.0\times10^{-13}}\\
&=6.3\times10^{-7}\text{ mol L}^{-1}.
\end{aligned}
\]

Its neutral pH is therefore

\[
\text{pH}
=-\log_{10}(6.3\times10^{-7})
=6.20.
\]

So a neutral solution can have pH below 7.

Nothing has gone wrong. Both \([\mathrm{H}^{+}]\) and \([\ce{OH-}]\) have increased equally.

The statement “pH below 7 means acidic” is therefore a convenient \(25^\circ\text{C}\) rule, not the deepest definition of acidity.

In the same way,

\[
\text{pH}+\text{pOH}=14.00
\]

is specifically the \(25^\circ\text{C}\) version of the more general relationship

\[
\text{pH}+\text{pOH}=\text{p}K_w,
\]

where

\[
\text{p}K_w=-\log_{10}K_w.
\]

11Common mistakes worth catching early

Treating pH as a linear scale

pH 2 is not “twice as acidic” as pH 4 in terms of hydrogen ion concentration.

The difference is two pH units, so the ratio is

\[
10^2=100.
\]

The pH 2 solution has 100 times the \([\mathrm{H}^{+}]\).

Using \(14-\text{pH}\) without checking the temperature

At \(25^\circ\text{C}\), this is correct:

\[
\text{pOH}=14.00-\text{pH}.
\]

If a question gives a different \(K_w\), use that information instead of automatically inserting 14.

Pressing the calculator buttons in the wrong direction

If you are given a concentration, use a logarithm.

If you are given pH or pOH, undo the logarithm using a power of ten.

For example:

\[
[\mathrm{H}^{+}]=10^{-4.27},
\]

not

\[
-\log_{10}(4.27).
\]

Thinking dilution can make an acid basic

Diluting an acid with pure water reduces its excess hydrogen ion concentration and moves it towards neutral.

It does not keep marching through neutral into the basic region just because a simple concentration calculation produces an absurdly small \([\mathrm{H}^{+}]\). Close to neutrality, water’s own ionisation can no longer be ignored.

This boundary case becomes important in Question 4 below.

Assuming the usual pH scale is an absolute physical boundary

The familiar 0-14 scale is extremely useful for ordinary dilute aqueous solutions at about \(25^\circ\text{C}\). It is not a mathematical rule saying pH can never fall below 0 or rise above 14.

For HSC work, use the conditions and model given in the question rather than treating 0 and 14 as unbreakable walls.

12Calculated pH and measured pH are different ideas

A calculation predicts pH from a chemical model and given concentrations. An experiment measures pH from a real solution.

Those two values can differ if the chemical assumptions are incomplete, concentrations are inaccurate, temperature matters, or experimental equipment has limitations.

If you are working with indicators, probes, calibration, or experimental reliability, see Measuring pH of Acids and Bases: HSC Chemistry Guide.

13Questions and solutions

Question 1

At \(25^\circ\text{C}\), a solution has

\[
[\mathrm{H}^{+}]=2.5\times10^{-3}\text{ mol L}^{-1}.
\]

Calculate:

a. the pH
b. the pOH
c. \([\ce{OH-}]\)
d. whether the solution is acidic, basic, or neutral.

Solution 1

The solution has pH \(2.60\), pOH \(11.40\), \([\ce{OH-}]=4.0\times10^{-12}\text{ mol L}^{-1}\), and is acidic.

For part a, use

\[
\text{pH}=-\log_{10}[\mathrm{H}^{+}].
\]

Substituting,

\[
\begin{aligned}
\text{pH}
&=-\log_{10}(2.5\times10^{-3})\\
&=2.60.
\end{aligned}
\]

For part b, at \(25^\circ\text{C}\),

\[
\begin{aligned}
\text{pOH}
&=14.00-\text{pH}\\
&=14.00-2.60\\
&=11.40.
\end{aligned}
\]

For part c,

\[
\begin{aligned}
[\ce{OH-}]
&=10^{-11.40}\\
&=4.0\times10^{-12}\text{ mol L}^{-1}.
\end{aligned}
\]

You can check this using \(K_w\):

\[
\begin{aligned}
[\ce{OH-}]
&=\frac{1.0\times10^{-14}}{2.5\times10^{-3}}\\
&=4.0\times10^{-12}\text{ mol L}^{-1}.
\end{aligned}
\]

For part d,

\[
[\mathrm{H}^{+}]>[\ce{OH-}],
\]

so the solution is acidic.

The common trap is to treat \(2.5\times10^{-3}\) itself as something that can be compared directly with the pH scale. Concentration must first be converted using the logarithm.

Question 2

Two acidic solutions at \(25^\circ\text{C}\) have pH values of \(3.40\) and \(5.90\).

a. How many times greater is \([\mathrm{H}^{+}]\) in the pH \(3.40\) solution?

b. Equal volumes of the two solutions are mixed. Assuming no additional chemical reaction occurs, calculate the approximate pH of the mixture.

Solution 2

The pH \(3.40\) solution has about \(316\) times the hydrogen ion concentration, and the equal-volume mixture has pH approximately \(3.70\).

For part a, the pH difference is

\[
5.90-3.40=2.50.
\]

A pH difference of \(2.50\) corresponds to a concentration ratio of

\[
10^{2.50}=316.
\]

So the lower-pH solution has about \(316\) times the \([\mathrm{H}^{+}]\).

For part b, first convert each pH to concentration:

\[
\begin{aligned}
[\mathrm{H}^{+}]_1
&=10^{-3.40}\\
&=3.98\times10^{-4}\text{ mol L}^{-1},
\end{aligned}
\]

and

\[
\begin{aligned}
[\mathrm{H}^{+}]_2
&=10^{-5.90}\\
&=1.26\times10^{-6}\text{ mol L}^{-1}.
\end{aligned}
\]

Because equal volumes are mixed, the final concentration is the average of the two concentrations:

\[
\begin{aligned}
[\mathrm{H}^{+}]_{\text{final}}
&=\frac{3.98\times10^{-4}+1.26\times10^{-6}}{2}\\
&=2.00\times10^{-4}\text{ mol L}^{-1}.
\end{aligned}
\]

Therefore,

\[
\begin{aligned}
\text{pH}
&=-\log_{10}(2.00\times10^{-4})\\
&=3.70.
\end{aligned}
\]

The tempting route is to average the two pH values:

\[
\frac{3.40+5.90}{2}=4.65.
\]

That fails because pH is logarithmic. The concentrations mix, not the logarithms of those concentrations.

Question 3

A \(25.0\text{ mL}\) sample of a strong monoprotic acid has pH \(1.70\) at \(25^\circ\text{C}\). It is diluted to a total volume of \(400.0\text{ mL}\).

Assuming complete ionisation and that water’s contribution to \([\mathrm{H}^{+}]\) is negligible, calculate the final pH.

Solution 3

The final pH is approximately \(2.90\).

First convert the initial pH into hydrogen ion concentration:

\[
\begin{aligned}
[\mathrm{H}^{+}]_{\text{initial}}
&=10^{-1.70}\\
&=1.995\times10^{-2}\text{ mol L}^{-1}.
\end{aligned}
\]

Now calculate the initial amount of \(\mathrm{H}^{+}\):

\[
\begin{aligned}
n(\mathrm{H}^{+})
&=cV\\
&=(1.995\times10^{-2})(0.0250)\\
&=4.99\times10^{-4}\text{ mol}.
\end{aligned}
\]

Dilution does not change this amount. The final volume is \(0.4000\text{ L}\), so

\[
\begin{aligned}
[\mathrm{H}^{+}]_{\text{final}}
&=\frac{4.99\times10^{-4}}{0.4000}\\
&=1.25\times10^{-3}\text{ mol L}^{-1}.
\end{aligned}
\]

Therefore,

\[
\begin{aligned}
\text{pH}_{\text{final}}
&=-\log_{10}(1.25\times10^{-3})\\
&=2.90.
\end{aligned}
\]

The pH has risen from \(1.70\) to \(2.90\), so the solution has become less acidic.

A tempting shortcut is to multiply the pH itself by the dilution factor. That has no physical basis. Dilution changes concentration first, and pH must then be recalculated from that concentration.

There is also a useful check. The volume increased by a factor of

\[
\frac{400.0}{25.0}=16.
\]

So \([\mathrm{H}^{+}]\) fell by a factor of 16. Because

\[
\log_{10}(16)\approx1.20,
\]

the pH should rise by about \(1.20\):

\[
1.70+1.20=2.90.
\]

The two methods agree.

Question 4

At \(25^\circ\text{C}\), two strong acid solutions have initial pH values:

• solution A: pH \(2.00\)
• solution B: pH \(6.00\)

Each solution is separately diluted by a factor of \(1000\) using pure water.

A student argues:

“A 1000-fold dilution lowers \([\mathrm{H}^{+}]\) by \(1000\), so each pH increases by 3. Solution A becomes pH 5.00 and solution B becomes pH 9.00.”

Evaluate the student’s prediction. For solution B, account for the ionisation of water.

Solution 4

The student’s prediction is a good approximation for solution A but physically wrong for solution B. Solution A becomes approximately pH \(5.00\), while solution B approaches pH \(7\) from the acidic side rather than becoming pH \(9.00\).

For solution A,

\[
[\mathrm{H}^{+}]_{\text{initial}}=10^{-2}=1.0\times10^{-2}\text{ mol L}^{-1}.
\]

A 1000-fold dilution gives an acid-derived concentration of approximately

\[
\frac{1.0\times10^{-2}}{1000}
=1.0\times10^{-5}\text{ mol L}^{-1}.
\]

This is much larger than the roughly \(10^{-7}\text{ mol L}^{-1}\) hydrogen ion concentration associated with water itself.

So ignoring water ionisation is reasonable:

\[
\text{pH}\approx5.00.
\]

For solution B,

\[
[\mathrm{H}^{+}]_{\text{initial}}=10^{-6}\text{ mol L}^{-1}.
\]

The student’s calculation would give an acid-derived concentration after dilution of

\[
C=1.0\times10^{-9}\text{ mol L}^{-1}.
\]

If we blindly converted that into pH, we would obtain pH \(9.00\). That would mean diluting an acid with neutral water had made it basic.

That is the warning sign.

Near neutrality, water’s own ionisation can no longer be neglected.

Let the total equilibrium hydrogen ion concentration be \(x\). The diluted strong acid supplies an excess positive-ion concentration of

\[
C=1.0\times10^{-9}\text{ mol L}^{-1}.
\]

Charge balance gives

\[
[\mathrm{H}^{+}]=C+[\ce{OH-}].
\]

Since

\[
[\mathrm{H}^{+}][\ce{OH-}]=K_w,
\]

we have

\[
x=C+\frac{K_w}{x}.
\]

Multiplying by \(x\),

\[
x^2-Cx-K_w=0.
\]

Substituting \(C=1.0\times10^{-9}\) and \(K_w=1.0\times10^{-14}\),

\[
x^2-(1.0\times10^{-9})x-1.0\times10^{-14}=0.
\]

Using the positive solution of the quadratic,

\[
\begin{aligned}
x
&=\frac{1.0\times10^{-9}
+\sqrt{(1.0\times10^{-9})^2+4(1.0\times10^{-14})}}{2}\\
&\approx1.005\times10^{-7}\text{ mol L}^{-1}.
\end{aligned}
\]

Therefore,

\[
\begin{aligned}
\text{pH}
&=-\log_{10}(1.005\times10^{-7})\\
&\approx6.998.
\end{aligned}
\]

The solution is still very slightly acidic.

The tempting method fails because it assumes all hydrogen ions come from the acid. That is fine when the acid contribution is far larger than \(10^{-7}\text{ mol L}^{-1}\), but it breaks down near neutrality.

The deeper constraint is physical: adding neutral water to an acidic solution can move it towards neutrality, but dilution alone cannot carry it through neutrality and make it basic.

Question 5

Two aqueous solutions are measured under different temperature conditions.

Solution X is at \(25^\circ\text{C}\) and has pH \(6.80\).

Solution Y is at a higher temperature where

\[
K_w=4.0\times10^{-13},
\]

and has pH \(6.40\).

A student makes two claims:

1. “Solution Y has the lower pH, so it must be more acidic than solution X.”
2. “Both pH values are below 7, so both solutions are acidic.”

For each solution, calculate \([\mathrm{H}^{+}]\) and the relevant \([\ce{OH-}]\), classify it as acidic, basic, or neutral, and evaluate both claims.

Solution 5

Solution X is acidic, but solution Y is basic even though Y has both the lower pH and the greater hydrogen ion concentration. Both student claims are therefore incorrect.

For solution X,

\[
\begin{aligned}
[\mathrm{H}^{+}]_{\text{X}}
&=10^{-6.80}\\
&=1.58\times10^{-7}\text{ mol L}^{-1}.
\end{aligned}
\]

At \(25^\circ\text{C}\),

\[
K_w=1.0\times10^{-14}.
\]

Therefore,

\[
\begin{aligned}
[\ce{OH-}]_{\text{X}}
&=\frac{1.0\times10^{-14}}{1.58\times10^{-7}}\\
&=6.33\times10^{-8}\text{ mol L}^{-1}.
\end{aligned}
\]

Since

\[
1.58\times10^{-7}>6.33\times10^{-8},
\]

solution X is acidic.

Now consider solution Y.

Its pH is \(6.40\), so

\[
\begin{aligned}
[\mathrm{H}^{+}]_{\text{Y}}
&=10^{-6.40}\\
&=3.98\times10^{-7}\text{ mol L}^{-1}.
\end{aligned}
\]

At this temperature,

\[
K_w=4.0\times10^{-13}.
\]

Therefore,

\[
\begin{aligned}
[\ce{OH-}]_{\text{Y}}
&=\frac{4.0\times10^{-13}}{3.98\times10^{-7}}\\
&=1.01\times10^{-6}\text{ mol L}^{-1}.
\end{aligned}
\]

So

\[
[\ce{OH-}]_{\text{Y}}>[\mathrm{H}^{+}]_{\text{Y}},
\]

which means solution Y is basic.

We can check this another way by finding the neutral pH at Y’s temperature.

For a neutral solution,

\[
[\mathrm{H}^{+}]=[\ce{OH-}]=\sqrt{K_w}.
\]

Therefore,

\[
\begin{aligned}
[\mathrm{H}^{+}]_{\text{neutral}}
&=\sqrt{4.0\times10^{-13}}\\
&=6.32\times10^{-7}\text{ mol L}^{-1}.
\end{aligned}
\]

The neutral pH is

\[
\begin{aligned}
\text{pH}_{\text{neutral}}
&=-\log_{10}(6.32\times10^{-7})\\
&=6.20.
\end{aligned}
\]

Solution Y has pH \(6.40\), which is above the neutral pH of \(6.20\) at that temperature. It is therefore basic.

The first claim fails because comparing pH values tells us which solution has the larger \([\mathrm{H}^{+}]\), but acidity versus basicity is determined by the balance between \([\mathrm{H}^{+}]\) and \([\ce{OH-}]\) under the solution’s conditions. Y really does have more \(\mathrm{H}^{+}\) than X:

\[
3.98\times10^{-7}>1.58\times10^{-7},
\]

yet Y has even more \(\ce{OH-}\).

The second claim fails because “pH below 7 means acidic” assumes \(25^\circ\text{C}\). The more general definition is:

\[
\text{acidic: }[\mathrm{H}^{+}]>[\ce{OH-}]
\]

\[
\text{neutral: }[\mathrm{H}^{+}]=[\ce{OH-}]
\]

\[
\text{basic: }[\mathrm{H}^{+}]<[\ce{OH-}].
\]

The subtle point is that pH measures hydrogen ion concentration. It does not, by itself and without the temperature-dependent value of \(K_w\), completely determine where neutrality lies.

14What this lets you understand next

The useful chain is now complete:

\[
[\mathrm{H}^{+}]
\longleftrightarrow
\text{pH}
\longleftrightarrow
\text{pOH}
\longleftrightarrow
[\ce{OH-}].
\]

At \(25^\circ\text{C}\), \(K_w=1.0\times10^{-14}\) ties the two ion concentrations together, and pH + pOH = 14.00 ties the logarithmic scales together.

The next important step is to stop assuming that an acid’s concentration automatically equals \([\mathrm{H}^{+}]\). For strong acids that approximation often works. For weak acids and bases, the extent of ionisation must be determined from equilibrium. That is where pH calculations begin connecting directly to \(K_a\), \(K_b\), equilibrium position, and acid strength.