Ionisation of Water, the pH Scale and pOH

Water Ionises Itself

Pure water conducts electricity. The conductance is minute, and a sensitive bridge is needed to detect it at all, but it is not zero. Charge cannot move through a liquid of neutral molecules, so pure water must contain ions.

Water is amphiprotic: the same molecule can donate a proton and can accept one. In pure water there is nothing else for a water molecule to react with, so it reacts with another water molecule. One acts as the acid and hands over a proton; the other acts as the base and takes it.

$$\mathrm{H_2O(l)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{OH^-(aq)}$$

Reading the equation from left to right, the first $\mathrm{H_2O}$ is the acid and the second is the base; $\mathrm{H_3O^+}$ is the conjugate acid of the second and $\mathrm{OH^-}$ is the conjugate base of the first. The process goes by three names in different books — self-ionisation, autoprotolysis and autoionisation — and they all mean this one proton transfer.

The equilibrium constant for the transfer is

$$K = \frac{[\mathrm{H_3O^+}][\mathrm{OH^-}]}{[\mathrm{H_2O}]}$$

Water here is a pure liquid in vast excess. Its concentration is fixed by its own density and does not change measurably when a trace of it ionises, so it is absorbed into the constant on the left. What remains is a new constant with a name of its own.

Key Point (Definition): The ionic product of water, $K_w$, is the product of the molar concentrations of hydronium and hydroxide ions in water at a given temperature: $$K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]$$ At 298 K, $K_w = 1.0 \times 10^{-14}$.

Experiment fixes the value. Careful conductance measurements on the purest water obtainable give $[\mathrm{H_3O^+}] = 1.0 \times 10^{-7}\ \mathrm{mol\,L^{-1}}$ at 298 K. The proton transfer produces one $\mathrm{OH^-}$ for every $\mathrm{H_3O^+}$, and pure water starts with neither, so the two concentrations must be equal:

$$[\mathrm{H_3O^+}] = [\mathrm{OH^-}] = 1.0 \times 10^{-7}\ \mathrm{mol\,L^{-1}}$$

$$K_w = (1.0 \times 10^{-7})(1.0 \times 10^{-7}) = 1.0 \times 10^{-14}\ \mathrm{M^2} \quad \text{at 298 K}$$

The units $\mathrm{M^2}$ follow from the definition, but $K_w$ is normally quoted as a bare number, because each concentration is understood to be divided by the standard concentration of $1\ \mathrm{M}$.

Two pieces of shorthand appear everywhere and both are safe to use. A bare proton does not exist in water — it is always attached to at least one water molecule — but $\mathrm{H^+}$ is written for $\mathrm{H_3O^+}$ to save space, so $K_w = [\mathrm{H^+}][\mathrm{OH^-}]$ means the same thing. And the equation is often condensed to $2\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{OH^-(aq)}$.

[JEE/NEET] $K_w$ is an equilibrium constant, so it depends on temperature and on nothing else. Adding acid, adding alkali, adding salt or diluting the solution does not change its value. Those changes move the two concentrations in opposite directions; their product stays put.

Only Two Molecules in a Billion

The concentration of $\mathrm{H_3O^+}$ in pure water can be got straight from $K_w$ without any experimental input beyond the constant itself. Pure water is electrically neutral and the only ions present come from the self-ionisation, so the two concentrations are equal. Writing each as $x$,

$$x \times x = 1.0 \times 10^{-14} \qquad x = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}\ \mathrm{mol\,L^{-1}}$$

That figure needs comparing with the water itself. The density of pure water is $1000\ \mathrm{g\,L^{-1}}$ and its molar mass is $18.0\ \mathrm{g\,mol^{-1}}$, so

$$[\mathrm{H_2O}] = \frac{1000\ \mathrm{g\,L^{-1}}}{18.0\ \mathrm{g\,mol^{-1}}} = 55.55\ \mathrm{mol\,L^{-1}}$$

The fraction that has ionised is then

$$\frac{1.0 \times 10^{-7}}{55.55} = 1.8 \times 10^{-9} \approx 2 \text{ in } 10^{9}$$

About two water molecules in a billion are ionised at any instant. The equilibrium lies overwhelmingly on the side of undissociated water, which is exactly what an equilibrium constant of $10^{-14}$ says. It also explains why water is such a poor conductor while still not being an insulator.

The number of ions is small in fraction but not small in count. One litre of water holds roughly $3.3 \times 10^{25}$ molecules, of which about $6 \times 10^{16}$ are ionised at any moment — enough particles to carry a measurable current, and more than enough to control the chemistry of everything dissolved in the water.

$K_w$ is not a property of pure water alone. The self-ionisation equilibrium is present in every aqueous solution, so the relation holds everywhere.

Key Point: In any aqueous solution at 298 K, $[\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$. Hydronium and hydroxide ions coexist in every solution; adding acid does not remove hydroxide ions, it only pushes their concentration down so that the product is unchanged.

A $0.1\ \mathrm{M}$ solution of $\mathrm{HCl}$ has $[\mathrm{H_3O^+}] = 0.1\ \mathrm{M}$, and therefore $[\mathrm{OH^-}] = 10^{-14}/10^{-1} = 10^{-13}\ \mathrm{M}$. Hydroxide has not vanished; it has been suppressed by a factor of a million relative to pure water.

The same relation gives the classification of solutions, and this classification is the one that never breaks.

Solution	Condition	At 298 K
Acidic	$[\mathrm{H_3O^+}] > [\mathrm{OH^-}]$	$[\mathrm{H_3O^+}] > 10^{-7}\ \mathrm{M}$
Neutral	$[\mathrm{H_3O^+}] = [\mathrm{OH^-}]$	$[\mathrm{H_3O^+}] = 10^{-7}\ \mathrm{M}$
Basic	$[\mathrm{H_3O^+}] < [\mathrm{OH^-}]$	$[\mathrm{H_3O^+}] < 10^{-7}\ \mathrm{M}$

The middle column is the definition. The right-hand column is a consequence that is true at 298 K and at no other temperature, for the reason taken up next.

$K_w$ Rises with Temperature, and the Neutral Point Moves

Breaking an $\mathrm{O-H}$ bond costs energy. The self-ionisation of water is endothermic, with $\Delta H \approx +57.3\ \mathrm{kJ\,mol^{-1}}$ for the forward change. Le Chatelier's principle then settles the temperature behaviour at once: supplying heat to an endothermic equilibrium drives it forward, so hot water contains more $\mathrm{H_3O^+}$ and more $\mathrm{OH^-}$ than cold water, and $K_w$ is larger.

$T$ / K	$K_w$	$\mathrm{p}K_w$	pH of pure water
273	$0.114 \times 10^{-14}$	14.94	7.47
283	$0.292 \times 10^{-14}$	14.53	7.27
298	$1.008 \times 10^{-14}$	14.00	7.00
310	$2.70 \times 10^{-14}$	13.57	6.78
323	$5.48 \times 10^{-14}$	13.26	6.63
373	$51.3 \times 10^{-14}$	12.29	6.14

Read the last column carefully. Pure water at 373 K has a pH of 6.14. Nothing has been added to it. It contains exactly as many hydroxide ions as hydronium ions, because the only source of either is the same proton transfer. Pure water at 373 K is neutral, and its pH is 6.14.

Key Point: Neutral means $[\mathrm{H_3O^+}] = [\mathrm{OH^-}]$. It does not mean pH 7. The neutral pH equals $\tfrac{1}{2}\mathrm{p}K_w$, which is 7.00 only at 298 K. Above 298 K the neutral pH is below 7; below 298 K it is above 7.

The arithmetic behind the last column is one line. In pure water $[\mathrm{H_3O^+}] = \sqrt{K_w}$, so

$$\mathrm{pH_{neutral}} = -\log\sqrt{K_w} = \tfrac{1}{2}\,\mathrm{p}K_w$$

At 310 K, $K_w = 2.7 \times 10^{-14}$, so $\mathrm{p}K_w = 13.57$ and neutral water has pH $6.78$.

This is the single most-tested trap in the whole of ionic equilibrium, and it is worth stating in the three ways an examiner can phrase it.

• Water heated from 298 K to 373 K shows a falling pH. It does not become acidic. The falling pH and the falling neutral point move together.
• A solution of pH 6.9 at 310 K is basic, because the neutral point at 310 K is 6.78. The same pH at 298 K would be acidic.
• Human blood is at 310 K and has pH 7.4. Measured against the neutral value 6.78 at that temperature, blood is distinctly basic — more so than the number 7.4 suggests when compared against 7.

[JEE Main] A question that gives a value of $K_w$ different from $10^{-14}$ is telling you the temperature is not 298 K. Every relation containing the number 14 must then be rebuilt from $\mathrm{p}K_w$.

One thing does not change with temperature: the definition of acidic and basic. A solution is acidic when it holds more hydronium than hydroxide, whatever the thermometer reads. The pH number that marks the boundary is what moves.

The pH Scale and pOH

Concentrations such as $1.0 \times 10^{-7}$ and $3.8 \times 10^{-3}$ are awkward to compare and clumsy to plot. S. P. L. Sorensen replaced them with the logarithm of the concentration, taken negative so that ordinary solutions get small positive numbers.

Strictly the definition uses the activity $a_{\mathrm{H^+}}$ of the hydrogen ion, a dimensionless quantity. In dilute solutions — below about $0.01\ \mathrm{M}$ — the activity is equal in magnitude to the molarity, and the working definition is

$$\mathrm{pH} = -\log a_{\mathrm{H^+}} = -\log\left\{\frac{[\mathrm{H^+}]}{\mathrm{mol\,L^{-1}}}\right\}$$

Key Point (Definition): $\mathrm{pH} = -\log_{10}[\mathrm{H_3O^+}]$, where the concentration is in $\mathrm{mol\,L^{-1}}$. pH itself has no units.

The lower-case $\mathrm{p}$ is a general operator meaning "minus the logarithm to base ten", and it is applied to whatever follows it. Three uses appear constantly:

$$\mathrm{pOH} = -\log[\mathrm{OH^-}] \qquad \mathrm{p}K_w = -\log K_w \qquad \mathrm{p}K_a = -\log K_a$$

Applying the operator to the ionic product links pH and pOH. Starting from $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]$ and taking $-\log$ of both sides,

$$-\log K_w = -\log[\mathrm{H_3O^+}] - \log[\mathrm{OH^-}]$$

$$\mathrm{p}K_w = \mathrm{pH} + \mathrm{pOH}$$

At 298 K, $K_w = 10^{-14}$ and $\mathrm{p}K_w = 14.00$, giving the relation everybody quotes.

Key Point: $\mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_w = 14.00$ at 298 K. At any other temperature the sum equals $\mathrm{p}K_w$ at that temperature — 13.57 at 310 K, 12.29 at 373 K.

Two quick applications show the machinery working. A $10^{-2}\ \mathrm{M}$ solution of $\mathrm{HCl}$ has $[\mathrm{H^+}] = 10^{-2}\ \mathrm{M}$, so $\mathrm{pH} = 2$. A solution of $\mathrm{NaOH}$ with $[\mathrm{OH^-}] = 10^{-4}\ \mathrm{M}$ has $[\mathrm{H_3O^+}] = 10^{-14}/10^{-4} = 10^{-10}\ \mathrm{M}$, so $\mathrm{pH} = 10$; equivalently $\mathrm{pOH} = 4$ and $\mathrm{pH} = 14 - 4 = 10$.

The scale is not fenced in at 0 and 14. Those limits come from the accidental values $[\mathrm{H^+}] = 1\ \mathrm{M}$ and $[\mathrm{OH^-}] = 1\ \mathrm{M}$, and stronger solutions run past them. A $10\ \mathrm{M}$ solution of a strong acid has $\mathrm{pH} = -\log 10 = -1$; concentrated hydrochloric acid sits near $-1.0$ and a saturated solution of sodium hydroxide near $15$. Negative pH values are unusual but not forbidden.

Measurement matches the arithmetic in two grades of precision. pH paper carries strips that take different colours at the same pH, and reading the combination fixes the pH over the range 1 to 14 to about $\pm 0.5$. A pH meter measures the pH-dependent electrical potential of the solution and resolves it to about $0.001$.

One pH Unit Is a Factor of Ten

Because the scale is logarithmic, differences in pH are ratios in concentration.

$$\text{pH falls by 1} \;\Longleftrightarrow\; [\mathrm{H_3O^+}] \text{ rises } 10 \text{ times}$$

A change of 2 units is a factor of 100, a change of 3 units a factor of 1000. Lemon juice at pH 2.2 and black coffee at pH 5.0 differ by 2.8 units, so lemon juice holds about $10^{2.8} \approx 630$ times more hydronium ion. Gastric juice at pH 1.2 and milk at pH 6.8 are 5.6 units apart, so the gastric juice holds about four hundred thousand times more hydronium ion, $10^{5.6} \approx 4 \times 10^{5}$ — a gap that no linear scale would ever fit on one page.

The same compression is why small pH differences in biology matter so much. Blood held at 7.4 rather than 7.0 differs in hydronium concentration by a factor of 2.5, and blood outside the range 7.0 to 7.8 is fatal.

Fluid	pH	Fluid	pH
Saturated $\mathrm{NaOH}$ solution	$\approx 15$	Black coffee	5.0
$0.1\ \mathrm{M}$ $\mathrm{NaOH}$	13	Tomato juice	$\approx 4.2$
Lime water	10.5	Soft drinks, vinegar	$\approx 3.0$
Milk of magnesia	10	Lemon juice	$\approx 2.2$
Egg white, sea water	7.8	Gastric juice	$\approx 1.2$
Human blood	7.4	$1\ \mathrm{M}$ $\mathrm{HCl}$	$\approx 0$
Milk	6.8	Concentrated $\mathrm{HCl}$	$\approx -1.0$
Human saliva	6.4

Reading the table downwards on the right and upwards on the left gives the whole range of ordinary aqueous chemistry: sixteen pH units, sixteen powers of ten in hydronium concentration.

[NEET] Comparing acidity is a subtraction, never a division. A solution of pH 2 is not "three times more acidic" than one of pH 6; it is $10^{4}$ times more acidic. Dividing pH values has no meaning at all.

pH of Strong Acids and Strong Bases

A strong acid is completely ionised in aqueous solution, so every mole of acid delivers its ionisable protons to the water. The hydronium concentration is the acid concentration multiplied by the number of protons the formula can release.

$$[\mathrm{H_3O^+}] = n \times c_{\text{acid}} \qquad \mathrm{pH} = -\log[\mathrm{H_3O^+}]$$

For a strong base the route runs through pOH.

$$[\mathrm{OH^-}] = n \times c_{\text{base}} \qquad \mathrm{pOH} = -\log[\mathrm{OH^-}] \qquad \mathrm{pH} = 14 - \mathrm{pOH} \;\; \text{(298 K)}$$

The multiplier $n$ is where marks are lost, so it is worth tabulating.

Solute	Ionisation	$n$	Illustration
$\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$, $\mathrm{HNO_3}$, $\mathrm{HClO_4}$	one proton	1	$0.01\ \mathrm{M}$ gives $[\mathrm{H_3O^+}] = 0.01\ \mathrm{M}$
$\mathrm{H_2SO_4}$	two protons	2	$0.005\ \mathrm{M}$ gives $[\mathrm{H_3O^+}] = 0.01\ \mathrm{M}$
$\mathrm{NaOH}$, $\mathrm{KOH}$	one hydroxide	1	$0.01\ \mathrm{M}$ gives $[\mathrm{OH^-}] = 0.01\ \mathrm{M}$
$\mathrm{Ca(OH)_2}$, $\mathrm{Ba(OH)_2}$, $\mathrm{Sr(OH)_2}$	two hydroxides	2	$0.005\ \mathrm{M}$ gives $[\mathrm{OH^-}] = 0.01\ \mathrm{M}$

Key Point: The stoichiometric factor multiplies the concentration, before the logarithm is taken. It never multiplies or divides the pH. For $0.005\ \mathrm{M}$ $\mathrm{H_2SO_4}$ the answer is $\mathrm{pH} = -\log(0.01) = 2.00$, not $-\log(0.005) = 2.30$ and not $2.30/2$.

A refinement about sulphuric acid is worth knowing even though the Class 11 treatment ignores it. Only the first proton is given up completely; the second comes from $\mathrm{HSO_4^-}$, which is a moderately strong acid with $K_{a2} = 1.2 \times 10^{-2}$, not a strong one. Treating both protons as fully ionised is an approximation, and it is the approximation intended whenever a question says "assume complete ionisation".

Calcium hydroxide raises a different practical point. It is only sparingly soluble, so a "$0.005\ \mathrm{M}$ solution" means that much has actually dissolved; whatever dissolves is completely ionised into $\mathrm{Ca^{2+}}$ and two $\mathrm{OH^-}$.

Three worked lines fix the pattern.

$0.02\ \mathrm{M}$ $\mathrm{HNO_3}$: $[\mathrm{H_3O^+}] = 0.02\ \mathrm{M}$, $\mathrm{pH} = 2 - \log 2 = 2 - 0.301 = 1.70$.

$0.05\ \mathrm{M}$ $\mathrm{KOH}$: $[\mathrm{OH^-}] = 0.05\ \mathrm{M}$, $\mathrm{pOH} = 2 - \log 5 = 2 - 0.699 = 1.30$, $\mathrm{pH} = 12.70$.

$0.002\ \mathrm{M}$ $\mathrm{Ba(OH)_2}$: $[\mathrm{OH^-}] = 0.004\ \mathrm{M}$, $\mathrm{pOH} = 3 - \log 4 = 3 - 0.602 = 2.40$, $\mathrm{pH} = 11.60$.

Four logarithms carry almost every strong-acid question in an exam hall: $\log 2 = 0.301$, $\log 3 = 0.477$, $\log 5 = 0.699$, $\log 7 = 0.845$. From these, $\log 4 = 0.602$, $\log 6 = 0.778$, $\log 8 = 0.903$ and $\log 9 = 0.954$ follow by addition.

Going Backwards: Concentration from pH

Reversing the definition needs an antilogarithm.

$$[\mathrm{H_3O^+}] = 10^{-\mathrm{pH}} \qquad [\mathrm{OH^-}] = 10^{-\mathrm{pOH}} = \frac{K_w}{[\mathrm{H_3O^+}]}$$

An integer pH is immediate: $\mathrm{pH} = 5$ gives $[\mathrm{H_3O^+}] = 10^{-5}\ \mathrm{mol\,L^{-1}}$.

A non-integer pH needs the exponent split into a whole number and a positive decimal, because tables of antilogarithms only handle the decimal part. The rule is to write the negative exponent as a negative integer plus a positive fraction.

$$\mathrm{pH} = 4.50 \;\Rightarrow\; [\mathrm{H_3O^+}] = 10^{-4.50} = 10^{0.50} \times 10^{-5} = 3.16 \times 10^{-5}\ \mathrm{mol\,L^{-1}}$$

$$\mathrm{pH} = 3.72 \;\Rightarrow\; [\mathrm{H_3O^+}] = 10^{-3.72} = 10^{0.28} \times 10^{-4} = 1.9 \times 10^{-4}\ \mathrm{mol\,L^{-1}}$$

The common wreck is to read the decimal part as the mantissa of the answer and write $4.5 \times 10^{-5}$ for a pH of 4.50. The correct value $3.16 \times 10^{-5}$ is smaller, and the two differ by more than 40 per cent.

A sanity check costs nothing. A pH between 4 and 5 must give a concentration between $10^{-5}$ and $10^{-4}$, and closer to which end depends on which pH it is nearer. Since 4.50 sits in the middle, $3.16 \times 10^{-5}$ is right and anything outside $10^{-5}$ to $10^{-4}$ is wrong.

Once $[\mathrm{H_3O^+}]$ is known, $[\mathrm{OH^-}]$ follows from $K_w$ in one division, or from $\mathrm{pOH} = 14 - \mathrm{pH}$ in one subtraction at 298 K. For $\mathrm{pH} = 4.50$: $\mathrm{pOH} = 9.50$ and $[\mathrm{OH^-}] = 10^{-9.50} = 3.16 \times 10^{-10}\ \mathrm{mol\,L^{-1}}$.

For a strong monobasic acid the concentration found this way is also the concentration of the acid, since ionisation is complete. For a weak acid it is not — the acid concentration is much larger than the hydronium concentration, and recovering it needs $K_a$.

Very Dilute Acids: When Water's Own Ions Cannot Be Ignored

Applying $\mathrm{pH} = -\log c$ to a $1.0 \times 10^{-8}\ \mathrm{M}$ solution of $\mathrm{HCl}$ gives $\mathrm{pH} = 8$. That answer is not merely inaccurate; it is impossible.

Adding an acid to water can only raise the hydronium concentration. A pH of 8 means $[\mathrm{H_3O^+}] = 10^{-8}\ \mathrm{M}$, which is less than the $10^{-7}\ \mathrm{M}$ present in the pure water before the acid was added, and it means $[\mathrm{OH^-}] = 10^{-6}\ \mathrm{M}$, a hundred times the hydronium concentration. A solution of hydrochloric acid cannot be basic.

The mistake is in the assumption, not the arithmetic. Writing $[\mathrm{H_3O^+}] = c$ silently assumes the acid supplies all the hydronium ion. At $c = 0.1\ \mathrm{M}$ that is a superb assumption: the water contributes at most $10^{-7}\ \mathrm{M}$ out of $0.1\ \mathrm{M}$, one part in a million. At $c = 10^{-8}\ \mathrm{M}$ the water is the major supplier, and the acid is the small correction.

Both sources have to be counted. Take $x$ as the hydroxide concentration in the final solution, all of which came from water.

$$\mathrm{HCl(aq)} + \mathrm{H_2O(l)} \rightarrow \mathrm{H_3O^+(aq)} + \mathrm{Cl^-(aq)} \qquad \text{gives } 10^{-8}\ \mathrm{M}$$

$$2\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{OH^-(aq)} \qquad \text{gives } x$$

The solution must be electrically neutral, so the total hydronium concentration is

$$[\mathrm{H_3O^+}] = 10^{-8} + x$$

Substituting into $K_w$:

$$(10^{-8} + x)(x) = 10^{-14}$$

$$x^2 + 10^{-8}x - 10^{-14} = 0$$

$$x = \frac{-10^{-8} + \sqrt{(10^{-8})^2 + 4 \times 10^{-14}}}{2} = 9.5 \times 10^{-8}\ \mathrm{mol\,L^{-1}}$$

So $[\mathrm{OH^-}] = 9.5 \times 10^{-8}\ \mathrm{M}$, $\mathrm{pOH} = 7.02$, and

$$\mathrm{pH} = 14 - 7.02 = 6.98$$

Checking directly: $[\mathrm{H_3O^+}] = 10^{-8} + 9.5 \times 10^{-8} = 1.05 \times 10^{-7}\ \mathrm{M}$, and $-\log(1.05 \times 10^{-7}) = 6.98$. The two routes agree.

Key Point: A $10^{-8}\ \mathrm{M}$ solution of $\mathrm{HCl}$ has $\mathrm{pH} = 6.98$ — just below 7, faintly acidic, exactly as adding a trace of acid to water should be. It is never 8.

Solving the quadratic in general form gives a single expression covering every strong monobasic acid:

$$[\mathrm{H_3O^+}] = \frac{c + \sqrt{c^2 + 4K_w}}{2}$$

$c$ of $\mathrm{HCl}$ / $\mathrm{mol\,L^{-1}}$	true pH	naive $-\log c$
$10^{-4}$	4.00	4.00
$10^{-6}$	6.00	6.00
$10^{-7}$	6.79	7.00
$10^{-8}$	6.98	8.00
$10^{-9}$	7.00	9.00

The pattern is a working rule. For $c$ above about $10^{-6}\ \mathrm{M}$ the water contributes nothing worth keeping and $\mathrm{pH} = -\log c$ is safe. Between $10^{-6}$ and $10^{-8}\ \mathrm{M}$ the quadratic is compulsory. Below that, the acid is swamped and the pH creeps back towards 7 from underneath, never crossing it.

The mirror image holds for a very dilute strong base. A $10^{-8}\ \mathrm{M}$ solution of $\mathrm{NaOH}$ has $\mathrm{pOH} = 6.98$ and therefore $\mathrm{pH} = 7.02$ — just above 7, never 6.

[JEE Main] Two checks catch this every time. An acid solution must have pH below the neutral value; a base solution must have pH above it. If a calculation puts a strong acid above 7 or a strong base below 7, the water's own ionisation was dropped.

Question 1: pH of a dilute strong acid

Calculate the pH and $[\mathrm{OH^-}]$ of a $1.0 \times 10^{-3}\ \mathrm{M}$ solution of $\mathrm{HNO_3}$ at 298 K.

Answer:

Nitric acid is a strong monobasic acid, so it is completely ionised and one mole gives one mole of $\mathrm{H_3O^+}$.

$[\mathrm{H_3O^+}] = 1.0 \times 10^{-3}\ \mathrm{mol\,L^{-1}}$

$\mathrm{pH} = -\log(1.0 \times 10^{-3}) = 3.00$

For the hydroxide concentration I divide $K_w$ by the hydronium concentration.

$[\mathrm{OH^-}] = \dfrac{1.0 \times 10^{-14}}{1.0 \times 10^{-3}} = 1.0 \times 10^{-11}\ \mathrm{mol\,L^{-1}}$

Ans: $\mathrm{pH} = 3.00$, $[\mathrm{OH^-}] = 1.0 \times 10^{-11}\ \mathrm{mol\,L^{-1}}$

Question 2: pH of a soft drink

The hydrogen ion concentration in a sample of soft drink is $3.8 \times 10^{-3}\ \mathrm{M}$. Find its pH.

Answer:

I take the negative logarithm and split the product.

$\mathrm{pH} = -\log(3.8 \times 10^{-3}) = -\{\log 3.8 + \log 10^{-3}\}$

$= -\{0.58 + (-3.0)\} = 2.42$

The value lies below 7, so the drink is acidic, which matches the carbonic and phosphoric acid in it.

Ans: $\mathrm{pH} = 2.42$

Question 3: A dibasic strong acid

Calculate the pH of $0.005\ \mathrm{M}$ $\mathrm{H_2SO_4}$ at 298 K, assuming both protons are completely ionised.

Answer:

Each formula unit supplies two protons, so I multiply the concentration by two before taking any logarithm.

$[\mathrm{H_3O^+}] = 2 \times 0.005 = 0.01\ \mathrm{mol\,L^{-1}}$

$\mathrm{pH} = -\log(1.0 \times 10^{-2}) = 2.00$

Ans: $\mathrm{pH} = 2.00$ Watch out: Using $-\log(0.005)$ gives $2.30$ and forgets the second proton. Dividing $2.30$ by two gives $1.15$ and applies the factor to the wrong quantity. The two multiplies the concentration.

Question 4: pH of a strong monoacidic base

Calculate the pH of $0.02\ \mathrm{M}$ $\mathrm{NaOH}$ at 298 K.

Answer:

Sodium hydroxide is completely ionised and gives one hydroxide ion per formula unit.

$[\mathrm{OH^-}] = 0.02 = 2 \times 10^{-2}\ \mathrm{mol\,L^{-1}}$

$\mathrm{pOH} = -\log(2 \times 10^{-2}) = 2 - 0.301 = 1.70$

$\mathrm{pH} = 14.00 - 1.70 = 12.30$

Ans: $\mathrm{pH} = 12.30$ Watch out: Stopping at $1.70$ reports the pOH as the pH and turns a strong alkali into a strong acid.

Question 5: pH of a strong diacidic base

Calculate the pH of a $0.005\ \mathrm{M}$ solution of $\mathrm{Ca(OH)_2}$ at 298 K.

Answer:

Calcium hydroxide releases two hydroxide ions per formula unit, so the hydroxide concentration is twice the salt concentration.

$[\mathrm{OH^-}] = 2 \times 0.005 = 0.01\ \mathrm{mol\,L^{-1}}$

$\mathrm{pOH} = -\log(1.0 \times 10^{-2}) = 2.00$

$\mathrm{pH} = 14.00 - 2.00 = 12.00$

Ans: $\mathrm{pH} = 12.00$ Watch out: Ignoring the second hydroxide gives $\mathrm{pOH} = 2.30$ and $\mathrm{pH} = 11.70$.

Question 6: Concentration from a non-integer pH

A solution has $\mathrm{pH} = 4.50$ at 298 K. Find $[\mathrm{H_3O^+}]$ and $[\mathrm{OH^-}]$.

Answer:

I take the antilogarithm, splitting the exponent into a whole number and a positive decimal.

$[\mathrm{H_3O^+}] = 10^{-4.50} = 10^{0.50} \times 10^{-5} = 3.16 \times 10^{-5}\ \mathrm{mol\,L^{-1}}$

Then $\mathrm{pOH} = 14.00 - 4.50 = 9.50$.

$[\mathrm{OH^-}] = 10^{-9.50} = 3.16 \times 10^{-10}\ \mathrm{mol\,L^{-1}}$

A check: the product is $3.16 \times 10^{-5} \times 3.16 \times 10^{-10} = 1.0 \times 10^{-14}$, as it must be.

Ans: $[\mathrm{H_3O^+}] = 3.16 \times 10^{-5}\ \mathrm{mol\,L^{-1}}$, $[\mathrm{OH^-}] = 3.16 \times 10^{-10}\ \mathrm{mol\,L^{-1}}$ Watch out: Writing $4.5 \times 10^{-5}$ copies the decimal part of the pH into the mantissa. That is not what an antilogarithm does.

Question 7: A basic solution

The pH of a solution is $9.70$ at 298 K. Calculate $[\mathrm{H^+}]$ and $[\mathrm{OH^-}]$.

Answer:

$[\mathrm{H^+}] = 10^{-9.70} = 10^{0.30} \times 10^{-10} = 2.0 \times 10^{-10}\ \mathrm{mol\,L^{-1}}$

$[\mathrm{OH^-}] = \dfrac{1.0 \times 10^{-14}}{2.0 \times 10^{-10}} = 5.0 \times 10^{-5}\ \mathrm{mol\,L^{-1}}$

Hydroxide exceeds hydronium by a large factor, so the solution is basic, consistent with a pH above 7.

Ans: $[\mathrm{H^+}] = 2.0 \times 10^{-10}\ \mathrm{mol\,L^{-1}}$, $[\mathrm{OH^-}] = 5.0 \times 10^{-5}\ \mathrm{mol\,L^{-1}}$

Question 8: Neutral water at 373 K

At 373 K the ionic product of water is $5.13 \times 10^{-13}$. Find the pH of pure water at this temperature and state whether the water is acidic, basic or neutral.

Answer:

Pure water contains equal concentrations of the two ions, so each equals $\sqrt{K_w}$.

$[\mathrm{H_3O^+}] = \sqrt{5.13 \times 10^{-13}} = 7.16 \times 10^{-7}\ \mathrm{mol\,L^{-1}}$

$\mathrm{pH} = -\log(7.16 \times 10^{-7}) = 7 - 0.855 = 6.14$

Alternatively $\mathrm{p}K_w = -\log(5.13 \times 10^{-13}) = 12.29$ and the neutral pH is half of that, $6.145$.

Since $[\mathrm{H_3O^+}] = [\mathrm{OH^-}]$, the water is neutral.

Ans: $\mathrm{pH} = 6.14$; the water is neutral Watch out: A pH of 6.14 looks acidic only because 7 has been memorised as the neutral value. Neutrality is the equality of the two ion concentrations, and 7 is the neutral pH at 298 K alone.

Question 9: Mixing a strong acid and a strong base

$50\ \mathrm{mL}$ of $0.2\ \mathrm{M}$ $\mathrm{HCl}$ is mixed with $50\ \mathrm{mL}$ of $0.1\ \mathrm{M}$ $\mathrm{NaOH}$ at 298 K. Calculate the pH of the mixture.

Answer:

I work in millimoles, since the volumes are in millilitres.

$\mathrm{H^+}$ taken: $50 \times 0.2 = 10\ \mathrm{mmol}$

$\mathrm{OH^-}$ taken: $50 \times 0.1 = 5\ \mathrm{mmol}$

Neutralisation removes $5\ \mathrm{mmol}$ of each, leaving $5\ \mathrm{mmol}$ of $\mathrm{H^+}$ in a total volume of $100\ \mathrm{mL}$.

$[\mathrm{H_3O^+}] = \dfrac{5\ \mathrm{mmol}}{100\ \mathrm{mL}} = 0.05\ \mathrm{mol\,L^{-1}}$

$\mathrm{pH} = -\log(5 \times 10^{-2}) = 2 - 0.699 = 1.30$

Ans: $\mathrm{pH} = 1.30$ Watch out: Forgetting that the volume doubles gives $[\mathrm{H_3O^+}] = 0.1\ \mathrm{M}$ and $\mathrm{pH} = 1.00$.

Question 10: A very dilute strong acid

Calculate the pH of a $1.0 \times 10^{-8}\ \mathrm{M}$ solution of $\mathrm{HCl}$ at 298 K.

Answer:

At this concentration the acid supplies less hydronium ion than the water does, so both sources must be counted.

$2\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{OH^-(aq)}$, with $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 10^{-14}$

I let $x = [\mathrm{OH^-}]$, which comes entirely from the water. The acid contributes $10^{-8}\ \mathrm{M}$ of hydronium and the water contributes $x$, so

$[\mathrm{H_3O^+}] = 10^{-8} + x$

$(10^{-8} + x)(x) = 10^{-14}$

$x^2 + 10^{-8}x - 10^{-14} = 0$

$x = 9.5 \times 10^{-8}\ \mathrm{mol\,L^{-1}}$

$\mathrm{pOH} = -\log(9.5 \times 10^{-8}) = 7.02$, so $\mathrm{pH} = 14.00 - 7.02 = 6.98$.

Ans: $\mathrm{pH} = 6.98$ Watch out: The naive $-\log(10^{-8}) = 8$ says that adding acid to water made it basic. Any answer above 7 for an acid solution at 298 K is wrong by inspection.

Question 11: Reading $K_w$ off a measured pH

At a certain temperature the pH of pure water is $6.50$. Calculate $K_w$ and $[\mathrm{OH^-}]$ at that temperature.

Answer:

The water is pure, so the two ion concentrations are equal.

$[\mathrm{H_3O^+}] = 10^{-6.50} = 3.16 \times 10^{-7}\ \mathrm{mol\,L^{-1}} = [\mathrm{OH^-}]$

$K_w = (3.16 \times 10^{-7})^2 = 1.0 \times 10^{-13}$

The value exceeds $10^{-14}$, so this temperature is above 298 K.

Ans: $K_w = 1.0 \times 10^{-13}$, $[\mathrm{OH^-}] = 3.16 \times 10^{-7}\ \mathrm{mol\,L^{-1}}$ Watch out: Reporting $K_w = 10^{-6.5}$ skips the squaring. $K_w$ is the product of two concentrations, not one of them.

What to Carry Forward

The whole section runs on one equilibrium and one operator.

Relation	Where it holds
$K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]$	every aqueous solution, at the stated temperature
$K_w = 1.0 \times 10^{-14}$	298 K only
$\mathrm{pH} = -\log[\mathrm{H_3O^+}]$, $\mathrm{pOH} = -\log[\mathrm{OH^-}]$	always
$\mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_w$	always
$\mathrm{pH} + \mathrm{pOH} = 14.00$	298 K only
neutral pH $= \tfrac{1}{2}\mathrm{p}K_w$	always; equals 7.00 at 298 K only

Five errors account for most of the marks lost here.

Treating pH 7 as the definition of neutral. Neutral is $[\mathrm{H_3O^+}] = [\mathrm{OH^-}]$; the number 7 belongs to 298 K.

Applying the stoichiometric factor to the pH instead of the concentration, so $0.005\ \mathrm{M}$ $\mathrm{H_2SO_4}$ comes out as $2.30$ or $1.15$ rather than $2.00$.

Reporting pOH as pH for an alkali, which converts a pH of 12.30 into 1.70.

Copying the decimal part of a pH into the mantissa of the concentration, giving $4.5 \times 10^{-5}$ where $3.16 \times 10^{-5}$ is meant.

Writing pH 8 for $10^{-8}\ \mathrm{M}$ $\mathrm{HCl}$ instead of 6.98, by ignoring the hydronium ion the water itself supplies.

The next section takes the same $K_w$ and the same logarithms into weak acids and weak bases, where ionisation is partial and $K_a$ and $K_b$ decide how far it goes.