Equilibrium · Lesson 8 of 14

Water, pH, and Measuring Acidity

“Use water’s ionization and the logarithmic pH scale to interpret acidic, neutral, and basic solutions.”

Learning Objectives

• Write water’s ionic-product expression and use its temperature dependence. • Convert between pH, pOH and ion concentrations in suitable dilute solutions. • Explain why very dilute acid calculations must include water ionization. • Interpret indicator colours, pH paper and pH-meter measurements.

Water supplies a small concentration of ions

Water can donate a proton to another water molecule. One becomes hydroxide and the other hydronium. In pure water the two ions are produced in equal amounts, so their equilibrium concentrations are equal. Only a tiny fraction of water molecules are ionized, but those ions matter enormously in acid–base chemistry.

Water autoionizationLaTeX

2 H_{2} O (l) ⇌ H_{3} O^{+} (aq) + O H^{-} (aq), K_{w} = a_{H_{3} O^{+}} a_{O H^{-}}

The pure-liquid water activity is taken as one. In dilute solutions, the conventional numerical expression is Kw = [H3O+][OH−] with concentrations expressed in moles per litre.

At 298 K, $K_{w} \approx 1.0 \times 1 0^{- 14}$ , giving $[H_{3} O^{+}] = [O H^{-}] = 1.0 \times 1 0^{- 7} M$ in pure water. Liquid water itself is about 55.5 M, so the ionized fraction is only about $1.8 \times 1 0^{- 9}$ . Adding acid raises hydronium and lowers hydroxide; adding base does the reverse. At fixed temperature their product remains Kw at equilibrium.

Solution character	Relative concentrations	At 298 K in the dilute approximation
Acidic	[H3O+] > [OH−]	[H3O+] > $1 0^{- 7}$ M
Neutral	[H3O+] = [OH−]	[H3O+] = $1 0^{- 7}$ M
Basic	[H3O+] < [OH−]	[H3O+] < $1 0^{- 7}$ M

A logarithmic scale makes large ranges manageable

A hundredfold concentration change can be awkward to compare as a string of zeros. The pH scale compresses it into two units. Formally, pH is minus the base-ten logarithm of hydrogen-ion activity. Activity is dimensionless and accounts for how a species behaves relative to its standard state. In sufficiently dilute solutions we approximate it using concentration divided by one mole per litre.

pH and related scalesLaTeX

pH = - lo g_{10} a_{H^{+}} \approx - lo g_{10} (\frac{[ H _{3} O ^{+} ]}{c ^{\circ}}), pOH \approx - lo g_{10} (\frac{[ O H ^{-} ]}{c ^{\circ}})

Here c° = 1 mol per litre. The reverse conversion is [H3O+] approximately equal to c° multiplied by 10 to the power −pH.

A fall of one pH unit corresponds to a tenfold rise in hydrogen-ion activity; a fall of two units means a hundredfold rise. A pH of 3 is not “twice as acidic” as pH 6. The scale can extend below zero or above fourteen in concentrated solutions, although simple concentration calculations become less accurate there.

The sum ruleLaTeX

pH + pOH = p K_{w}, p K_{w} = - lo g_{10} K_{w}

At 298 K, pKw is approximately 14. At other temperatures use the appropriate Kw rather than automatically using 14.

Taking minus logarithms of the product Kw gives the sum of pH and pOH. Neutral water always has equal hydronium and hydroxide activities, so its neutral pH is pKw/2. Neutrality is therefore a balance condition, not a universal requirement that pH equal seven.

Example — A soft drink

Problem
A dilute-model soft drink has hydrogen-ion concentration $3.8 \times 1 0^{- 3} M$ . Calculate its pH.

1.Use $pH \approx - lo g_{10} (3.8 \times 1 0^{- 3})$ .
2.Separate the logarithm: $lo g_{10} 3.8 - 3 = 0.580 - 3 = - 2.420$ .
3.The pH is 2.42. At 298 K this is acidic because hydronium exceeds hydroxide.

Example — A strong-base solution

Problem
Assume complete dissociation of 0.0050 M NaOH at 298 K and neglect activity corrections. Find pH.

1.Each NaOH unit supplies one hydroxide ion, so $[O H^{-}] = 0.0050 M$ .
2.Then $pOH = - lo g_{10} (0.0050) = 2.301$ .
3.At 298 K, $pH = 14.000 - 2.301 = 11.699$ . Alternatively, $[H_{3} O^{+}] = 1 0^{- 14} /0.0050 = 2.0 \times 1 0^{- 12} M$ .

When the acid is more dilute than water’s own ions

For 0.010 M HCl, the acid contribution overwhelms the hydronium initially supplied by water, so a simple strong-acid calculation is adequate. For a nominal acid concentration of only ten to the power minus eight M, ignoring water would incorrectly predict a basic solution. We must satisfy the ionic product and electrical neutrality together.

Example — Very dilute hydrochloric acid

Problem
Find the pH of $1.0 \times 1 0^{- 8} M$ HCl at 298 K, assuming complete dissociation and ideal dilute behaviour.

1.Let $C = 1.0 \times 1 0^{- 8} M$ and let $x = [O H^{-}]$ . Charge balance requires $[H_{3} O^{+}] = C + x$ because chloride contributes concentration C of negative charge.
2.The water product gives $x (C + x) = 1 0^{- 14}$ , or $x^{2} + C x - 1 0^{- 14} = 0$ .
3.The positive root is $x = 9.5125 \times 1 0^{- 8} M$ . Therefore $[H_{3} O^{+}] = 1.05125 \times 1 0^{- 7} M$ .
4.Thus pH = 6.978, slightly acidic. Do not interpret x as water contributing equal final amounts of both ions independently; the final concentrations must satisfy charge balance and Kw together.

Example — Neutral water at another temperature

Problem
At 310 K, $K_{w} = 2.7 \times 1 0^{- 14}$ . Calculate the pH of neutral water.

1.Neutrality gives equal hydronium and hydroxide, so $[H_{3} O^{+}] = 2.7 \times 1 0^{- 14} = 1.643 \times 1 0^{- 7} M$ .
2.Equivalently, $p K_{w} = 14 - lo g_{10} 2.7 = 13.5686$ .
3.Neutral pH is half this value: 6.7843. The water is neutral despite having pH below seven.

Indicators turn the scale into a visible observation

An acid–base indicator has differently coloured protonated and deprotonated forms. Its equilibrium shifts as hydronium activity changes. pH paper combines indicators whose colour pattern estimates pH, while a calibrated pH meter measures an electrical response related to hydrogen-ion activity. A digital display with many digits does not guarantee equal accuracy: calibration, temperature and the sample all matter.

An indicator reports an equilibrium shift— The two indicator forms have different colours. Increasing hydronium shifts the indicator towards its protonated form.

Sample	Approximate illustrative pH	Interpretation
Concentrated HCl	−1	Activity corrections are important
1 M HCl	0	Strongly acidic
Gastric juice	1.2	Acidic
Lemon juice	2.2	Acidic
Soft drinks / vinegar	About 3	Varies by sample
Tomato juice	4.2	Acidic
Black coffee	5.0	Acidic
Saliva	6.4	Varies by sample
Milk	6.8	Slightly acidic near room temperature
Blood	7.4	Closely regulated biologically
Egg white / seawater	About 7.8	Mildly basic
Milk of magnesia	10	Basic
Limewater	10.5	Basic
0.1 M NaOH	13	Basic
Saturated NaOH	About 15	Concentrated; activity matters

For an investigation, compare labelled samples of fruit or vegetable juice, a soft drink and available water using the same pH paper. Record sample identity, temperature and colour estimate. If a meter is available, calibrate it and compare readings rather than treating either instrument as automatically exact. Observe indicator colours in solutions of known pH and use an appropriate indicator in a supervised acid–base titration. The changing colour reflects an equilibrium and its transition range, not the disappearance of every ion at once.

Common mistake

Use pH + pOH = 14 only when the temperature justifies it. A pH below seven is not sufficient by itself to establish acidity at every temperature. For concentrated samples, pH is based on activity rather than simply a molarity number.

Quick check

A decrease of two pH units corresponds to hydrogen-ion activity becoming

Quick check

At 298 K a solution with pOH = 3 has pH

Quick check

Neutral water at a different temperature must have

Quick check

Why is pH = 8 incorrect for $1 0^{- 8}$ M HCl at 298 K?

Quick check

An acid–base indicator changes colour because

Quick check

Which is the formal quantity in the definition of pH?

Practice Problems

Calculate pH for a dilute solution with hydrogen-ion concentration $2.0 \times 1 0^{- 4}$ M.
Calculate hydrogen-ion concentration corresponding to pH 3.76 in the dilute approximation.
Assuming full dissolution and dissociation, find pH for 0.0020 M Ca(OH)2 at 298 K.
Explain why adding a tiny amount of strong acid to pure water cannot make it basic.
A water sample has pH 6.80 at a temperature where neutral pH is 6.60. Is it acidic or basic? Explain.
Plan a pH-paper and pH-meter comparison, stating what you would measure and what differences could mean.

Key Takeaways

Water’s ionic product links hydronium and hydroxide at each temperature. pH is logarithmic and formally uses activity; dilute-solution concentration is an approximation. Neutral pH depends on temperature, and very dilute acid or base calculations must include water.

Previous · Lesson 7

Ionic Equilibrium and Models of Acids and Bases

Next · Lesson 9

Weak Acids and Their Ionization

Water supplies a small concentration of ions

A logarithmic scale makes large ranges manageable

When the acid is more dilute than water’s own ions

Indicators turn the scale into a visible observation

Quiz

Practice Problems

Key Takeaways