MHT-CET Chemistry · Ionic Equilibria

pH, pOH and the Ionic Product of Water

Water self-ionises so that the product of its hydrogen- and hydroxide-ion concentrations is fixed; pH and pOH are the logarithmic measures of those concentrations, and together they add up to 14 at 25 degrees C.

All Ionic Equilibria notes MHT-CET Chemistry strategy

Why this matters

This is one of the most reliable scoring blocks in MHT-CET Chemistry — around two dozen PYQs, almost all one- or two-step plug-ins. Nearly every question reduces to the same routine: find the ion concentration, take a negative log, and use pH + pOH = 14. The recurring traps are a strong acid or base that furnishes more than one ion per formula unit (dibasic, diacidic), a weak acid or base where you must first multiply by the degree of dissociation, and the special case of an extremely dilute strong acid where the pH creeps back toward 7. Learn the four formulas cold — Kw, pH, pOH, and pH + pOH = 14 — and you can attempt every question here on sight.

Concept 1 of 4

Ionic product of water, Kw

Intuition

Even pure water conducts a tiny bit of electricity because a few molecules split into H+ and OH- ions. At any temperature the product of these two concentrations is a fixed number called Kw. At 25 degrees C that number is 10 to the minus 14, so if you know one ion concentration you can always get the other by dividing.

Definition

The ionic product of water:

Water self-ionises: $2\,\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-$ .
The product $K_w = [\text{H}^+][\text{OH}^-]$ is constant at a given temperature; at $25\,^{\circ}\text{C}$ it equals $10^{-14}$ .
In pure (neutral) water the two ions are equal: $[\text{H}^+] = [\text{OH}^-] = 10^{-7}\,\text{M}$ at $25\,^{\circ}\text{C}$ .
$K_w$ is temperature-dependent — self-ionisation is endothermic, so heating water raises $K_w$ (and lowers the neutral pH below 7), while the product of the two ions still stays constant at that new temperature.
Rearranged: $[\text{OH}^-] = \dfrac{K_w}{[\text{H}^+]}$ and $[\text{H}^+] = \dfrac{K_w}{[\text{OH}^-]}$ .

Ionic product of water

K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \quad (25\,^{\circ}\text{C})

K_wionic product of water (mol^2 L^-2)
[\text{H}^+]hydrogen-ion (hydronium) concentration (mol/L)
[\text{OH}^-]hydroxide-ion concentration (mol/L)

Worked example

A solution has a hydrogen-ion concentration of

2 \times 10^{-3}\,\text{M}

at 25 degrees C. Find its hydroxide-ion concentration.

At $25\,^{\circ}\text{C}$ , $[\text{H}^+][\text{OH}^-] = K_w = 10^{-14}$ .
$[\text{OH}^-] = \dfrac{K_w}{[\text{H}^+]} = \dfrac{10^{-14}}{2 \times 10^{-3}}$ .
$= \dfrac{1}{2} \times 10^{-14+3} = 0.5 \times 10^{-11} = 5 \times 10^{-12}\,\text{M}$ .

Answer:

[\text{OH}^-] = 5 \times 10^{-12}\,\text{M}

Practice this conceptself-check · 4 quick reps

Try it yourself

A solution contains

0.05\,\text{M}

hydrogen ions. What is its hydroxide-ion concentration at 25 degrees C?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Value of Kw for water at 25 degrees C?
2.
[H+] in pure neutral water at 25 degrees C?
3.
If [H+] = 10^-3 M, find [OH-] at 25 degrees C.
4.
Does Kw increase or decrease when water is heated?

From the bank · past-year question

Example 1Ionic EquilibriaEASY

What is the concentration of

OH^-

ion in a solution containing

0.05\text{ M } H^+

ions?

[Q73 · 11th May Shift 1 · 2024]

Kw = 10 to the minus 14 only at 25 degrees C

The value

K_w = 10^{-14}

holds **only at

25\,^{\circ}\text{C}

**. Self-ionisation is endothermic, so at higher temperatures

K_w

is larger and neutral water has a pH below 7 — even though it is still neutral (

[\text{H}^+] = [\text{OH}^-]

Divide into Kw, do not subtract

To get one ion from the other, **divide

K_w

by the known concentration** — do not subtract exponents casually.

[\text{OH}^-] = K_w/[\text{H}^+]

; mixing up which ion you started with flips the answer between acidic and basic.

Drill 1 more on ionic product of water, kw

Concept 2 of 4

pH, pOH and the relation pH + pOH = 14

Intuition

Ion concentrations are awkward tiny numbers like 10 to the minus 4, so we take the negative base-10 log to get a friendly 0-to-14 scale. pH measures the H+ side, pOH measures the OH- side, and because their product Kw is fixed, the two scales are locked together: pH + pOH = 14 at 25 degrees C. You can also go backwards — raise 10 to the minus pH to recover the concentration.

Definition

The logarithmic pH/pOH scale:

$\text{pH} = -\log[\text{H}^+]$ and $\text{pOH} = -\log[\text{OH}^-]$ (base-10 logs).
Taking $-\log$ of $K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$ gives $\text{pH} + \text{pOH} = 14$ at $25\,^{\circ}\text{C}$ .
The relationship is inverse: a rise of one pH unit means $[\text{H}^+]$ has fallen by a factor of 10.
To recover a concentration from pH: $[\text{H}^+] = 10^{-\text{pH}}$ , and likewise $[\text{OH}^-] = 10^{-\text{pOH}}$ .
Useful log values: $\log 2 = 0.301$ , $\log 3 = 0.477$ , $\log 5 = 0.699$ .

pH, pOH and their sum

\text{pH} = -\log[\text{H}^+], \quad \text{pOH} = -\log[\text{OH}^-], \quad \text{pH} + \text{pOH} = 14

\text{pH}negative log of hydrogen-ion concentration
\text{pOH}negative log of hydroxide-ion concentration
[\text{H}^+]hydrogen-ion concentration (mol/L)

Worked example

A solution has

[\text{H}^+] = 4.62 \times 10^{-4}\,\text{M}

. Find its pH.

$\text{pH} = -\log(4.62 \times 10^{-4}) = 4 - \log 4.62$ .
$\log 4.62 \approx 0.665$ .
$\text{pH} = 4 - 0.665 = 3.34$ .

Answer:

\text{pH} = 3.34

Practice this conceptself-check · 5 quick reps

Try it yourself

The pH of a vinegar sample is 3.76. Calculate its hydrogen-ion concentration in mol dm^-3.

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

1.
pH changes from 4 to 5. How does [H3O+] change?
2.
pH of a solution with [H+] = 2.2 x 10^-6 M?
3.
pOH of a solution is 11. Find [H+].
4.
[OH-] if pOH = 4.94?
5.
Write the relation between pH and pOH at 25 degrees C.

From the bank · past-year question

Example 2Ionic EquilibriaMODERATE

The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it in

oldm^{- 3}

[Q81 · 19 April Shift I · 2025]

pH + pOH = 14 only at 25 degrees C

The sum

\text{pH} + \text{pOH} = 14

comes from

K_w = 10^{-14}

, which is a **

25\,^{\circ}\text{C}

value**. At other temperatures

K_w

differs, so the sum is no longer exactly 14. In every MHT-CET numerical it is 14 — but the conceptual questions test whether you know why.

Higher pH means LOWER concentration

Because pH is a negative log, a bigger pH means a smaller

[\text{H}^+]

. Going from pH 4 to pH 5 the acidity falls 10-fold — the concentration decreases by 10 times, it does not increase.

Drill 6 more on ph, poh and the relation ph + poh = 14

Concept 3 of 4

pH of strong acids and strong bases

Intuition

A strong acid or base is fully dissociated, so the ion concentration is just the concentration you put in — times how many H+ or OH- ions each formula unit releases. Diprotic acids like H2SO4 give two H+, and bases like Ba(OH)2 give two OH-, so remember to double before taking the log. For a base, find pOH first, then subtract from 14.

Definition

Strong acids/bases dissociate completely:

For a strong monoprotic acid, $[\text{H}^+] = c$ ; take $\text{pH} = -\log c$ directly.
Multiply by the number of ionisable ions: a strong dibasic/diprotic acid gives $[\text{H}^+] = 2c$ (e.g. $\text{H}_2\text{SO}_4$ ); a diacidic base like $\text{Ba(OH)}_2$ gives $[\text{OH}^-] = 2c$ .
For a strong base, first find $\text{pOH} = -\log[\text{OH}^-]$ , then $\text{pH} = 14 - \text{pOH}$ .
Convert grams to molarity when needed: $c = \dfrac{\text{mass}/\text{molar mass}}{\text{volume in litres}}$ .

Strong acid / strong base

[\text{H}^+] = Z\,c \;\Rightarrow\; \text{pH} = -\log(Z\,c); \qquad \text{pH} = 14 - \text{pOH}

cmolar concentration of the acid or base
Znumber of H+ (or OH-) furnished per formula unit
\text{pOH}= -log[OH-], for a base

Worked example

Calculate the pH of a

0.005\,\text{M}

NaOH solution at 25 degrees C.

NaOH is a strong monoacidic base: $[\text{OH}^-] = 0.005 = 5 \times 10^{-3}\,\text{M}$ .
$\text{pOH} = -\log(5 \times 10^{-3}) = 3 - \log 5 = 3 - 0.699 = 2.30$ .
$\text{pH} = 14 - 2.30 = 11.70$ .

Answer:

\text{pH} = 11.7

Practice this conceptself-check · 5 quick reps

Try it yourself

Calculate the pH of a

0.05\,\text{M}

\text{H}_2\text{SO}_4

solution at 25 degrees C. (

\text{H}_2\text{SO}_4

is a strong diprotic acid.)

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

1.
pH of 0.01 M H2SO4 (strong diprotic)?
2.
pH of 1.36 x 10^-2 M perchloric acid (HClO4)?
3.
pOH of a millimolar Ba(OH)2 solution?
4.
pH of 0.002 M KOH?
5.
pH of 1 x 10^-4 M strong monoacidic base?

From the bank · past-year question

Example 3Ionic EquilibriaMODERATE

4 gram of NaOH is added in water to form 500 mL solution at 298 K . What is pH of solution? (Molar mass of

NaOH = 40\text{ }g{\text{ }mol}^{- 1}

)

[Q67 · 20 April Shift II · 2025]

Double for dibasic / diacidic

\text{H}_2\text{SO}_4

and other diprotic acids give two H+ per formula unit, so

[\text{H}^+] = 2c

, not

c

. Likewise

\text{Ba(OH)}_2

gives

[\text{OH}^-] = 2c

. For

0.01\,\text{M } \text{H}_2\text{SO}_4

the pH is

1.7

(from

0.02\,\text{M H}^+

), not

2.0

For a base, do not forget the 14 - pOH step

A strong base gives you

[\text{OH}^-]

directly, so you naturally compute pOH. The question almost always wants pH — finish with

\text{pH} = 14 - \text{pOH}

. Stopping at pOH is the most common careless loss of marks here.

Drill 8 more on ph of strong acids and strong bases

Concept 4 of 4

pH of weak acids and weak bases

Intuition

A weak acid or base only partly dissociates, so the ion concentration is a fraction of what you added. If the question gives a degree of dissociation (a percentage), just multiply: ion concentration = alpha times concentration. If instead it gives the dissociation constant Ka, use the square-root formula [H+] = sqrt(Ka times c). Everything after that is the same negative-log routine as before.

Definition

Weak electrolytes dissociate partially:

Given a degree of dissociation $\alpha$ (a percentage $\div 100$ ): $[\text{H}^+] = \alpha c$ for a weak acid, or $[\text{OH}^-] = \alpha c$ for a weak base.
Given a dissociation constant $K_a$ : $[\text{H}^+] = \sqrt{K_a\,c}$ , which also gives $\text{pH} = \tfrac12\!\left(pK_a - \log c\right)$ where $pK_a = -\log K_a$ .
The two are linked by Ostwald's dilution law $\alpha = \sqrt{K_a/c}$ , so $[\text{H}^+] = \alpha c = \sqrt{K_a c}$ .
For a weak base, form $\text{pOH} = -\log[\text{OH}^-]$ first, then $\text{pH} = 14 - \text{pOH}$ .
A weak dibasic acid still furnishes 2 ionisable H+, so $[\text{H}^+] = \alpha c Z$ with $Z = 2$ .

Weak acid / base

[\text{H}^+] = \alpha c = \sqrt{K_a\,c} \;\Rightarrow\; \text{pH} = \tfrac12\!\left(pK_a - \log c\right)

\alphadegree of dissociation (percentage / 100)
cmolar concentration of the weak electrolyte
K_aacid dissociation constant

Worked example

Calculate the pH of a

0.1\,\text{M}

weak monobasic acid whose dissociation constant

K_a

1 \times 10^{-5}

$[\text{H}^+] = \sqrt{K_a\,c} = \sqrt{(1 \times 10^{-5})(0.1)} = \sqrt{1 \times 10^{-6}}$ .
$[\text{H}^+] = 1 \times 10^{-3}\,\text{M}$ .
$\text{pH} = -\log(10^{-3}) = 3$ .

Answer:

\text{pH} = 3

Practice this conceptself-check · 4 quick reps

Try it yourself

Calculate the pH of a centimolar (

0.01\,\text{M}

) solution of a monoacidic weak base that is 10% dissociated.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
pH of 0.02 M monobasic acid that is 2% dissociated?
2.
pH of a weak dibasic acid, 2% dissociated in M/100 solution?
3.
pH of NaOH-type base, 2% dissociated in 0.01 M solution?
4.
Ostwald's dilution law for a weak acid: alpha = ?

From the bank · past-year question

Example 4Ionic EquilibriaMODERATE

Calculate

[\text{H}_3\text{O}^+]

in 0.02 M solution of monobasic acid if dissociation constant is

1.8\times10^{-5}

[Q56 · 3rd May 2nd Shift · 2023]

Apply the degree of dissociation before the log

For a weak electrolyte the reacting ion is only

\alpha c

, not the full concentration

c

. For

0.02\,\text{M}

acid at 2% dissociation,

[\text{H}^+] = 0.02 \times 0.02 = 4 \times 10^{-4}\,\text{M}

(pH

3.4

) — using the full

0.02\,\text{M}

gives a wrong pH of

1.7

sqrt(Ka c), not Ka c

When a dissociation constant is given,

[\text{H}^+] = \sqrt{K_a\,c}

— take the square root of the product. Forgetting the root (using

K_a c

itself) makes the concentration far too small and the pH far too high.

A weak dibasic acid still furnishes 2 H+

The 'weak' label controls

\alpha

; the 'dibasic' label controls the ion count. Keep both:

[\text{H}^+] = \alpha c Z

with

Z = 2

. For 2% dissociation in

M/100

[\text{H}^+] = 0.02 \times 0.01 \times 2 = 4 \times 10^{-4}\,\text{M}

, pH

3.398

Drill 5 more on ph of weak acids and weak bases

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Ionic product of water, Kw
Ionic product of water
$K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \quad (25\,^{\circ}\text{C})$
pH, pOH and the relation pH + pOH = 14
pH, pOH and their sum
$\text{pH} = -\log[\text{H}^+], \quad \text{pOH} = -\log[\text{OH}^-], \quad \text{pH} + \text{pOH} = 14$
pH of strong acids and strong bases
Strong acid / strong base
$[\text{H}^+] = Z\,c \;\Rightarrow\; \text{pH} = -\log(Z\,c); \qquad \text{pH} = 14 - \text{pOH}$
pH of weak acids and weak bases
Weak acid / base
$[\text{H}^+] = \alpha c = \sqrt{K_a\,c} \;\Rightarrow\; \text{pH} = \tfrac12\!\left(pK_a - \log c\right)$

Watch out for (9)

Kw = 10 to the minus 14 only at 25 degrees C
→ Ionic product of water, Kw
Divide into Kw, do not subtract
→ Ionic product of water, Kw
pH + pOH = 14 only at 25 degrees C
→ pH, pOH and the relation pH + pOH = 14
Higher pH means LOWER concentration
→ pH, pOH and the relation pH + pOH = 14
Double for dibasic / diacidic
→ pH of strong acids and strong bases
For a base, do not forget the 14 - pOH step
→ pH of strong acids and strong bases
Apply the degree of dissociation before the log
→ pH of weak acids and weak bases
sqrt(Ka c), not Ka c
→ pH of weak acids and weak bases
A weak dibasic acid still furnishes 2 H+
→ pH of weak acids and weak bases

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Ionic EquilibriaMODERATE

What is the pH of

10^{-8}\,\text{M HCl}

solution?

[Q92 · 10th May Shift 1 · 2023]

Example 2Ionic EquilibriaEASY

What is the pH of solution containing

4.62\times10^{-4}\ \text{M H}^+

ions?

[Q52 · 13th May Shift 2 · 2024]

Example 3Ionic EquilibriaEASY

What is the pOH of millimolar solution of

\text{Ba(OH)}_2

[Q64 · 9th May Shift 1 · 2023]

Example 4Ionic EquilibriaMODERATE

What is the value of pH of a NaOH solution that dissociates

2\%

in its 0.01 M solution?

[Q90 · 21 April Shift I · 2025]

Example 5Ionic EquilibriaEASY

Calculate the [OH⁻] if pOH of solution is 4.94

[Q100 · 4th May Shift 2 · 2023]

Drill every past-year question on this subtopic

24 questions from the bank — paginated, with cart and Word-export support.

Drill the 24 questions

Related notes

Acids, bases and salts — the pH scale →