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MHT-CET Chemistry · Ionic Equilibria

Ionic Equilibrium: Ka, Kb and Degree of Dissociation

A weak acid or base only partly splits into ions; the fraction that splits is the degree of dissociation, and Ostwald's dilution law ties it to the dissociation constant so you can find Ka, Kb, the ion concentration or the solution's concentration from one another.

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Why this matters

This is the arithmetic heart of Ionic Equilibria in MHT-CET Chemistry and one of its most reliable scoring blocks — almost every PYQ is a one-line plug-in of the same relation Ka = c times alpha squared, asked in four disguises: find Ka or Kb, find percent dissociation, find the H+ or OH- concentration, or find the solution's concentration. Learn the single formula and its rearrangements, keep percent versus fraction straight, and you can attempt every question here on sight.

Concept 1 of 4

Degree of dissociation and percent dissociation

Intuition

A weak acid or base is lazy: only a small slice of the molecules break into ions, the rest stay whole. The degree of dissociation, written alpha, is exactly that slice — the fraction of molecules that have split. Multiply alpha by 100 and you get percent dissociation; that single conversion is what most of these questions hinge on.

Definition

Degree of dissociation and percent dissociation:

  • Degree of dissociation α\alpha = fraction of the dissolved acid/base that has actually ionised, α=moles dissociatedmoles taken\alpha = \dfrac{\text{moles dissociated}}{\text{moles taken}}. It lies between 0 and 1.
  • Percent dissociation =α×100= \alpha \times 100. So 2%2\% means α=0.02=2×10−2\alpha = 0.02 = 2\times 10^{-2}, and 0.05%0.05\% means α=5×10−4\alpha = 5\times 10^{-4}.
  • For a weak acid α\alpha can also be found from the constant and concentration: α=Kac\alpha = \sqrt{\dfrac{K_a}{c}} (from Ostwald's law, next concept).
  • α\alpha is a pure number (no units); [H+][\text{H}^+], KaK_a and cc all carry units.

Percent dissociation and alpha from Ka

% dissociation=α×100α=Kac\text{\% dissociation} = \alpha \times 100 \qquad \alpha = \sqrt{\dfrac{K_a}{c}}
  • \alphadegree of dissociation (fraction ionised, 0 to 1)
  • K_aacid dissociation constant
  • cinitial molar concentration of the acid

Worked example

A weak acid HX has dissociation constant Ka=4×10−5K_a = 4 \times 10^{-5}. Find its percent dissociation in a 0.01 M solution.
  1. Find α\alpha from Ostwald's law: α=Kac=4×10−510−2\alpha = \sqrt{\dfrac{K_a}{c}} = \sqrt{\dfrac{4\times 10^{-5}}{10^{-2}}}.
  2. =4×10−3=40×10−4=6.32×10−2= \sqrt{4\times 10^{-3}} = \sqrt{40\times 10^{-4}} = 6.32\times 10^{-2}.
  3. Percent dissociation =α×100=6.32%= \alpha \times 100 = 6.32\%.
Answer:α≈6.3×10−2\alpha \approx 6.3\times 10^{-2}, i.e. about 6.3%6.3\% dissociation.
Practice this conceptself-check · 4 quick reps

Try it yourself

A weak monobasic acid has Ka=1.0×10−5K_a = 1.0 \times 10^{-5}. What is its percent dissociation in 0.1 M solution?

Practice — Level 1 (4 reps)

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

  1. 1.
    Convert a degree of dissociation α=0.015\alpha = 0.015 into percent dissociation.
  2. 2.
    Percent dissociation is 0.05%. Write α\alpha in scientific notation.
  3. 3.
    A 0.04 M acid has Ka=3.2×10−4K_a = 3.2\times 10^{-4}. Find α\alpha.
  4. 4.
    Does α\alpha carry any unit?

From the bank · past-year question

Example 1Ionic EquilibriaEASY
The degree of dissociation of 0.01 M solution of NH4OH\text{NH}_4\text{OH} is 4.2×10−24.2 \times 10^{-2}. What is the percent dissociation of NH4OH\text{NH}_4\text{OH}?

[Q80 · 15th May Shift 2 · 2023]

Percent is 100 times the fraction

α\alpha (the fraction) and percent dissociation differ by a factor of 100. If a question quotes 1.2%1.2\%, use α=1.2×10−2\alpha = 1.2\times 10^{-2} in the formula, not 1.21.2. Reading 0.05%0.05\% as α=0.05\alpha = 0.05 is the single commonest slip here.

alpha from Ka needs the DIVISION form

To get α\alpha from KaK_a use α=Ka/c\alpha = \sqrt{K_a/c} — the constant is divided by concentration under the root. Writing α=Ka⋅c\alpha = \sqrt{K_a \cdot c} confuses it with the [H+][\text{H}^+] formula and gives a wildly wrong answer.
Drill 5 more on degree of dissociation and percent dissociation

Concept 2 of 4

Ostwald's dilution law: Ka and Kb from alpha and concentration

Intuition

Ostwald's dilution law is the workhorse formula of this whole subtopic. Write the equilibrium for a weak acid, plug the equilibrium amounts into the Ka expression, and everything collapses to Ka = c times alpha squared once alpha is small. The identical relation with Kb works for a weak base — one formula covers both.

Definition

Ostwald's dilution law (weak monobasic acid HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-):

  • Exact form: Ka=cα21−αK_a = \dfrac{c\alpha^2}{1-\alpha}.
  • When the acid is weak, α\alpha is tiny so 1−α≈11-\alpha \approx 1, giving the working form Ka≈cα2K_a \approx c\alpha^2.
  • Rearranged: α=Kac\alpha = \sqrt{\dfrac{K_a}{c}} and c=Kaα2c = \dfrac{K_a}{\alpha^2}.
  • For a weak base BOH⇌B++OH−\text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^-, the identical law holds with KbK_b: Kb≈cα2K_b \approx c\alpha^2.
  • KaK_a and KbK_b are what stay constant as you dilute; α\alpha is what changes.

Ostwald's dilution law

Ka=cα21−α≈cα2(Kb≈cα2 for a base)K_a = \dfrac{c\alpha^2}{1-\alpha} \approx c\alpha^2 \qquad (K_b \approx c\alpha^2 \text{ for a base})
  • K_a, K_bacid / base dissociation constant
  • cinitial molar concentration
  • \alphadegree of dissociation

Worked example

A weak monobasic acid is 3%3\% dissociated in its 0.05 M solution. Calculate its dissociation constant.
  1. Convert percent to fraction: α=3%=3×10−2\alpha = 3\% = 3\times 10^{-2}.
  2. Use the working form Ka=cα2K_a = c\alpha^2 (α\alpha is small).
  3. Ka=0.05×(3×10−2)2=5×10−2×9×10−4K_a = 0.05 \times (3\times 10^{-2})^2 = 5\times 10^{-2} \times 9\times 10^{-4}.
  4. =45×10−6=4.5×10−5= 45\times 10^{-6} = 4.5\times 10^{-5}.
Answer:Ka=4.5×10−5K_a = 4.5\times 10^{-5}.
Practice this conceptself-check · 4 quick reps

Try it yourself

A weak monoacidic base dissociates to 2%2\% in 0.1 M solution. Calculate KbK_b.

Practice — Level 1 (4 reps)

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

  1. 1.
    Acetic acid is 1.2%1.2\% dissociated in 0.01 M solution. Find KaK_a.
  2. 2.
    A weak base is 5%5\% dissociated in 0.01 M solution. Find KbK_b.
  3. 3.
    Rearrange Ostwald's law (small α\alpha) to make c the subject.
  4. 4.
    Ka=1.8×10−5K_a = 1.8\times 10^{-5} and α=0.03\alpha = 0.03. Find the concentration.

From the bank · past-year question

Example 2Ionic EquilibriaEASY
Calculate the value of dissociation constant of weak monoacidic base if it dissociates to 2%2\% in 0.1 M solution?

[Q89 · 22 April Shift I · 2025]

Square the alpha, not just alpha

It is Ka=cα2K_a = c\alpha^2, so the fraction is SQUARED. For α=2×10−2\alpha = 2\times 10^{-2}, α2=4×10−4\alpha^2 = 4\times 10^{-4} — forgetting the square leaves you a factor of α\alpha (often 10−210^{-2} or more) too big.

Use the small-alpha approximation only when it is small

Ka≈cα2K_a \approx c\alpha^2 drops the (1−α)(1-\alpha) denominator, which is safe only when α≪1\alpha \ll 1 (a few percent). If a problem gives a large α\alpha or explicitly wants the exact value, use Ka=cα21−αK_a = \dfrac{c\alpha^2}{1-\alpha} instead.
Drill 11 more on ostwald's dilution law: ka and kb from alpha and concentration

Concept 3 of 4

Ion concentration of a weak acid or base

Intuition

Once you know how big a fraction ionised (alpha) and how much acid you started with (c), the amount of H+ actually floating around is just their product: c times alpha. If instead you are handed Ka and c, the same H+ concentration comes straight out as the square root of their product.

Definition

Ion concentration of a weak acid/base:

  • Directly from α\alpha: [H+]=cα[\text{H}^+] = c\alpha (weak acid) and [OH−]=cα[\text{OH}^-] = c\alpha (weak base).
  • From the constant and concentration: substitute α=Ka/c\alpha = \sqrt{K_a/c} to get [H+]=cKa/c=Ka c[\text{H}^+] = c\sqrt{K_a/c} = \sqrt{K_a\,c}.
  • Likewise for a base: [OH−]=Kb c[\text{OH}^-] = \sqrt{K_b\,c}.
  • Note the two square-root forms differ: α=Ka/c\alpha = \sqrt{K_a/c} (divide) but [H+]=Ka c[\text{H}^+] = \sqrt{K_a\,c} (multiply).

Hydrogen / hydroxide ion concentration

[H+]=cα=Ka c[OH−]=cα=Kb c[\text{H}^+] = c\alpha = \sqrt{K_a\,c} \qquad [\text{OH}^-] = c\alpha = \sqrt{K_b\,c}
  • [\text{H}^+]hydrogen (hydronium) ion concentration
  • [\text{OH}^-]hydroxide ion concentration
  • cinitial concentration of acid / base
  • \alphadegree of dissociation
  • K_a, K_bdissociation constant

Worked example

A monobasic acid is 0.04%0.04\% dissociated in a 0.05 M solution. Find [H3O+][\text{H}_3\text{O}^+].
  1. Convert percent to fraction: α=0.04%=4×10−4\alpha = 0.04\% = 4\times 10^{-4}.
  2. Use [H3O+]=cα[\text{H}_3\text{O}^+] = c\alpha.
  3. =0.05×4×10−4=5×10−2×4×10−4= 0.05 \times 4\times 10^{-4} = 5\times 10^{-2} \times 4\times 10^{-4}.
  4. =2.0×10−5 mol L−1= 2.0\times 10^{-5}\ \text{mol L}^{-1}.
Answer:[H3O+]=2.0×10−5 mol L−1[\text{H}_3\text{O}^+] = 2.0\times 10^{-5}\ \text{mol L}^{-1}.
Practice this conceptself-check · 4 quick reps

Try it yourself

A monoacidic base is 3%3\% ionised in its 0.04 M solution. Find [OH−][\text{OH}^-].

Practice — Level 1 (4 reps)

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

  1. 1.
    0.001 M acetic acid has α=0.134\alpha = 0.134. Find [H3O+][\text{H}_3\text{O}^+].
  2. 2.
    Give the formula for [OH−][\text{OH}^-] of a weak base in terms of KbK_b and c.
  3. 3.
    Which uses multiply and which uses divide: [H+][\text{H}^+] vs α\alpha from KaK_a?
  4. 4.
    A weak acid: Ka=10−6K_a = 10^{-6}, c=0.01 Mc = 0.01\ \text{M}. Find [H+][\text{H}^+].

From the bank · past-year question

Example 3Ionic EquilibriaEASY
What is the concentration of H3O+\text{H}_3\text{O}^+ ion in mol L−1^{-1} of 0.001 M acetic acid (α=0.134)(\alpha = 0.134)?

[Q60 · 9th May Shift 2 · 2024]

c times alpha, not c times alpha squared

The ION concentration is [H+]=cα[\text{H}^+] = c\alpha (first power of α\alpha). It is the CONSTANT Ka=cα2K_a = c\alpha^2 that squares α\alpha. Mixing the two — using cα2c\alpha^2 for the H+ concentration — is a frequent trap.

Multiply under the root for [H+]

[H+]=Ka c[\text{H}^+] = \sqrt{K_a\,c} multiplies KaK_a by cc, whereas α=Ka/c\alpha = \sqrt{K_a/c} divides. Same square root, opposite operation — check which quantity the question asks for before you decide.
Drill 3 more on ion concentration of a weak acid or base

Concept 4 of 4

Ka x Kb = Kw, relative strength and the effect of dilution

Intuition

A weak acid and its conjugate base are a team: the stronger one partner is at giving up its proton, the weaker the other is at taking it back — and their two constants always multiply to the same fixed number, Kw. Dilution, meanwhile, always pushes a weak electrolyte to ionise MORE, so alpha rises as you add water even though Ka stays put.

Definition

Conjugate pair, relative strength and dilution:

  • For a conjugate acid-base pair: Ka×Kb=Kw=1.0×10−14K_a \times K_b = K_w = 1.0\times 10^{-14} at 298 K. Note it is a product, not a sum.
  • A larger KaK_a means a stronger acid. Relative strength of two acids of the same concentration =α1α2=Ka1Ka2= \dfrac{\alpha_1}{\alpha_2} = \sqrt{\dfrac{K_{a1}}{K_{a2}}}.
  • Effect of dilution: from α=Ka/c\alpha = \sqrt{K_a/c}, lowering cc (adding water) increases α\alpha. Diluting a weak electrolyte raises its degree of dissociation.
  • KaK_a itself does not change with dilution or concentration — only with temperature. Only α\alpha responds to dilution.

Conjugate-pair relation and relative strength

Ka×Kb=Kwα1α2=Ka1Ka2K_a \times K_b = K_w \qquad \dfrac{\alpha_1}{\alpha_2} = \sqrt{\dfrac{K_{a1}}{K_{a2}}}
  • K_aacid dissociation constant of an acid
  • K_bbase dissociation constant of its conjugate base
  • K_wionic product of water, 1.0×10−141.0\times 10^{-14} at 298 K
  • K_{a1}, K_{a2}constants of the two acids being compared

Worked example

The dissociation constant of a weak acid is Ka=2.5×10−5K_a = 2.5\times 10^{-5}. Find KbK_b of its conjugate base at 298 K.
  1. For a conjugate pair, Ka×Kb=Kw=1.0×10−14K_a \times K_b = K_w = 1.0\times 10^{-14}.
  2. Kb=KwKa=1.0×10−142.5×10−5K_b = \dfrac{K_w}{K_a} = \dfrac{1.0\times 10^{-14}}{2.5\times 10^{-5}}.
  3. =0.4×10−9=4.0×10−10= 0.4\times 10^{-9} = 4.0\times 10^{-10}.
Answer:Kb=4.0×10−10K_b = 4.0\times 10^{-10}.
Practice this conceptself-check · 4 quick reps

Try it yourself

Two weak acids have Ka1=4×10−6K_{a1} = 4\times 10^{-6} and Ka2=1×10−6K_{a2} = 1\times 10^{-6} at equal concentration. What is their relative strength?

Practice — Level 1 (4 reps)

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

  1. 1.
    For a conjugate pair Ka×Kb=?K_a \times K_b = ? at 298 K.
  2. 2.
    On diluting a weak acid, does its degree of dissociation increase or decrease?
  3. 3.
    Does KaK_a change when you dilute the solution?
  4. 4.
    Ka=5×10−9K_a = 5\times 10^{-9}, α=5×10−4\alpha = 5\times 10^{-4}. Find the concentration.

From the bank · past-year question

Example 4Ionic EquilibriaMODERATE
Calculate the concentration of weak monobasic acid if its degree of dissociation and dissociation constant are 5.0×10−45.0\times10^{-4} and 5.0×10−95.0\times10^{-9} respectively.

[Q92 · 4th May Shift 1 · 2023]

Ka times Kb equals Kw — a product, not a sum

For a conjugate pair the constants MULTIPLY to KwK_w: Ka×Kb=10−14K_a \times K_b = 10^{-14}. Writing Ka+Kb=KwK_a + K_b = K_w is wrong. To get one constant from the other, divide KwK_w by the known constant.

Dilution raises alpha but leaves Ka fixed

Adding water lowers cc, and since α=Ka/c\alpha = \sqrt{K_a/c} a smaller cc gives a LARGER α\alpha — the acid ionises more. But KaK_a is unchanged: it only shifts with temperature, never with concentration.
Drill 1 more on ka x kb = kw, relative strength and the effect of dilution

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)

Watch out for (8)

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
Weak acid HX has dissociation constant 1×10−51\times10^{-5}. Calculate the percent dissociation in its 0.1 M solution.

[Q58 · 10th May Shift 2 · 2024]

Example 2Ionic EquilibriaMODERATE
A weak base is 5%5\% dissociated in its 0.01 M solution. Calculate the dissociation constant.

[Q61 · 23 April Shift I · 2025]

Example 3Ionic EquilibriaMODERATE
Calculate [H3O+][\text{H}_3\text{O}^+] of a monobasic acid if it is 0.04% dissociated in 0.05 M solution.

[Q62 · 2nd May Shift 2 · 2023]

Example 4Ionic EquilibriaEASY
Dissociation constant and degree of dissociation of weak acid are 1.8×10−51.8\times10^{-5} and 0.03 respectively. What will be the concentration of solution of weak acid?

[Q92 · 9th May Shift 1 · 2023]

Example 5Ionic EquilibriaMODERATE
Calculate 'α\alpha' for 0.1 M acetic acid [Ka=1.0×10−5][K_a=1.0\times10^{-5}]

[Q86 · 3rd May Shift 2 · 2023]

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