MHT-CET Chemistry · Formula sheet

Ionic Equilibria formulas

18 formulas, 6 reference tables and 47 common traps for MHT-CET Chemistry Ionic Equilibria, grouped by subtopic.

Theories of Acids and Bases

Conjugate acid-base pairs

Conjugate base from an acid

\text{Acid} \;\rightleftharpoons\; \text{Conjugate base} + \text{H}^+

$\text{Acid}$ proton donor (the species with the extra H+)
$\text{Conjugate base}$ what remains after the acid loses one H+
$\text{H}^+$ the single proton that distinguishes the pair

The three theories: Arrhenius, Bronsted-Lowry and Lewis

Theory	Acid is	Base is	Example acid / base
Arrhenius	Gives $\text{H}^+$ in water	Gives $\text{OH}^-$ in water	$\text{HCl}$ / $\text{NaOH}$
Bronsted-Lowry	Proton $(\text{H}^+)$ donor	Proton $(\text{H}^+)$ acceptor	$\text{HCl}$ / $\text{NH}_3$ A Bronsted base ACCEPTS a proton — this is why $\text{NH}_3$ 'acts as a base when reacted with water' (it takes an $\text{H}^+$ to become $\text{NH}_4^+$ ).
Lewis	Electron-pair acceptor	Electron-pair donor	$\text{BF}_3$ / $\text{NH}_3$ $\text{BCl}_3$ is a Lewis acid but NOT a Bronsted acid — it accepts an electron pair yet has no proton to donate.

Each later theory contains the earlier one; a Lewis acid is the most general kind.

Amphoteric species

Species	Amphoteric?	Why
$\text{H}_2\text{O}$	Yes	Gives $\text{OH}^-$ (acid) and takes $\text{H}^+$ to form $\text{H}_3\text{O}^+$ (base) Water is the bank's default answer for 'which species is amphoteric'.
$\text{HCO}_3^-$	Yes	Loses $\text{H}^+$ to $\text{CO}_3^{2-}$ or gains $\text{H}^+$ to $\text{H}_2\text{CO}_3$
$\text{HCl}$	No	Only donates a proton (acid only)
$\text{NaOH}$	No	Only gives $\text{OH}^-$ (base only)
$\text{CH}_3\text{COOH}$	No	Acts only as an acid (donates a proton)

Amphoteric = can be either acid or base; water and

\text{HCO}_3^-

are the standard examples.

Common traps

Match the activity to the right theory

'Donate a pair of electrons' is the Lewis base definition. Do not confuse it with 'accept

\text{H}^+

' (that is a Bronsted base) or 'donate

\text{OH}^-

' (that is an Arrhenius base). A Lewis base gives an electron pair, not a proton or a hydroxide ion.

Lewis acid vs Bronsted acid

\text{BCl}_3

is a Lewis acid but not a Bronsted acid — it accepts an electron pair but has no

\text{H}^+

to donate.

\text{HNO}_3

and

\text{HSO}_4^-

are both Lewis and Bronsted acids because they can donate a proton too.

Conjugate base of a strong acid is weak

The conjugate base of a strong acid like

\text{HClO}_4

\text{ClO}_4^-

, which is a very weak base. In general, the stronger the acid, the weaker its conjugate base — don't expect

\text{ClO}_4^-

to behave like a strong base.

Pick species differing by ONE proton — not a random pair

\text{HCl}+\text{NH}_3 \rightleftharpoons \text{NH}_4^+ + \text{Cl}^-

, the pair is

\text{NH}_4^+/\text{NH}_3

, NOT

\text{NH}_4^+/\text{HCl}

\text{Cl}^-/\text{NH}_4^+

. A conjugate pair must be the same core species before and after losing one

\text{H}^+

Water is amphoteric — acetic acid is not

\text{H}_2\text{O}

is amphoteric because it both donates and accepts protons.

\text{CH}_3\text{COOH}

is a distractor here: it can lose a proton to a stronger base, but it does not readily accept one, so it acts only as an acid and is not amphoteric.

Ionic Equilibrium: Ka, Kb and Degree of Dissociation

Learn this subtopic in the notes

Degree of dissociation and percent dissociation

Percent dissociation and alpha from Ka

\text{\% dissociation} = \alpha \times 100 \qquad \alpha = \sqrt{\dfrac{K_a}{c}}

$\alpha$ degree of dissociation (fraction ionised, 0 to 1)
$K_a$ acid dissociation constant
$c$ initial molar concentration of the acid

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

Ostwald's dilution law

K_a = \dfrac{c\alpha^2}{1-\alpha} \approx c\alpha^2 \qquad (K_b \approx c\alpha^2 \text{ for a base})

$K_a, K_b$ acid / base dissociation constant
$c$ initial molar concentration
$\alpha$ degree of dissociation

Ion concentration of a weak acid or base

Hydrogen / hydroxide ion concentration

[\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
$c$ initial concentration of acid / base
$\alpha$ degree of dissociation
$K_a, K_b$ dissociation constant

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

Conjugate-pair relation and relative strength

K_a \times K_b = K_w \qquad \dfrac{\alpha_1}{\alpha_2} = \sqrt{\dfrac{K_{a1}}{K_{a2}}}

$K_a$ acid dissociation constant of an acid
$K_b$ base dissociation constant of its conjugate base
$K_w$ ionic product of water, $1.0\times 10^{-14}$ at 298 K
$K_{a1}, K_{a2}$ constants of the two acids being compared

Common traps

Percent is 100 times the fraction

\alpha

(the fraction) and percent dissociation differ by a factor of 100. If a question quotes

1.2\%

, use

\alpha = 1.2\times 10^{-2}

in the formula, not

1.2

. Reading

0.05\%

\alpha = 0.05

is the single commonest slip here.

alpha from Ka needs the DIVISION form

To get

\alpha

from

K_a

use

\alpha = \sqrt{K_a/c}

— the constant is divided by concentration under the root. Writing

\alpha = \sqrt{K_a \cdot c}

confuses it with the

[\text{H}^+]

formula and gives a wildly wrong answer.

Square the alpha, not just alpha

It is

K_a = c\alpha^2

, so the fraction is SQUARED. For

\alpha = 2\times 10^{-2}

\alpha^2 = 4\times 10^{-4}

— forgetting the square leaves you a factor of

\alpha

(often

10^{-2}

or more) too big.

Use the small-alpha approximation only when it is small

K_a \approx c\alpha^2

drops the

(1-\alpha)

denominator, which is safe only when

\alpha \ll 1

(a few percent). If a problem gives a large

\alpha

or explicitly wants the exact value, use

K_a = \dfrac{c\alpha^2}{1-\alpha}

instead.

c times alpha, not c times alpha squared

The ION concentration is

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

(first power of

\alpha

). It is the CONSTANT

K_a = c\alpha^2

that squares

\alpha

. Mixing the two — using

c\alpha^2

for the H+ concentration — is a frequent trap.

Multiply under the root for [H+]

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

multiplies

K_a

c

, whereas

\alpha = \sqrt{K_a/c}

divides. Same square root, opposite operation — check which quantity the question asks for before you decide.

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

For a conjugate pair the constants MULTIPLY to

K_w

K_a \times K_b = 10^{-14}

. Writing

K_a + K_b = K_w

is wrong. To get one constant from the other, divide

K_w

by the known constant.

Dilution raises alpha but leaves Ka fixed

Adding water lowers

c

, and since

\alpha = \sqrt{K_a/c}

a smaller

c

gives a LARGER

\alpha

— the acid ionises more. But

K_a

is unchanged: it only shifts with temperature, never with concentration.

pH, pOH and the Ionic Product of Water

Learn this subtopic in the notes

Ionic product of water, Kw

Ionic product of water

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

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

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

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

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}

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

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)

$\alpha$ degree of dissociation (percentage / 100)
$c$ molar concentration of the weak electrolyte
$K_a$ acid dissociation constant

Common traps

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.

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.

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.

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

Salt Hydrolysis

Learn this subtopic in the notes

Hydrolysis constant, degree of hydrolysis and pH

Hydrolysis constant, degree and salt pH

K_h = \dfrac{K_w}{K_a}\qquad h = \sqrt{\dfrac{K_h}{c}}\qquad \text{pH} = 7 + \tfrac{1}{2}\left(pK_a + \log c\right)

$K_h$ hydrolysis constant of the salt
$K_w$ ionic product of water ( $10^{-14}$ at 298 K)
$K_a$ ionisation constant of the weak parent acid
$K_b$ ionisation constant of the weak parent base
$h$ degree of hydrolysis (fraction hydrolysed)
$c$ molar concentration of the salt

The four salt types and their solution pH

Salt type	Example salt	Ion that hydrolyses	Solution / pH
Strong acid + strong base	$\text{NaCl}$ , $\text{KNO}_3$	None	Neutral, $\text{pH} = 7$ These salts are NOT hydrolysed — both ions come from strong parents and do not react with water.
Strong acid + weak base	$\text{NH}_4\text{Cl}$ , $\text{CuSO}_4$	Cation	Acidic, $\text{pH} < 7$
Weak acid + strong base	$\text{CH}_3\text{COONa}$ , $\text{Na}_2\text{CO}_3$	Anion	Basic, $\text{pH} > 7$
Weak acid + weak base	$\text{CH}_3\text{COONH}_4$ , $\text{NH}_4\text{CN}$	Both ions	Depends on $K_a$ vs $K_b$ $\text{NH}_4\text{CN}$ is basic because HCN ( $K_a \approx 4\times10^{-10}$ ) is a much weaker acid than $\text{NH}_4\text{OH}$ ( $K_b \approx 1.8\times10^{-5}$ ) is a base, so $K_b > K_a$ .

The pH is set by the WEAKER parent: weak base → acidic, weak acid → basic, both strong → neutral.

Which ion hydrolyses — classifying a given salt

Salt	Weak parent	Ion that hydrolyses	Litmus effect
$\text{CuCl}_2$	Weak base $\text{Cu(OH)}_2$	$\text{Cu}^{2+}$ (cation)	Acidic — blue litmus turns red
$\text{NH}_4\text{NO}_3$	Weak base $\text{NH}_4\text{OH}$	$\text{NH}_4^{+}$ (cation)	Acidic — blue litmus turns red
$\text{CH}_3\text{COONa}$	Weak acid $\text{CH}_3\text{COOH}$	$\text{CH}_3\text{COO}^{-}$ (anion)	Basic — red litmus turns blue
$\text{KCN}$	Weak acid HCN	$\text{CN}^{-}$ (anion)	Basic — red litmus turns blue
$\text{NaNO}_3$	None (both strong)	Neither ion	Neutral — no litmus change $\text{NaNO}_3$ , NaCl and KCl are neutral — they are classic 'no change' distractors in litmus questions.

Weak-base cation → acidic (blue→red); weak-acid anion → basic (red→blue).

Common traps

A strong-acid + strong-base salt does NOT hydrolyse

Salts like

\text{NaCl}

\text{KNO}_3

and

\text{Na}_2\text{SO}_4

come from a strong acid AND a strong base, so neither ion reacts with water. The solution stays neutral — there is no hydrolysis at all. In a 'which is NOT hydrolysed' question, this is the answer.

Match the salt to the RIGHT parents

\text{Na}_2\text{SO}_4

is a favourite distractor in 'salt of strong acid and weak base' questions: it is actually strong acid (

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

) + strong base (NaOH), so it is neutral, NOT acidic. Always write out both parents before classifying.

Only the ion of the WEAKER partner hydrolyses

The spectator ion (from the strong parent) does nothing. In

\text{CH}_3\text{COONa}

the

\text{Na}^{+}

is inert; it is the acetate anion (from weak

\text{CH}_3\text{COOH}

) that grabs

\text{H}^{+}

from water and leaves

\text{OH}^{-}

behind. Never let the strong-parent ion drive the pH.

Weak-base cation → acidic, not basic

Students sometimes assume an ammonium or copper salt is basic 'because it came from a base'. The opposite is true: a cation from a weak base (

\text{NH}_4^{+}

\text{Cu}^{2+}

) hydrolyses to give

\text{H}^{+}

, so the solution is acidic and turns blue litmus red.

Divide $K_w$ by the WEAK parent's constant

For a weak-acid salt use

K_h = K_w/K_a

; for a weak-base salt use

K_h = K_w/K_b

. Picking the wrong constant (or using the salt's own 'K') gives a wrong

K_h

and hence a wrong

h

. A smaller

K_a

(weaker acid) means a LARGER

K_h

and more hydrolysis.

The degree of hydrolysis carries a square root

h = \sqrt{K_h/c}

, not

K_h/c

. If

K_h/c = 10^{-8}

, then

h = 10^{-4}

, not

10^{-8}

. Take the square root at the end, exactly as in weak-acid degree-of-dissociation problems.

Buffer Solutions and the Henderson-Hasselbalch Equation

Learn this subtopic in the notes

Henderson-Hasselbalch equation — pH of an acidic buffer

Henderson-Hasselbalch (acidic buffer)

\text{pH} = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}

$pK_a$ acid dissociation exponent, $= -\log K_a$
$[\text{salt}]$ concentration of the conjugate base (the salt)
$[\text{acid}]$ concentration of the weak acid

Equal salt and acid — pH equals pKa

Buffer with equal salt and acid

[\text{salt}] = [\text{acid}] \;\Rightarrow\; \text{pH} = pK_a = -\log K_a

$K_a$ acid dissociation constant of the weak acid
$pK_a$ $= -\log K_a$

Basic buffers — the pOH form and converting to pH

Henderson-Hasselbalch (basic buffer)

\text{pOH} = pK_b + \log\dfrac{[\text{salt}]}{[\text{base}]} \qquad \text{pH} = 14 - \text{pOH}

$pK_b$ base dissociation exponent, $= -\log K_b$
$[\text{salt}]$ concentration of the salt (conjugate acid)
$[\text{base}]$ concentration of the weak base

What a buffer is and how to recognise one

Buffer type	Components	Example
Acidic buffer (pH < 7)	Weak acid + salt of that acid with a strong base	$\text{CH}_3\text{COOH} + \text{CH}_3\text{COONa}$ The salt supplies the conjugate base (acetate). A strong acid + salt is NOT a buffer.
Basic buffer (pH > 7)	Weak base + salt of that base with a strong acid	$\text{NH}_4\text{OH} + \text{NH}_4\text{Cl}$ The salt supplies the conjugate acid (ammonium). Note the components: weak base + its salt with a strong acid.
Blood buffer	Carbonic acid + its salt (bicarbonate)	$\text{H}_2\text{CO}_3 / \text{HCO}_3^-$ The bicarbonate buffer holds human blood pH near 7.4 — a frequently asked recall item.

A buffer always pairs a weak partner with its conjugate (from the salt).

Common traps

A strong acid + its salt is NOT a buffer

A buffer needs a weak acid (or weak base).

\text{HCl} + \text{NaCl}

has no weak partner to soak up added base, so it cannot resist pH change. Only weak-acid/salt or weak-base/salt pairs buffer.

Match the salt to the right partner

An acidic buffer's salt comes from the weak acid + strong base (giving the conjugate base). A basic buffer's salt comes from the weak base + strong acid (giving the conjugate acid). Mixing acetic acid with ammonium chloride is NOT a buffer — the salt is not the conjugate of the acid.

Ratio is salt over acid — don't invert it

The most common wrong answer inverts the ratio to

\log([\text{acid}]/[\text{salt}])

, which flips the sign of the log term and lands on the decoy option. Keep it

\log\dfrac{[\text{salt}]}{[\text{acid}]}

: more salt raises the pH. (Only in the

[\text{H}^+] = K_a\,[\text{acid}]/[\text{salt}]

form does acid go on top.)

Use concentrations directly — no volume conversion

When equal volumes are mixed, both the salt and acid are diluted by the same factor, so the ratio is unchanged. Plug the given molarities straight in; converting to moles first is extra work that changes nothing.

Equal concentrations means the log term is zero

When

[\text{salt}] = [\text{acid}]

\log\dfrac{[\text{salt}]}{[\text{acid}]} = \log 1 = 0

, so

\text{pH} = pK_a

. Don't waste time on the log — just compute

-\log K_a

. If the stem gives

K_a

rather than

pK_a

, converting it is the whole job.

Find pOH first, then subtract from 14

pK_b + \log([\text{salt}]/[\text{base}])

gives the pOH, not the pH. A basic buffer has pH

> 7

, so quoting the pOH (a number below 7) as the pH is the classic decoy. Always finish with

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

Use pKb for a base, pKa for an acid

Don't plug a weak base's data into the acidic form. A basic buffer (weak base + its salt) uses

pK_b

and the pOH equation; an acidic buffer uses

pK_a

and the pH equation directly.

Solubility Product (Ksp)

Learn this subtopic in the notes

Solubility product expression

General solubility product

K_{sp} = [A^{y+}]^x\,[B^{x-}]^y \qquad (A_x B_y \rightleftharpoons x\,A^{y+} + y\,B^{x-})

$K_{sp}$ solubility product (constant at a given temperature)
$[A^{y+}]$ molar concentration of the cation
$[B^{x-}]$ molar concentration of the anion
$x, y$ number of cations and anions in the formula (their exponents)

Solubility of a 1:1 (AB) salt: Ksp = S squared

AB salt: solubility and solubility product

K_{sp} = S^2 \qquad\Longleftrightarrow\qquad S = \sqrt{K_{sp}}

$S$ molar solubility (mol dm^-3), equals each ion concentration
$K_{sp}$ solubility product of the 1:1 salt

Solubility of AB2, A2B and A2B3 salts

AB2 / A2B salt: solubility and solubility product

K_{sp} = 4S^3 \qquad\Longleftrightarrow\qquad S = \sqrt[3]{\dfrac{K_{sp}}{4}}

$S$ molar solubility (mol dm^-3)
$K_{sp}$ solubility product of the AB2 / A2B salt
$4$ the factor 2^2 from the doubly-produced ion (2S)^2

Ksp from pH and from mass solubility

Molar solubility from mass solubility

S = \dfrac{m}{M} \qquad\text{and}\qquad [\text{OH}^-] = 10^{-\text{pOH}},\ \ \text{pOH} = 14 - \text{pH}

$S$ molar solubility (mol dm^-3)
$m$ mass solubility (g dm^-3)
$M$ molar mass of the salt (g mol^-1)
$[\text{OH}^-]$ hydroxide-ion concentration from the pH

Common ion effect on solubility

Solubility in presence of a common ion

S = \dfrac{K_{sp}}{C}

$S$ molar solubility of the salt in the common-ion solution
$K_{sp}$ solubility product of the salt
$C$ concentration of the common ion (from the added strong electrolyte)

Ksp in terms of solubility, by salt type

Salt type	Dissociation	Ksp in terms of S	Example salt
AB (1:1)	$AB \rightleftharpoons A^+ + B^-$	$K_{sp} = S^2$	AgCl, AgBr, CaCO3, NiS Most common type in the bank. Recover S by a single square root: $S = \sqrt{K_{sp}}$ .
AB2 or A2B (1:2)	$AB_2 \rightleftharpoons A^{2+} + 2B^-$	$K_{sp} = 4S^3$	PbI2, PbCl2, Ag2CrO4, Ba(OH)2 Recover S by $S = \sqrt[3]{K_{sp}/4}$ — divide by 4 first, then take the cube root.
AB3 or A3B (1:3)	$AB_3 \rightleftharpoons A^{3+} + 3B^-$	$K_{sp} = 27S^4$	Fe(OH)3-type, AlCl3-type Recover S by $S = \left(\dfrac{K_{sp}}{27}\right)^{1/4}$ .
A2B3 or A3B2 (2:3)	$A_2B_3 \rightleftharpoons 2A^{3+} + 3B^{2-}$	$K_{sp} = 108\,S^5$	Ca3(PO4)2, Al2(SO4)3 Factor is $2^2 \times 3^3 = 4 \times 27 = 108$ . Recover S by $S = \left(\dfrac{K_{sp}}{108}\right)^{1/5}$ .

The numerical factor is the product of each coefficient raised to its own power; the exponent on S is the total number of ions produced.

Common traps

Raise each ion to its own coefficient

For

\text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}

, the silver-ion concentration is squared:

K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}]

. Dropping the exponent (writing

[\text{Ag}^+][\text{CrO}_4^{2-}]

) is the most common slip.

The solid is left out

The solubility product involves only the aqueous ions. The concentration of the pure solid salt is taken as constant (activity 1) and never appears in

K_{sp}

AB2 is 4S cubed, not S squared

For

AB_2 \rightleftharpoons A^{2+} + 2B^-

[B^-] = 2S

, so

K_{sp} = (S)(2S)^2 = 4S^3

. Treating it as

S^2

(the AB formula) is the classic mistake — the coefficient 2 both squares AND multiplies in the factor 4.

Match the root to the exponent on S

From

K_{sp} = 4S^3

you take a cube root (after dividing by 4); from

K_{sp} = 108S^5

a fifth root. Taking a square root out of habit gives a wildly wrong S.

Take the square root — do not report Ksp as the solubility

For AgBr with

K_{sp} = 4.9 \times 10^{-13}

, the solubility is

\sqrt{K_{sp}} = 7.0 \times 10^{-7}

, NOT

4.9 \times 10^{-13}

. The distractor equal to

K_{sp}

itself is always offered — remember to root it.

Handle the power correctly under the root

Rewrite the mantissa so the exponent is even:

\sqrt{4.9 \times 10^{-13}} = \sqrt{49 \times 10^{-14}} = 7 \times 10^{-7}

. Splitting

10^{-13}

49 \times 10^{-14}

keeps the arithmetic clean.

Divide by the factor BEFORE taking the root

For

K_{sp} = 4S^3

, first isolate

S^3 = K_{sp}/4

, THEN cube-root. Cube-rooting

K_{sp}

directly (forgetting the 4) gives an answer too large by

\sqrt[3]{4} \approx 1.59

Group the power of ten into a multiple of the root

To cube-root

1.08 \times 10^{-7} / 4 = 2.7 \times 10^{-8}

, rewrite it as

27 \times 10^{-9}

; then

\sqrt[3]{27} = 3

and

\sqrt[3]{10^{-9}} = 10^{-3}

, giving

3 \times 10^{-3}

. Choosing a clean exponent makes the root exact.

Metal-ion concentration is HALF the hydroxide in M(OH)2

For

\text{Ba(OH)}_2 \rightleftharpoons \text{Ba}^{2+} + 2\text{OH}^-

, two hydroxides come from each barium, so

[\text{Ba}^{2+}] = \tfrac{1}{2}[\text{OH}^-]

. Using

[\text{Ba}^{2+}] = [\text{OH}^-]

doubles the answer.

Convert grams to moles before using Ksp

K_{sp}

relations use molar solubility. Plugging a mass solubility (g dm

^{-3}

) straight into

K_{sp} = S^2

is wrong — divide by the molar mass first to get S in mol dm

^{-3}

A common ion LOWERS solubility

The common ion effect always DECREASES the solubility of a sparingly soluble salt (and suppresses a weak acid's ionization). Expecting more dissolving is the classic error — Le Chatelier pushes the equilibrium the other way.

Use the common-ion concentration, not the square root

In pure water

S = \sqrt{K_{sp}}

for a 1:1 salt, but with a common ion at concentration

C

you use

S = K_{sp}/C

. The two answers differ by orders of magnitude — check whether a common ion is present before choosing the formula.

Ionic Equilibria formulas

Theories of Acids and Bases

Conjugate acid-base pairs

The three theories: Arrhenius, Bronsted-Lowry and Lewis

Amphoteric species

Common traps

Ionic Equilibrium: Ka, Kb and Degree of Dissociation

Degree of dissociation and percent dissociation

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

Ion concentration of a weak acid or base

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

Common traps

pH, pOH and the Ionic Product of Water

Ionic product of water, Kw

pH, pOH and the relation pH + pOH = 14

pH of strong acids and strong bases

pH of weak acids and weak bases

Common traps

Salt Hydrolysis

Hydrolysis constant, degree of hydrolysis and pH

The four salt types and their solution pH

Which ion hydrolyses — classifying a given salt

Common traps

Buffer Solutions and the Henderson-Hasselbalch Equation

Henderson-Hasselbalch equation — pH of an acidic buffer

Equal salt and acid — pH equals pKa

Basic buffers — the pOH form and converting to pH

What a buffer is and how to recognise one

Common traps

Solubility Product (Ksp)

Solubility product expression

Solubility of a 1:1 (AB) salt: Ksp = S squared

Solubility of AB2, A2B and A2B3 salts

Ksp from pH and from mass solubility

Common ion effect on solubility

Ksp in terms of solubility, by salt type

Common traps

More MHT-CET Chemistry formula sheets