Quick Recap — Chemical Equilibrium

  • Dynamic equilibrium: forward and backward rates are equal; both reactions continue.
  • Law of mass action: for aA+bB⇌cC+dDaA+bB\rightleftharpoons cC+dD, Kc=[C]c[D]d[A]a[B]bK_c=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}; KpK_p uses partial pressures.
  • Kp=Kc(RT)ΔngK_p=K_c(RT)^{\Delta n_g}; pure solids and liquids are omitted; KK depends only on temperature.
  • Le Chatelier: the system shifts to oppose a change (added reactant →\to forward; higher pressure →\to fewer gas moles; catalyst →\to no shift).
  • Reaction quotient QQ: Q<KQ<K forward, Q=KQ=K equilibrium, Q>KQ>K backward.

Beyond-NCERT JEE Formulae

These push past the NCERT recap into the shortcuts JEE Main actually rewards. Each block tells you when to use it and flags the trap examiners like to set.

1. Gas equilibria: KpK_p, KcK_c and the quotient QQ

When to use: a gaseous reaction that hands you one equilibrium constant and asks for the other, or a snapshot of concentrations and asks which way it moves.

  • Switch constants with Kp=Kc(RT)ΔngK_p=K_c(RT)^{\Delta n_g}, where Δng\Delta n_g is (gaseous product moles) minus (gaseous reactant moles). Take R = 0.0821 L atm/(mol K) whenever KpK_p is wanted in atm.
  • Fix the direction from the reaction quotient QQ (same expression as KK, but current values): Q<KQ<K moves forward, Q=KQ=K is equilibrium, and Q>KQ>K moves backward.
  • For a one-into-two dissociation such as PCl5⇌PCl3+Cl2PCl_5\rightleftharpoons PCl_3+Cl_2 at total pressure P, the degree of dissociation obeys Kp=α2P1−α2K_p=\dfrac{\alpha^2 P}{1-\alpha^2}, which rearranges to α=KpKp+P\alpha=\sqrt{\dfrac{K_p}{K_p+P}}.

[JEE Tip] The sign of Δng\Delta n_g sets the size gap: ammonia synthesis has Δng=−2\Delta n_g=-2, making KpK_p much SMALLER than KcK_c. Solids and liquids never enter Δng\Delta n_g, and an inert gas added at constant volume shifts nothing.

2. Ionic equilibrium: the master conversions

When to use: any acid-base sum where you hop between [H+], [OH-], pH and pOH, or between an acid and its conjugate base.

  • Definition and water link: pH=−log⁡[H+]pH=-\log[H^+] and, at 25 C, pH+pOH=14pH+pOH=14 because Kw=[H+][OH−]=10−14K_w=[H^+][OH^-]=10^{-14}.
  • Conjugate pairs obey KaKb=KwK_a K_b=K_w, hence pKa+pKb=14pK_a+pK_b=14; a strong acid owns a feeble conjugate base, and the reverse.

[JEE Tip] For a very dilute strong acid (about 10−610^{-6} mol/L or less) you cannot ignore water's own [H+]; the pH then creeps toward 7 from below and can NEVER exceed 7 for an acid.

3. Weak acids and bases: Ostwald dilution law

When to use: a single weak monoprotic acid (or base) of known KaK_a (or KbK_b) and concentration C, valid while the degree of dissociation stays small.

  • Hydrogen-ion concentration and dissociation: [H+]=KaC[H^+]=\sqrt{K_a C} with α=KaC\alpha=\sqrt{\dfrac{K_a}{C}}; for a weak base swap KbK_b in to reach [OH-].
  • Because α\alpha scales as 1/C1/\sqrt{C}, dilution raises the fraction dissociated yet lowers [H+], so a diluted weak acid is MORE ionised but LESS acidic.

[JEE Tip] The KaC\sqrt{K_a C} shortcut assumes C is at least about 100 times KaK_a (so α\alpha stays under roughly 5 percent). When KaCK_a C is not tiny, fall back to the full Ka=Cα21−αK_a=\dfrac{C\alpha^2}{1-\alpha} or you overstate [H+].

4. Buffers: the Henderson-Hasselbalch equation

When to use: a weak acid with its salt (acidic buffer) or a weak base with its salt (basic buffer); the quickest route to a pH sitting near pKa.

  • Acidic buffer: pH=pKa+log⁡[salt][acid]pH=pK_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}.
  • Basic buffer: pOH=pKb+log⁡[salt][base]pOH=pK_b+\log\dfrac{[\text{salt}]}{[\text{base}]}, then convert with pH = 14 - pOH.
  • Buffer capacity peaks when [salt] equals [acid]; there the log term vanishes and pH = pKa.

[JEE Tip] Only the salt-to-acid RATIO appears, so diluting a buffer barely moves its pH: a classic distractor. Equal concentrations of salt and acid pin pH exactly at pKa.

5. Solubility product, common ion and hydrolysis

When to use: a sparingly soluble salt: converting KspK_{sp} to solubility s, or judging how a shared ion suppresses dissolving.

  • For a salt AxByA_xB_y the product is Ksp=(xs)x(ys)yK_{sp}=(xs)^x(ys)^y: type AB gives Ksp=s2K_{sp}=s^2, type A2BA_2B or AB2AB_2 gives Ksp=4s3K_{sp}=4s^3, and type AB3AB_3 gives Ksp=27s4K_{sp}=27s^4.
  • Common-ion effect: with the shared ion fixed by an added strong electrolyte (say 0.1 mol/L), hold that concentration constant inside KspK_{sp} and solve for the tiny leftover solubility.
  • Salt hydrolysis pH: a salt of weak acid and strong base is basic, pH=7+12(pKa+log⁡C)pH=7+\dfrac{1}{2}(pK_a+\log C); a salt of weak base and strong acid is acidic, pH=7−12(pKb+log⁡C)pH=7-\dfrac{1}{2}(pK_b+\log C).

[JEE Tip] A precipitate appears only once the ionic product QQ tops KspK_{sp}. When two solutions are mixed, always recompute each ion from the DILUTED volumes first: using the stock concentrations is the intended mistake.

Solved Examples - Beyond-NCERT Formulae

Example 1 - Convert Kc to Kp with a negative mole change. For N2(g)+3H2(g)⇌2NH3(g)N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g) the value Kc=0.50K_c=0.50 (mol/L)^-2 at 400 K. Find KpK_p (take R = 0.0821 L atm/(mol K)).

  • Count gas moles: Δng=2−(1+3)=−2\Delta n_g=2-(1+3)=-2.
  • Apply the bridge Kp=Kc(RT)ΔngK_p=K_c(RT)^{\Delta n_g} with RT=0.0821×400=32.84RT=0.0821\times400=32.84.

Kp=0.50×(32.84)−2=0.501078.5=4.64×10−4K_p=0.50\times(32.84)^{-2}=\dfrac{0.50}{1078.5}=4.64\times10^{-4}

Answer: Kp=4.64×10−4K_p=4.64\times10^{-4} atm^-2. The negative Δng\Delta n_g drives KpK_p far below KcK_c.

Example 2 - Which way does the reaction run (Q vs K)? For H2(g)+I2(g)⇌2HI(g)H_2(g)+I_2(g)\rightleftharpoons 2HI(g), Kc=50K_c=50 at 700 K. A vessel holds [H2] = 0.20 mol/L, [I2] = 0.20 mol/L and [HI] = 0.40 mol/L. Predict the direction.

  • Build the quotient with the same form as the constant:

Q=[HI]2[H2][I2]=(0.40)2(0.20)(0.20)=0.160.04=4Q=\dfrac{[HI]^2}{[H_2][I_2]}=\dfrac{(0.40)^2}{(0.20)(0.20)}=\dfrac{0.16}{0.04}=4

  • Compare: Q=4Q=4 is well below Kc=50K_c=50.

Answer: Since Q<KcQ<K_c, the reaction runs FORWARD, making more HI until QQ climbs to 50.

Example 3 - Degree of dissociation from Kp. PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g) reaches equilibrium at a total pressure of 1.6 atm with Kp=0.90K_p=0.90 atm. Find the degree of dissociation.

  • One mole gives two, so Kp=α2P1−α2K_p=\dfrac{\alpha^2 P}{1-\alpha^2}, i.e. α=KpKp+P\alpha=\sqrt{\dfrac{K_p}{K_p+P}}.
  • Substitute P = 1.6 atm and Kp=0.90K_p=0.90 atm:

α=0.900.90+1.6=0.902.5=0.36=0.60\alpha=\sqrt{\dfrac{0.90}{0.90+1.6}}=\sqrt{\dfrac{0.90}{2.5}}=\sqrt{0.36}=0.60

Answer: α=0.60\alpha=0.60, so PCl5PCl_5 is 60 percent dissociated at this pressure.

Example 4 - pH and dissociation of a weak acid (Ostwald). Find the pH and degree of dissociation of 0.10 mol/L acetic acid, Ka=1.8×10−5K_a=1.8\times10^{-5}.

  • Hydrogen-ion concentration from the Ostwald form:

[H+]=KaC=(1.8×10−5)(0.10)=1.8×10−6=1.34×10−3[H^+]=\sqrt{K_a C}=\sqrt{(1.8\times10^{-5})(0.10)}=\sqrt{1.8\times10^{-6}}=1.34\times10^{-3}

  • Then pH=−log⁡(1.34×10−3)=2.87pH=-\log(1.34\times10^{-3})=2.87.
  • Degree of dissociation:

α=KaC=1.8×10−50.10=1.8×10−4=1.34×10−2\alpha=\sqrt{\dfrac{K_a}{C}}=\sqrt{\dfrac{1.8\times10^{-5}}{0.10}}=\sqrt{1.8\times10^{-4}}=1.34\times10^{-2}

Answer: pH = 2.87 and α=1.34×10−2\alpha=1.34\times10^{-2} (about 1.3 percent dissociated), safely small so the shortcut is valid.

Example 5 - pH of an acidic buffer (Henderson). A buffer is 0.10 mol/L acetic acid (Ka=1.8×10−5K_a=1.8\times10^{-5}) plus 0.20 mol/L sodium acetate. Find the pH.

  • First pKa=−log⁡(1.8×10−5)=4.74pK_a=-\log(1.8\times10^{-5})=4.74.
  • Apply Henderson with [salt] = 0.20 and [acid] = 0.10:

pH=pKa+log⁡[salt][acid]=4.74+log⁡0.200.10=4.74+0.30=5.04pH=pK_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}=4.74+\log\dfrac{0.20}{0.10}=4.74+0.30=5.04

Answer: pH = 5.04. The excess acetate lifts the buffer about 0.30 unit above pKa.

Example 6 - pH of a basic buffer (Henderson via pOH). A buffer holds 0.20 mol/L ammonia (Kb=1.8×10−5K_b=1.8\times10^{-5}) and 0.10 mol/L ammonium chloride. Find the pH.

  • Here pKb=−log⁡(1.8×10−5)=4.74pK_b=-\log(1.8\times10^{-5})=4.74, with [salt] = 0.10 and [base] = 0.20.

pOH=pKb+log⁡[salt][base]=4.74+log⁡0.100.20=4.74−0.30=4.44pOH=pK_b+\log\dfrac{[\text{salt}]}{[\text{base}]}=4.74+\log\dfrac{0.10}{0.20}=4.74-0.30=4.44

  • Convert to pH: pH = 14 - 4.44.

Answer: pH = 9.56, correctly basic for an ammonia buffer.

Example 7 - Solubility from Ksp (A2B salt). Silver chromate Ag2CrO4Ag_2CrO_4 has Ksp=4×10−12K_{sp}=4\times10^{-12}. Find its molar solubility s in pure water.

  • Dissolving gives Ag2CrO4⇌2Ag++CrO42−Ag_2CrO_4\rightleftharpoons 2Ag^{+}+CrO_4^{2-}, so [Ag+] = 2s and [CrO4^2-] = s.
  • This is an A2BA_2B salt, so:

Ksp=(2s)2(s)=4s3=4×10−12  ⇒  s3=10−12K_{sp}=(2s)^2(s)=4s^3=4\times10^{-12}\;\Rightarrow\;s^3=10^{-12}

  • Take the cube root: s=1.0×10−4s=1.0\times10^{-4} mol/L.

Answer: s = 1.0×10−41.0\times10^{-4} mol/L, giving [Ag+] = 2.0×10−42.0\times10^{-4} mol/L and [CrO4^2-] = 1.0×10−41.0\times10^{-4} mol/L.

Example 8 - Common-ion effect on solubility. Find the solubility of Ag2CrO4Ag_2CrO_4 (Ksp=4×10−12K_{sp}=4\times10^{-12}) in 0.10 mol/L silver nitrate.

  • Silver nitrate fixes [Ag+] = 0.10 mol/L (it swamps the trace from the salt itself), while [CrO4^2-] = s'.

Ksp=[Ag+]2[CrO42−]=(0.10)2 s′=4×10−12K_{sp}=[Ag^{+}]^2[CrO_4^{2-}]=(0.10)^2\,s'=4\times10^{-12}

s′=4×10−121.0×10−2=4×10−10s'=\dfrac{4\times10^{-12}}{1.0\times10^{-2}}=4\times10^{-10}

Answer: s' = 4×10−104\times10^{-10} mol/L. The common ion cuts solubility from 1.0×10−41.0\times10^{-4} mol/L to 4×10−104\times10^{-10} mol/L, about 2.5×1052.5\times10^{5} times smaller.