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EAPCET Chemistry · Intermediate first year

Chemical Equilibrium and Acids-Bases

This chapter covers equilibrium constants, Le Chatelier’s principle, acid–base theories, pH, buffers and solubility product.

Subject
Chemistry
Taught in
Intermediate first year
Chemistry in the paper
40 of 160 questions

Key formulas and results

  • Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}
  • pH=−log⁡[HX+]\text{pH} = -\log[\ce{H+}]; pH+pOH=14\text{pH} + \text{pOH} = 14 at 25 °C
  • Kw=1×10−14K_w = 1 \times 10^{-14} at 25 °C
  • Buffer pH (Henderson): pH=pKa+log⁡([salt][acid])\text{pH} = \text{p}K_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right)
  • For a salt AB, Ksp=s2K_{sp} = s^2
  • Degree of dissociation from vapour density: α=D−d(n−1)d\alpha = \frac{D - d}{(n - 1)d}, where DD is the initial vapour density, dd the equilibrium vapour density and nn the moles of products from one mole of reactant
  • van ’t Hoff equation: log⁡K2K1=ΔH∘2.303R×(1T1−1T2)\log\frac{K_2}{K_1} = \frac{\Delta H^\circ}{2.303R} \times \left(\frac{1}{T_1} - \frac{1}{T_2}\right)
  • Ostwald dilution law: Ka=Cα21−α≈Cα2K_a = \frac{C\alpha^2}{1 - \alpha} \approx C\alpha^2, so α=KaC\alpha = \sqrt{\frac{K_a}{C}} and [HX+]=KaC[\ce{H+}] = \sqrt{K_a C}
  • pH of a weak acid: pH=12(pKa−log⁡C)\text{pH} = \tfrac{1}{2}(\text{p}K_a - \log C); pOH of a weak base: pOH=12(pKb−log⁡C)\text{pOH} = \tfrac{1}{2}(\text{p}K_b - \log C)
  • Salt of a weak acid and strong base: pH=12(pKw+pKa+log⁡C)\text{pH} = \tfrac{1}{2}(\text{p}K_w + \text{p}K_a + \log C)
  • Salt of a strong acid and weak base: pH=12(pKw−pKb−log⁡C)\text{pH} = \tfrac{1}{2}(\text{p}K_w - \text{p}K_b - \log C)
  • Salt of a weak acid and weak base: pH=12(pKw+pKa−pKb)\text{pH} = \tfrac{1}{2}(\text{p}K_w + \text{p}K_a - \text{p}K_b)
  • Basic buffer: pOH=pKb+log⁡([salt][base])\text{pOH} = \text{p}K_b + \log\left(\frac{[\text{salt}]}{[\text{base}]}\right)

What this chapter covers

  • Law of chemical equilibrium
  • Kp and Kc
  • Le Chatelier’s principle
  • Acid–base theories
  • pH and ionic product of water
  • Buffers and solubility product

How it is usually asked

  • Converting Kp and Kc
  • Effect of changes on an equilibrium
  • pH of acids, bases and buffers
  • Solubility from Ksp

Understanding Chemical Equilibrium and Acids-Bases

A reversible reaction reaches equilibrium when the forward and backward rates become equal; concentrations then stay constant, though both reactions continue. The equilibrium constant Kc is the ratio of product to reactant concentrations, each raised to its coefficient, with pure solids and liquids left out. Kp uses partial pressures, and Kp = Kc(RT)^Δn. K changes only with temperature; adding a catalyst speeds up both directions equally and does not change K.

Le Chatelier’s principle predicts how an equilibrium shifts when disturbed: it moves to oppose the change. Adding a reactant pushes it forward; raising pressure favours the side with fewer gas moles; raising temperature favours the endothermic direction and is the only change that alters K. Adding an inert gas at constant volume changes nothing, but at constant pressure it acts like lowering pressure.

For acids and bases, Arrhenius focuses on H⁺ and OH⁻ in water, Brønsted–Lowry on proton transfer (conjugate pairs), and Lewis on electron-pair acceptance. pH = −log[H⁺], and Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 25 °C. A buffer — a weak acid with its salt, or a weak base with its salt — resists pH change, with pH given by the Henderson equation. Ksp links solubility s to ion concentrations: Ksp = s² for AB, 4s³ for AB₂.

Worked example

EAPCET-style question

A buffer contains 0.20 M CHX3COONa\ce{CH3COONa} and 0.02 M CHX3COOH\ce{CH3COOH}. If pKa\text{p}K_a of acetic acid is 4.74, the pH of the buffer is:

  1. (1)3.74
  2. (2)4.74
  3. (3)5.74
  4. (4)6.74

Solution

  1. Use the Henderson equation for an acidic buffer: pH=pKa+log⁡([salt][acid])\text{pH} = \text{p}K_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right).
  2. [salt][acid]=0.200.02=10\frac{[\text{salt}]}{[\text{acid}]} = \frac{0.20}{0.02} = 10.
  3. log⁡10=1\log 10 = 1.
  4. pH=4.74+1=5.74\text{pH} = 4.74 + 1 = 5.74.
  5. 3.74 results from inverting the ratio — salt goes on top for an acidic buffer.

Answer: 5.74

Shortcuts and results to remember

  • Reversing a reaction inverts K, multiplying the equation by n raises K to the power n, and adding two reactions multiplies their K values.
  • Relative strength of two weak acids at the same concentration =Ka1Ka2= \sqrt{\frac{K_{a_1}}{K_{a_2}}}.
  • For a very dilute strong acid (below about 10−6 M10^{-6}\ \text{M}) include the 10−7 M10^{-7}\ \text{M} HX+\ce{H+} from water: 10−8 M10^{-8}\ \text{M} HCl has pH slightly below 7 (about 6.98), never 8.
  • Diluting a strong acid 10 times raises its pH by 1 (pHnew=pHold+log⁡V2V1\text{pH}_{\text{new}} = \text{pH}_{\text{old}} + \log\frac{V_2}{V_1}) as long as the pH stays well below 7.
  • On mixing a strong acid with a strong base, the excess ion concentration is N1V1−N2V2V1+V2\frac{N_1V_1 - N_2V_2}{V_1 + V_2}.

More practice questions

Practice question 1

The dissociation constant of acetic acid is 1.8×10−51.8 \times 10^{-5}. The concentration of acetate ion in a solution that is 0.01 M in acetic acid and 0.1 M in HCl is

  1. (1)1.8×10−5 M1.8 \times 10^{-5}\ \text{M}
  2. (2)1.8×10−6 M1.8 \times 10^{-6}\ \text{M}
  3. (3)1.8×10−7 M1.8 \times 10^{-7}\ \text{M}
  4. (4)4.2×10−4 M4.2 \times 10^{-4}\ \text{M}

Solution

  1. HCl is a strong acid, so [HX+]≈0.1 M[\ce{H+}] \approx 0.1\ \text{M} (common ion)
  2. The common ion suppresses acetic acid dissociation, so [CHX3COOH]≈0.01 M[\ce{CH3COOH}] \approx 0.01\ \text{M}
  3. Ka=[HX+][CHX3COOX−][CHX3COOH]K_a = \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}, so [CHX3COOX−]=Ka×0.010.1[\ce{CH3COO-}] = K_a \times \frac{0.01}{0.1}
  4. [CHX3COOX−]=1.8×10−5×0.1=1.8×10−6 M[\ce{CH3COO-}] = 1.8 \times 10^{-5} \times 0.1 = 1.8 \times 10^{-6}\ \text{M}

Answer: 1.8×10−6 M1.8 \times 10^{-6}\ \text{M}

Practice question 2

The pH of a 0.1 M solution of a weak monobasic acid with Ka=1×10−5K_a = 1 \times 10^{-5} is

  1. (1)2
  2. (2)3
  3. (3)4
  4. (4)5

Solution

  1. pH=12(pKa−log⁡C)\text{pH} = \tfrac{1}{2}(\text{p}K_a - \log C)
  2. pKa=5\text{p}K_a = 5 and log⁡0.1=−1\log 0.1 = -1
  3. pH=12(5+1)=3\text{pH} = \tfrac{1}{2}(5 + 1) = 3

Answer: 3

Common mistakes to avoid

  • Putting the solids or pure liquids such as CaCO₃(s) or H₂O(l) into the expression for K — they are left out.
  • Thinking a catalyst or a change in concentration alters K — only temperature changes the value of K.
  • Using Ksp = s² for every salt — for AB₂ such as CaF₂ it is 4s³, and for A₂B₃ it is 108s⁵.
  • Taking pH + pOH = 14 at all temperatures — it holds only at 25 °C, because Kw increases with temperature.

How to prepare it

Use only gaseous moles for Δn in Kp = Kc(RT)^Δn.

Once the formulas feel familiar, test them the way the exam does: practise Chemical Equilibrium and Acids-Bases questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Chemistry chapters

See every chapter in the EAPCET Chemistry syllabus.

Practise Chemical Equilibrium and Acids-Bases chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Chemical Equilibrium and Acids-Bases part of the EAPCET Chemistry syllabus?
Yes. Chemical Equilibrium and Acids-Bases is taught in Intermediate first year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Chemical Equilibrium and Acids-Bases for EAPCET?
Use only gaseous moles for Δn in Kp = Kc(RT)^Δn.
How many Chemistry questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.
What mistakes do students make in Chemical Equilibrium and Acids-Bases?
Putting the solids or pure liquids such as CaCO₃(s) or H₂O(l) into the expression for K — they are left out. Thinking a catalyst or a change in concentration alters K — only temperature changes the value of K. Using Ksp = s² for every salt — for AB₂ such as CaF₂ it is 4s³, and for A₂B₃ it is 108s⁵. Taking pH + pOH = 14 at all temperatures — it holds only at 25 °C, because Kw increases with temperature.