alleryAs per your syllabus, the pH of a 10⁻² M HCl solution (strong acid, fully dissociated) is closest to which value?
- 12
- 7
- 2
- 4
Show answer and solution
Correct answer: C. 2
syllabus example: 10⁻² M HCl → [H⁺] = 10⁻² ⇒ pH = 2.
AP Intermediate 1st Year Equilibrium — practice questions — page 2
Explore 78 free Equilibrium questions in AP Intermediate 1st Year Chemistry. Open each question or exercise to read its content, reference answers and available explanation.
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Showing questions 31–60 of 78.
As per your syllabus, the pH of a 10⁻² M HCl solution (strong acid, fully dissociated) is closest to which value?Read the answer and worked solution →As per your syllabus, the pH of a 10⁻⁴ M NaOH solution (strong base, fully dissociated) is closest to which value?Read the answer and worked solution →As per your syllabus, addition of CH₃COONa to a solution of CH₃COOH does what to the degree of ionisation of acetic acid?Read the answer and worked solution →As per your syllabus, the pH of an acidic buffer is given by which expression?Read the answer and worked solution →As per your syllabus, when [salt] = [acid] in a buffer, the pH of the buffer equals which quantity?Read the answer and worked solution →As per your syllabus, a buffer made from acetic acid and sodium acetate (pKa ≈ 4.76) at equimolar concentrations gives pH closest to which value?Read the answer and worked solution →As per your syllabus, on DILUTING a buffer solution, what happens to the buffer pH (to first approximation)?Read the answer and worked solution →As per your syllabus, for a sparingly soluble salt MX(s) ⇌ M⁺(aq) + X⁻(aq) with molar solubility S, the SOLUBILITY PRODUCT Ksp equals which?Read the answer and worked solution →As per your syllabus, for a salt M₂X(s) ⇌ 2 M⁺(aq) + X²⁻(aq) of molar solubility S, the SOLUBILITY PRODUCT Ksp equals which?Read the answer and worked solution →As per your syllabus, for a salt A₂X₃(s) ⇌ 2 A³⁺(aq) + 3 X²⁻(aq) of molar solubility S, the SOLUBILITY PRODUCT Ksp equals which?Read the answer and worked solution →As per your syllabus, comparing the ionic product Qsp with Ksp, PRECIPITATION of the salt occurs under which condition?Read the answer and worked solution →As per your syllabus, what is the molar SOLUBILITY S of barium sulphate at 298 K, given Ksp(BaSO₄) = 1.1 × 10⁻¹⁰?Read the answer and worked solution →In an exothermic reaction, as temperature INCREASES, the equilibrium constant Kc:Read the answer and worked solution →According to the Bronsted-Lowry theory, an acid is:Read the answer and worked solution →A solution with pH = 4 is:Read the answer and worked solution →A BUFFER solution is best described as:Read the answer and worked solution →A solution of NH4Cl (ammonium chloride) in water is:Read the answer and worked solution →If the solubility of AgCl is s mol/L in pure water, then its Ksp is:Read the answer and worked solution →Adding NaCl (sodium chloride) to a saturated AgCl solution will:Read the answer and worked solution →At dynamic equilibrium:Read the answer and worked solution →Equilibrium can be established in:Read the answer and worked solution →For N₂ + 3H₂ ⇌ 2NH₃, Kc equals:Read the answer and worked solution →Adding heat to an exothermic equilibrium shifts it:Read the answer and worked solution →Kw at 25°C equals:Read the answer and worked solution →A buffer solution:Read the answer and worked solution →For the gas-phase reaction $2\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{SO_3(g)}$, the value of $\Delta n$ (used in $K_p = K_c (RT)^{\Delta n}$) isRead the answer and worked solution →Match each acid to its conjugate base.Read the answer and worked solution →NEET · Chemistry · Le Chatelier — Shifts Right or LeftRead the answer and worked solution →NEET · Chemistry · Acid–Base pH — Gap FillRead the answer and worked solution →NEET · Chemistry · True or False — EquilibriumRead the answer and worked solution →
Worked practice questions
As per your syllabus, the pH of a 10⁻⁴ M NaOH solution (strong base, fully dissociated) is closest to which value?
- 14
- 4
- 7
- 10
Show answer and solution
Correct answer: D. 10
syllabus example: [OH⁻] = 10⁻⁴ ⇒ pOH = 4 ⇒ pH = 14 − 4 = 10.
As per your syllabus, addition of CH₃COONa to a solution of CH₃COOH does what to the degree of ionisation of acetic acid?
- Leaves it unchanged
- Sets it to 100%
- Decreases it
- Increases it
Show answer and solution
Correct answer: C. Decreases it
Your syllabus: common-ion (acetate from salt) shifts CH₃COOH ⇌ H⁺ + CH₃COO⁻ to the left, suppressing ionisation.
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