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In a laboratory, 90 mL of 0.05 M HCl is mixed with 120 mL of 0.03 M NaOH. Write the balanced chemical equation and calculate the pH of the resulting solution.

hard 5 marks Class 7 · Science · Acids, Bases and Salts
Answer
2.37

The correct answer is 2.37 — Compute pH: pH = –log₁₀(0.0042857) = 2.368 ≈ 2.37.

Step-by-step solution

  1. Write the neutralisation reaction: .
  2. Calculate moles of acid: 0.090 L × 0.05 mol L⁻¹ = 0.0045 mol.
  3. Calculate moles of base: 0.120 L × 0.03 mol L⁻¹ = 0.0036 mol.
  4. Neutralise the base with an equal amount of acid: 0.0036 mol of H⁺ reacts with 0.0036 mol of OH⁻, forming 0.0036 mol of water.
  5. Remaining excess acid: 0.0045 mol – 0.0036 mol = 0.0009 mol H⁺.
  6. Total volume of solution: 90 mL + 120 mL = 210 mL = 0.210 L.
  7. Concentration of excess H⁺: 0.0009 mol ÷ 0.210 L = 0.0042857 M.
  8. Compute pH: pH = –log₁₀(0.0042857) = 2.368 ≈ 2.37.
Common mistake: using incorrect volume conversion or neglecting excess reactant

Concept tested: Stoichiometry & pH calculation

Published · How this solution was prepared and checked

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