Chemistry

1. 1.

   25.0 cm^3 of NaOH is neutralised by 20.0 cm^3 of 0.100 mol dm^-3 HCl. The reaction is 1:1. Calculate the NaOH concentration.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Convert each volume from cm^3 to dm^3 by dividing by 1000.
   2. Use the acid data to find the moles reacting, then transfer that quantity using the balanced equation.
   3. The same number of moles is spread over a larger volume of alkali, so its concentration is lower than the acid's.

   Marking points

   • Moles HCl = 0.100 x 0.0200 = 0.00200 mol.
   • Moles NaOH = 0.00200 mol by the 1:1 ratio.
   • Concentration = 0.00200/0.0250 = 0.0800 mol dm^-3.

   Examiner tip: Equal reacting moles do not imply equal concentrations when the volumes differ.

2. 2.

   For H2(g) + I2(g) <=> 2HI(g), equilibrium concentrations are [H2] = 0.20, [I2] = 0.10 and [HI] = 0.60 mol dm^-3. Write Kc, calculate its value and determine its units.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Stoichiometric coefficients become powers in the equilibrium expression, so the HI concentration is squared.
   2. The numerator is 0.36 and denominator 0.02, giving 18.
   3. Both numerator and denominator contain concentration squared. Their units therefore divide to one.

   Marking points

   • Kc = [HI]^2/([H2][I2]).
   • Substitution gives 0.60^2/(0.20 x 0.10).
   • Kc = 18.
   • The concentration units cancel, so this Kc has no units.

   Examiner tip: Use equilibrium concentrations, not initial concentrations, in Kc.

3. 3.

   A buffer contains 0.100 mol HA and 0.150 mol A^- with pKa = 4.76. Add 0.0100 mol HCl without changing volume significantly. Assuming complete reaction with A^-, calculate the new pH using pH = pKa + log10([A^-]/[HA]) and explain how buffering occurs.

   [5 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Apply the neutralisation reaction before the buffer equation: subtract 0.0100 from A^- and add it to HA.
   2. The new ratio is 1.2727; its base-ten logarithm is about 0.1047. Add this to 4.76.
   3. The initial pH was about 4.94, so it falls by about 0.08. A buffer resists change; it does not keep pH exactly constant.

   Marking points

   • A^- + H+ -> HA consumes added acid.
   • New amounts: A^- = 0.140 mol and HA = 0.110 mol.
   • The concentration ratio equals the mole ratio because both share the same volume.
   • pH = 4.76 + log10(0.140/0.110) = 4.86 to two decimal places.
   • A^- removes most added H+, so the pH changes relatively little while buffer capacity remains.

   Examiner tip: Do not use the original mole ratio after adding acid. Buffer capacity is finite.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.