Chemistry

Making hydrochloric acid using sulfuric acid and calcium chloride solution

Step-by-step chemistry solution: Making hydrochloric acid using sulfuric acid and calcium chloride solution

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1. What the student is asking (in plain language)

The student wants to prepare hydrochloric acid (HCl) in the lab by:

  1. Electrolysing a copper(II) sulfate solution in order to generate sulfuric acid (H₂SO₄).
  2. Adding that sulfuric acid to a calcium‑chloride solution so that calcium sulfate (CaSO₄) precipitates and the remaining liquid is “hydrochloric acid”.

The student is also wondering:

  • How “pure” the resulting HCl solution will be.
  • Whether any leftover calcium sulfate (or other salts) will interfere with later reactions, such as the preparation of copper(II) chloride.

2. Step‑by‑step analysis

Below we examine every chemical step, the equilibria involved, the practical yields, and the expected contaminants.

Step 1 – Electrolysis of CuSO₄·5H₂O solution

Half‑reaction (acidic medium) Electrode Expected product
Anode (oxidation) 2 H₂O → O₂ + 4 H⁺ + 4 e⁻ (E° ≈ +1.23 V) Oxygen gas, acidifies the solution
Cathode (reduction) Cu²⁺ + 2 e⁻ → Cu(s) (E° ≈ +0.34 V) Metallic copper deposits on the cathode
  • What actually happens?
    • The copper ions are reduced to copper metal; they do not become sulfuric acid.
    • Water oxidation supplies oxygen; the solution becomes slightly more acidic because H⁺ is produced, but the acid formed is simply dilute sulfuric acid that was already present (the CuSO₄ solution already contains the sulfate anion).
  • Conclusion: Electrolysis does not create a useful amount of new H₂SO₄. At best you end up with a dilute mixture of H₂SO₄, Cu²⁺ (remaining), and a small amount of copper metal.

Step 2 – Reaction of H₂SO₄ with CaCl₂

The intended net reaction is

[ \underbrace{\mathrm{CaCl_2 (aq) + H_2SO_4 (aq) \longrightarrow CaSO_4 (s) + 2\,HCl (aq)}}_{\text{overall stoichiometry}} ]

2.1 Thermodynamics & solubility

  • Calcium sulfate is sparingly soluble. Its solubility product is

[ K_{sp}(\text{CaSO}_4) = [\text{Ca}^{2+}][\text{SO}_4^{2-}] \approx 2.4\times10^{-5}\;(25^{\circ}\text{C}) ]

  • Hydrochloric acid is a strong acid; in water it dissociates completely:

[ \mathrm{HCl \rightarrow H^{+} + Cl^{-}} ]

  • The equilibrium for the overall reaction is essentially dictated by the low solubility of CaSO₄. As soon as the ionic product ([\text{Ca}^{2+}][\text{SO}4^{2-}]) exceeds (K{sp}), CaSO₄ precipitates and the reaction is driven forward.

2.2 How much HCl can you actually obtain?

Assume you start with 0.10 mol L⁻¹ solutions of both reagents (a typical laboratory concentration that is easy to prepare).

Initial moles (per litre) CaCl₂ H₂SO₄
0.10 mol 0.10 0.10
  • Stoichiometry: 1 mol CaCl₂ consumes 1 mol H₂SO₄ to give 2 mol HCl.
  • If both reagents are present in equal amounts, all of the calcium and sulfate will be used (to the extent allowed by the solubility of CaSO₄).

Maximum theoretical HCl concentration (ignoring solubility limits) =

[ [HCl]_{\text{max}} = 2 \times 0.10\; \text{mol L}^{-1}=0.20\; \text{mol L}^{-1} ]

0.20 M HCl corresponds to ≈ 0.73 % (w/w) HCl, far weaker than commercial “hydrochloric acid” (usually 6–12 M).

2.3 Why the solution will still contain other ions

Even after filtration of the solid CaSO₄, the filtrate contains:

Species Origin
Cl⁻ From CaCl₂ (desired)
H⁺ From HCl (desired)
SO₄²⁻ Any sulfate that remained dissolved (CaSO₄ is only partially removed)
Ca²⁺ Small amount that stays in solution because CaSO₄ does not precipitate completely
Cu²⁺ Left over from the original CuSO₄ solution (electrolysis does not remove it)
Trace metal ions (Fe, Ni, etc.) From electrodes or glassware if they corrode

The solubility of CaSO₄ at 25 °C is ≈ 0.015 M (≈ 1.5 g L⁻¹). Thus, even after filtration, the solution will still contain about 0.015 M Ca²⁺ and 0.015 M SO₄²⁻, i.e. ≈ 1.5 % of the original sulfate remains in solution.

Step 3 – Filtration and the resulting “hydrochloric acid”

  • Filtration removes the bulk of the solid CaSO₄, but micron‑sized particles can pass through unless you use a very fine filter (e.g., 0.2 µm membrane).
  • The filtrate is not pure HCl; it is an aqueous solution containing a mixture of chloride, sulfate, calcium, and residual copper.

Step 4 – Using the “HCl” to make CuCl₂

If you now add copper metal or copper(II) oxide to this mixture to precipitate CuCl₂, the sulfate ions will stay in solution and may affect crystal growth:

  • CaSO₄ (or dissolved Ca²⁺/SO₄²⁻) can co‑precipitate as a hydrated “jelly” that you observed previously.
  • Sulfate ions can complex Cu²⁺ (forming CuSO₄) and lower the amount of CuCl₂ that actually crystallises.
  • The presence of Cu²⁺ already in the solution (from the original CuSO₄) makes it difficult to know the exact stoichiometry of the CuCl₂ you finally obtain.

3. Final answer – What you can realistically expect

Aspect Expected outcome
HCl concentration ≈ 0.1–0.2 M (≈ 0.4–0.8 % w/w) – a very dilute acid
Purity Not pure. The solution will contain appreciable amounts of sulfate (≈ 0.015 M), calcium (≈ 0.015 M), and copper(II) ions carried over from the electrolysis step.
Solid residues Fine CaSO₄ particles that are hard to remove completely; they can form a gelatinous precipitate that interferes with later crystallisations.
Suitability for making CuCl₂ Poor. The co‑existing sulfate and calcium ions will hinder the growth of clean copper(II) chloride crystals and will introduce contaminants into the product.
Overall feasibility The route is inefficient and yields a low‑purity acid. A more reliable laboratory preparation of HCl is the Mannheim process: (\mathrm{NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl\,(g)}) followed by gas absorption, or simply purchasing commercial HCl.

4. Common mistakes & how to avoid them

Mistake Why it’s wrong How to correct it
Assuming electrolysis creates sulfuric acid Electrolysis of CuSO₄ only plates out copper; water oxidation gives O₂ and H⁺, not H₂SO₄. Use a genuine source of H₂SO₄ (commercial acid) or a different method (e.g., oxidation of SO₂).
Treating the CaCl₂ + H₂SO₄ reaction as a “complete conversion to HCl” CaSO₄ is only sparingly soluble; the equilibrium leaves a lot of sulfate and calcium in solution. Expect only a dilute HCl solution and account for residual sulfate/calcium.
Filtering and assuming the filtrate is pure HCl Fine CaSO₄ particles pass through typical filter paper, and dissolved ions remain. Use a 0.2 µm membrane filter and, if high purity is needed, perform ion‑exchange or distillation.
Neglecting the presence of Cu²⁺ Copper ions remain after electrolysis and will stay in the final solution. Remove copper by precipitation as Cu(OH)₂ (add NaOH) or by ion‑exchange before the acid‑generation step.
Trying to grow CuCl₂ crystals from a solution that still contains sulfate Sulfate interferes with crystal morphology and can produce a gelatinous precipitate. Either purify the HCl (e.g., by distillation) or use a different acid source (conc. HCl) for CuCl₂ preparation.
Expecting a high‑concentration HCl The stoichiometric limit is low because you start from dilute aqueous reagents; HCl is a gas at room temperature and does not stay dissolved at high concentrations without special equipment. Use a gas‑generating method (NaCl + conc. H₂SO₄) and absorb the gas in water to obtain stronger HCl solutions.

Bottom line:
The described two‑step route will give you a very dilute, contaminated hydrochloric‑acid solution that is unsuitable for most laboratory purposes, especially for preparing pure copper(II) chloride. If you need HCl, obtain it directly (or generate it by the classic NaCl + H₂SO₄ route) rather than trying to synthesize it from calcium chloride and a weakly produced sulfuric acid.

Original question: Making hydrochloric acid using sulfuric acid and calcium chloride solution on Chemistry Stack Exchange, licensed CC BY-SA.

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