PAG7: Producing Pure, Dry Salts

An appropriate insoluble oxide or carbonate can react with a dilute acid to form a soluble salt. For the copper oxide and sulfuric acid example, copper(II) sulfate and water form. Adding a suitable excess of the insoluble solid helps use up the acid; the remaining solid can subsequently be removed. An excess soluble alkali cannot be removed by the same filtration step.

Use the heating and addition arrangements in the approved method. Warming can help the reaction proceed, but boiling or vigorous addition is not the purpose. Record the appearance and the point at which further solid remains rather than assuming that the first colour change proves all acid has reacted. The intended salt determines which acid and base are suitable.

This is a suitable PAG7 production-and-separation example. The centre can select another appropriate salt preparation that covers the required measurements, reaction handling and separation techniques.

Filter to remove unreacted insoluble solid. The liquid that passes through is the filtrate and contains the dissolved salt; the residue is the retained solid. Filter paper does not remove the dissolved product from a true solution. Label the fractions so the product-containing liquid is not discarded.

Concentrate the filtrate using the specified gentle heating method, then allow suitable crystallisation. Cooling a sufficiently concentrated solution can produce crystals while some dissolved material remains in the mother liquor. Separate and dry the crystals as directed. For a hydrated salt, forcing complete dryness by strong heating can alter the product rather than merely removing surface water.

A wet crystal sample can give an inflated mass because surface solution or water is included. Contamination also affects mass and purity. Conversely, product may remain dissolved or be lost during transfers and separation, giving a lower recovered yield. A high mass is not sufficient evidence of a successful preparation.

When a theoretical yield is supplied or appropriately calculated, percentage yield = actual yield ÷ theoretical yield × 100. Compare like chemical forms: anhydrous and hydrated formulas have different formula masses. Explain a specific source of loss or excess mass rather than treating every departure as an unexplained experimental error.

Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.