Question

Separate samples of a solution of an unknown salt are treated with dilute solutions of \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH} .\) A precipitate forms in all three cases. Which of the following cations could be present in the unknown salt solution: \(\mathrm{K}^{+}, \mathrm{Pb}^{2+}, \mathrm{Ba}^{2+}.\)

Short answer

Since a precipitate was formed in all three cases when the unknown salt solution was treated with HBr, H2SO4, and NaOH, we can conclude that the cation present in the unknown salt solution is Pb2+. This is because Pb2+ forms insoluble products with HBr and H2SO4 and a slightly soluble product with NaOH (which forms a precipitate at higher concentrations), while K+ and Ba2+ do not form precipitates with all three reagents.

Step-by-step solution

Analyze K+ ion

When we have a salt with K+ ions and treat it with the mentioned reagents, the possible products are: - KBr: formed with HBr. - K2SO4: formed with H2SO4. - KOH: formed with NaOH. Now, let's check if any of these compounds form a precipitate, i.e., if they are insoluble in water.

Analyze Pb2+ ion

Similarly, for a salt containing Pb2+ ions, the possible products with the mentioned reagents are: - PbBr2: formed with HBr. - PbSO4: formed with H2SO4. - Pb(OH)2: formed with NaOH. Now, we'll check if these compounds form a precipitate.

Analyze Ba2+ ion

For a salt containing Ba2+ ions, the possible products with the mentioned reagents are: - BaBr2: formed with HBr. - BaSO4: formed with H2SO4. - Ba(OH)2: formed with NaOH. Now, let's check if these compounds form a precipitate.

Check solubility rules

According to the solubility rules: - All alkali metal (Group IA) compounds, including those with K+ ions, are soluble. Thus, KBr, K2SO4, and KOH are soluble and do not form precipitates. - Lead (Pb2+) compounds have some exceptions. PbBr2 and PbSO4 are insoluble, while Pb(OH)2 is slightly soluble (forming a precipitate at higher concentrations). - Barium (Ba2+) compounds are mostly soluble, but BaSO4 and Ba(OH)2 are insoluble, while BaBr2 is soluble.

Determine which cation forms precipitates with all reagents

Comparing the solubility of the products formed by the proposed cations with the reagents mentioned, we find: - K+ forms soluble products with all three reagents (no precipitate). - Pb2+ forms insoluble products with HBr and H2SO4 (precipitates) and a slightly soluble product with NaOH (precipitate at higher concentrations). - Ba2+ forms an insoluble product with H2SO4 and NaOH (precipitates) and a soluble product with HBr (no precipitate). As a precipitate was formed from the reaction with all the reagents, the unknown salt cation is Pb2+.

Key concepts

Understanding Solubility Rules

The solubility rules are a collection of guidelines that help predict whether an ionic compound will dissolve in water. These rules are based on empirical observations and can be used to quickly determine the outcome of mixing different substances to form new products, which is invaluable for researchers and students alike.

For example, it is known that all alkali metal compounds, such as those containing potassium ions (K+), are soluble in water. This means that when K+ salts are treated with common reagents like HBr, H2SO4, or NaOH, the resulting compounds will not precipitate; they will remain dissolved in the solution. Understanding these rules can significantly aid in the analysis of reaction outcomes and is particularly useful when working with unknown samples. Knowing that K+ compounds are soluble, we can confidently infer that the presence of a precipitate suggests the original solution likely did not contain potassium ions. This piece of information assists in narrowing down the list of possible cations in an unknown salt.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two soluble salts react in aqueous solution to form an insoluble solid called a precipitate. This solid subsequently falls out of solution, and can often be observed as a cloudy or grainy substance within the liquid. They are particularly useful in the field of qualitative analysis.

Each ion's activity during a precipitation reaction can be foretold using solubility rules. For instance, in the reaction with HBr, H2SO4, and NaOH with various ions, we can deduce which ions will likely form a precipitate. Lead (Pb2+) and barium (Ba2+) ions, for example, tend to form precipitates under certain conditions; lead(II) bromide (PbBr2) and barium sulfate (BaSO4) are known to be insoluble and will precipitate out of solution. The ability to predict such outcomes is essential for interpreting the results of chemical tests and analysing substances.

Qualitative Analysis

Qualitative analysis is a method used in chemistry to identify the elements, ions, or compounds present in a given sample without necessarily determining their quantity. The process involves a set of procedures that can include the use of solubility rules and observation of precipitation reactions.

When an analyst, such as a student or a researcher, receives an unknown sample, they systematically test the sample against various reagents to observe the chemical reactions that occur. The formation of precipitates is a crucial indicator in such analyses. For example, in the provided exercise, the occurrence of a precipitate with all three reagents (HBr, H2SO4, and NaOH) led to the conclusion that the original salt was likely to contain Pb2+ ions. This systematic approach is powerful in deductive reasoning and is a fundamental skill in chemistry for both problem-solving and real-world applications, especially in the detection and identification of compounds in a mixture.