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Chapter 4: Problem 53

Separate samples of a solution of an unknown soluble ionic compound are treated with \(\mathrm{KCl}, \mathrm{Na}_{2} \mathrm{SO}_{4}\), and \(\mathrm{NaOH}\). A precipitate forms only when \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added. Which cations could be present in the unknown soluble ionic compound?

Short Answer

Expert verified
Based on the solubility rules and the given information, the possible cations that could be present in the unknown soluble ionic compound are Ba虏鈦�, Ca虏鈦�, and Pb虏鈦� ions, as these cations form insoluble compounds when reacted with sulfate ions (SO鈧劼测伝).

Step by step solution

01

Review some solubility rules for ionic compounds

We need to know which cations form insoluble sulfates. Below are some common solubility rules for ionic compounds: 1. Most nitrate (NO鈧冣伝) salts are soluble. 2. Most salts containing alkali metals (Group 1 elements) and ammonium ion (NH鈧勨伜) are soluble. 3. Most chloride (Cl鈦�), bromide (Br鈦�), and iodide (I鈦�) salts are soluble. 4. Most sulfate (SO鈧劼测伝) salts are soluble. However, barium sulfate (BaSO鈧�), calcium sulfate (CaSO鈧�) and lead(II) sulfate (PbSO鈧�) are insoluble.
02

Identify possible cations in the unknown compound based on solubility rules

Since precipitate forms only when Na鈧係O鈧� is added to the unknown compound, we are looking for cations that form insoluble sulfates. According to the solubility rules, some possibilities include Ba虏鈦�, Ca虏鈦�, and Pb虏鈦� ions. Ba虏鈦�: Forms an insoluble compound with sulfate ions - BaSO鈧� Ca虏鈦�: Forms an insoluble compound with sulfate ions - CaSO鈧� Pb虏鈦�: Forms an insoluble compound with sulfate ions - PbSO鈧�
03

Conclusion

Based on the given information and solubility rules, the possible cations that could be present in the unknown soluble ionic compound are Ba虏鈦�, Ca虏鈦�, and Pb虏鈦� ions, as these cations form insoluble compounds when reacted with sulfate ions (SO鈧劼测伝).

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Most popular questions from this chapter

Calculate the sodium ion concentration when \(70.0 \mathrm{~mL}\) of \(3.0 \mathrm{M}\) sodium carbonate is added to \(30.0 \mathrm{~mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{UO}_{2}^{2+}\) f. \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) b. \(\mathrm{As}_{2} \mathrm{O}_{3} \quad\) g. \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) c. \(\mathrm{NaBiO}_{3}\) h. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) d. As i. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) e. \(\mathrm{HAsO}_{2}\)

Douglasite is a mineral with the formula \(2 \mathrm{KCl} \cdot \mathrm{FeCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\). Calculate the mass percent of douglasite in a \(455.0-\mathrm{mg}\) sample if it took \(37.20 \mathrm{~mL}\) of a \(0.1000 \mathrm{M} \mathrm{AgNO}_{3}\) solution to precipitate all the \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\). Assume the douglasite is the only source of chloride ion.

Consider the reaction of \(19.0 \mathrm{~g}\) of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and \(29.0 \mathrm{~g}\) of solid metal is present. Calculate the mass of each metal in the \(29.0-\mathrm{g}\) mixture.

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$ \begin{array}{l} \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) \\ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) \end{array} $$ A \(10.00-\mathrm{g}\) mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with \(156 \mathrm{~mL}\) of \(3.00 M\) silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If \(78.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) was added, what was the concentration of the \(\mathrm{HCl}\) ?
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