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Which of the following salts will be substantially more soluble in an \(\mathrm{HNO}_{3}\) solution than in pure water: (a) \(\mathrm{BaSO}_{4}\) (b) CuS (c) \(\mathrm{Cd}(\mathrm{OH})_{2}\) (d) \(\mathrm{PbF}_{2}\) (e) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Short Answer

Expert verified
Cd(OH)_2 will be more soluble in HNO_3 than in pure water.

Step by step solution

01

Understand the Concept

The solubility of salts in a solution can be affected by the presence of common ions or by pH. In this question, we need to identify which salts are more soluble in an acidic solution, specifically in nitric acid ( HNO_3 ). This commonly affects salts involving hydroxide (OH^-) or basic anions (S^2-, CO_3^2-), because they react with the acid.
02

Analyze the Solubility of Each Salt

We analyze each salt to see if its solubility will increase in an acidic solution compared to water: - (a) BaSO_4: Sulfate (SO_4^2-) does not react with H^+ ions, so there is no change in solubility. - (b) CuS: Sulfide (S^2-) reacts with H+ ions to form H_2S gas, increasing solubility. - (c) Cd(OH)_2: Hydroxide (OH^-) reacts with H+ ions to form water, increasing solubility. - (d) PbF_2: Fluoride (F^-) may react with H^+ to form HF, but this effect is less notable compared to others. - (e) Cu(NO_3)_2: Already soluble due to NO_3^- ions; acid does not significantly change its solubility.
03

Determine Which Salts React With Acid

Identify which of the salts have anions that react with H^+ ions: - CuS has S^2-, which reacts to form H_2S gas, increasing its solubility in HNO_3. - Cd(OH)_2 has OH^-, which reacts to form water, also increasing its solubility in HNO_3.
04

Select the Most Acid-Soluble Salt

Among the salts listed, Cd(OH)_2 will be substantially more soluble in an HNO_3 solution because it contains OH^-, which significantly increases solubility when reacting with H^+ ions. CuS also becomes more soluble but not as prominently as Cd(OH)_2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid-Base Reaction
Acid-base reactions involve the transfer of protons from the acid to the base. These reactions are crucial in modifying the solubility of various salts. When an acid or a base dissolves in water, it dissociates into its ionic components, and acids release hydrogen ions (H鈦). In an acidic solution, the presence of excess H鈦 ions can interact with basic anions such as hydroxide (OH鈦) or sulfide (S虏鈦).
For instance, when a salt like CuS is placed in an acidic solution, the sulfide ions (S虏鈦) react with H鈦 ions to form hydrogen sulfide (H鈧係), an insoluble gas that escapes the solution. This reaction increases the dissolved copper ions' concentration, enhancing the solubility of CuS.
  • The interaction between H鈦 and hydroxide ions (in salts like Cd(OH)鈧) also leads to the formation of water, facilitating increased solubility of the salt in the acidic medium.
  • As a principle, through these reactions, acids can dissolve compounds that are otherwise insoluble in water by converting anions into neutral molecules or gases.
Chemical Equilibrium
Chemical equilibrium occurs when the forward and reverse reactions occur at the same rate, and the concentrations of reactants and products remain constant. In the case of salt solubility, the equilibrium in question is the dissolution and precipitation of the salt.
For a salt like Cd(OH)鈧, placed in nitric acid ( HNO鈧 ), the equilibrium shifts. In water, a small amount of Cd(OH)鈧 dissolves, reaching an equilibrium. When H鈦 increases due to the presence of an acid, the reaction Cd(OH)鈧(s) 鈫 Cd虏鈦(aq) + 2OH鈦(aq) is shifted to the right by the reaction of H鈦 with OH鈦 to form water. This shift removes OH鈦 ions, increasing solubility as more Cd(OH)鈧 dissolves to restore equilibrium.
  • This principle is called Le Chatelier's Principle, which states that a system at equilibrium will adjust to counteract the effect of a change in conditions.
  • By understanding these shifts in equilibrium, we can predict how changing conditions, like the acidity of a solution, affect solubility.
Common Ion Effect
The common ion effect is a decrease in the solubility of a salt when a solution already contains an ion common to the salt. It is an application of Le Chatelier's Principle, influencing chemical reactions and equilibrium.
In our exercise, some salts react with added H鈦 ions affecting solubility. For example, introducing an acid to a solution containing Cd(OH)鈧 results in the common ion (OH鈦) being engaged in neutralization, creating water. This effect counters the common ion typically reducing solubility, by instead making ions from the salt less available to mix back into the solid form.
  • Although typically known to reduce solubility, in specific conditions, the common ion effect can be bypassed, as when the ion reacts with an added component (like H鈦) to remove itself from the equilibrium.
  • Understanding the common ion effect helps in strategizing ways to either increase or decrease solubility depending on desired outcomes in experimental and industrial processes.

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Most popular questions from this chapter

Lead(II) carbonate, \(\mathrm{PbCO}_{3}\), is one of the components of the passivating layer that forms inside lead pipes. (a) If the \(K_{i p}\) for \(\mathrm{PbCO}_{3}\) is \(7.4 \times 10^{-14}\) what is the molarity of \(\mathrm{Pb}^{2+}\) in a saturated solution of lead(II) carbonate? (b) What is the concentration in ppb of \(\mathrm{Pb}^{2+}\) ions in a saturated solution? (c) Will the solubility of \(\mathrm{PbCO}_{3}\) increase or decrease as the \(\mathrm{pH}\) is lowered? (d) The EPA threshold for acceptable levels of lead ions in water is 15 ppb. Does a saturated solution of lead(II) carbonate produce a solution that exceeds the EPA limit?

Predict whether the equivalence point of each of the following titrations is below, above, or at pH 7: (a) \(\mathrm{NaHCO}_{3}\) titrated with \(\mathrm{NaOH},(\mathbf{b}) \mathrm{NH}_{3}\) titrated with \(\mathrm{HCl},(\mathbf{c}) \mathrm{KOH}\) titrated with \(\mathrm{HBr}\).

(a) If the molar solubility of \(\mathrm{CaF}_{2}\) at \(35^{\circ} \mathrm{C}\) is \(1.24 \times 10^{-3} \mathrm{~mol} / \mathrm{L},\) what is \(K_{s p}\) at this temperature? (b) It is found that \(1.1 \times 10^{-2} \mathrm{~g} \mathrm{SrF}_{2}\) dissolves per \(100 \mathrm{~mL}\) of aqueous solution at \(25^{\circ} \mathrm{C}\). Calculate the solubility product for \(\mathrm{SrF}_{2} .(\mathbf{c})\) The \(K_{s p}\) of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-10} .\) What is the molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} ?\)

Derive an equation similar to the Henderson-Hasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid HA and a base B are mixed. The pH of the resulting solution is 9.2 . (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and B. (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and B? (c) What is the value of \(K_{h}\) for B?

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