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Chapter 4: Problem 68

Zinc hydroxide is insoluble in water but dissolves in aqueous nitric acid. Why? Write balanced total ionic and net ionic equations, showing nitric acid as it actually exists in water and the reaction as a proton-transfer process.

Short Answer

Expert verified
Nitric acid donates protons to Zn(OH)鈧, forming soluble Zn虏鈦 ions and water. Net ionic equation: Zn(OH)鈧(s) + 2H鈦(aq) 鈫 Zn虏鈦(aq) + 2H鈧侽(l).

Step by step solution

01

Identify the Insoluble Compound

Zinc hydroxide, Zn(OH)鈧, is noted as being insoluble in water.
02

Understand Dissolution in Acid

When Zn(OH)鈧 is added to nitric acid, it dissolves due to a chemical reaction where nitric acid donates protons (H鈦 ions) to react with the hydroxide ions (OH鈦) from Zn(OH)鈧.
03

Write the Balanced Total Ionic Equation

First, dissociate all strong electrolytes into their respective ions in aqueous solution: Zn(OH)鈧(s) + 2HNO鈧(aq) 鈫 Zn虏鈦(aq) + 2NO鈧冣伝(aq) + 2H鈧侽(l)Separate strong electrolytes into ions:Zn(OH)鈧(s) + 2H鈦(aq) + 2NO鈧冣伝(aq) 鈫 Zn虏鈦(aq) + 2NO鈧冣伝(aq) + 2H鈧侽(l)
04

Simplify to the Net Ionic Equation

Cancel out the spectator ions (NO鈧冣伝), which appear on both sides of the equation: Zn(OH)鈧(s) + 2H鈦(aq) 鈫 Zn虏鈦(aq) + 2H鈧侽(l)
05

Final Proton-Transfer Reaction

The balance and nature of the net ionic equation reveals that it is a proton-transfer process:Zn(OH)鈧(s) + 2H鈦(aq) 鈫 Zn虏鈦(aq) + 2H鈧侽(l)

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Key Concepts

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

Insolubility in Water
Zinc hydroxide, or Zn(OH)鈧, is a compound that is known to be insoluble in water. This means it does not dissolve when mixed with water; instead, it forms a solid precipitate.
The insolubility occurs because Zn(OH)鈧 forms strong ionic bonds that water molecules can't easily break apart.
Even though water is a good solvent for many ionic compounds, the attraction between the zinc and hydroxide ions in Zn(OH)鈧 is too strong for the water molecules to separate.
Therefore, Zn(OH)鈧 remains a solid in water.
Proton-Transfer Reaction
When Zn(OH)鈧 is added to nitric acid (HNO鈧), a proton-transfer reaction occurs.
Nitric acid is a strong acid, which means it easily donates protons (H鈦 ions).
In a proton-transfer reaction, these H鈦 ions react with the hydroxide ions (OH鈦) from Zn(OH)鈧.
This reaction is as follows:
  • The H鈦 ions from the nitric acid combine with the OH鈦 ions from the Zn(OH)鈧.
  • This forms water molecules (H鈧侽), which are neutral and liquid.
  • As a result, Zn(OH)鈧 dissolves because its ions are now part of the solution.
The reaction demonstrates how acids can dissolve certain insoluble compounds by donating protons to reactants.
Balanced Total Ionic Equation
To fully understand the reaction, we need to write balanced total and net ionic equations.
Let's start with the balanced chemical equation:

Zn(OH)鈧(s) + 2HNO鈧(aq) 鈫 Zn虏鈦(aq) + 2NO鈧冣伝(aq) + 2H鈧侽(l)

In this equation, Zn(OH)鈧 stays as a solid, nitric acid dissociates into its ions: H鈦 and NO鈧冣伝.
Next, we write the total ionic equation by separating the strong electrolytes:

Zn(OH)鈧(s) + 2H鈦(aq) + 2NO鈧冣伝(aq) 鈫 Zn虏鈦(aq) + 2NO鈧冣伝(aq) + 2H鈧侽(l)

Now we simplify to the net ionic equation by cancelling out the spectator ions, which don't change during the reaction (here, NO鈧冣伝):

Zn(OH)鈧(s) + 2H鈦(aq) 鈫 Zn虏鈦(aq) + 2H鈧侽(l)

This final equation shows the proton-transfer process and balances the ions involved in the reaction.

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Most popular questions from this chapter

The amount of ascorbic acid (vitamin \(\left.\mathrm{C}, \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\) in tablets is determined by reaction with bromine and then titration of the hydrobromic acid with standard base: $$ \begin{array}{l} \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}(a q)+\mathrm{Br}_{2}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}(a q)+2 \mathrm{HBr}(a q) \\ \mathrm{HBr}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaBr}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ A certain tablet is advertised as containing \(500 \mathrm{mg}\) of vitamin \(\mathrm{C}\). One tablet was dissolved in water and reacted with \(\mathrm{Br}_{2}\). The solution was then titrated with \(43.20 \mathrm{~mL}\) of \(0.1350 \mathrm{M} \mathrm{NaOH}\). Did the tablet contain the advertised quantity of vitamin C?

In a titration of \(\mathrm{HNO}_{3}\), you add a few drops of phenolphthalein indicator to \(50.00 \mathrm{~mL}\) of acid in a flask. You quickly add \(20.00 \mathrm{~mL}\) of \(0.0502 \mathrm{M} \mathrm{NaOH}\) but overshoot the end point, and the solution turns deep pink. Instead of starting over, you add \(30.00 \mathrm{~mL}\) of the acid, and the solution turns colorless. Then, it takes \(3.22 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) to reach the end point. (a) What is the concentration of the \(\mathrm{HNO}_{3}\) solution? (b) How many moles of \(\mathrm{NaOH}\) were in excess after the first addition?

If \(25.98 \mathrm{~mL}\) of \(0.1180 \mathrm{M}\) KOH solution reacts with \(52.50 \mathrm{~mL} .\) of \(\mathrm{CH}_{3} \mathrm{COOH}\) solution, what is the molarity of the acid solution?

To study a marine organism, a biologist prepares a \(1.00-\mathrm{kg}\) sample to simulate the ion concentrations in seawater. She mixes \(26.5 \mathrm{~g}\) of \(\mathrm{NaCl}, 2.40 \mathrm{~g}\) of \(\mathrm{MgCl}_{2}, 3.35 \mathrm{~g}\) of \(\mathrm{MgSO}_{4}, 1.20 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}, 1.05 \mathrm{~g}\) of \(\mathrm{KCl}, 0.315 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3},\) and \(0.098 \mathrm{~g}\) of \(\mathrm{NaBr}\) in distilled water. (a) If the density of the solution is \(1.025 \mathrm{~g} / \mathrm{cm}^{3}\). what is the molarity of each ion? (b) What is the total molarity of alkali metal ions? (c) What is the total molarity of alkaline earth metal ions? (d) What is the total molarity of anions?

Predict the product(s) and write a balanced equation for each of the following redox reactions: (a) \(\operatorname{Sr}(s)+\operatorname{Br}_{2}(l) \longrightarrow\) (b) \(\mathrm{Ag}_{2} \mathrm{O}(s) \stackrel{\Delta}{\longrightarrow}\) (c) \(\operatorname{Mn}(s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\)
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