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Chapter 14: Problem 1

Write a net ionic equation for the reaction between aqueous solutions of (a) ammonia and hydrofluoric acid. (b) perchloric acid and rubidium hydroxide. (c) sodium sulfite and hydriodic acid. (d) nitric acid and calcium hydroxide.

Short Answer

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Question: Write the net ionic equation for each reaction. a) A reaction between ammonia and hydrofluoric acid. b) A reaction between perchloric acid and rubidium hydroxide. c) A reaction between sodium sulfite and hydriodic acid. d) A reaction between nitric acid and calcium hydroxide. Answer: a) NH3(aq) + HF(aq) → NH4+(aq) + F-(aq) b) H+(aq) + OH-(aq) → H2O(l) c) SO3^2-(aq) + 2H+(aq) → H2O(l) + SO2(g) d) 2H+(aq) + 2OH-(aq) → 2H2O(l)

Step by step solution

01

Write the balanced molecular equation.

The balanced molecular equation for the reaction between ammonia (NH3) and hydrofluoric acid (HF) is: NH3(aq) + HF(aq) → NH4F(aq)
02

Write the total ionic equation.

To write the total ionic equation, break down the aqueous compounds into their respective ions: NH3(aq) + HF(aq) → NH4+(aq) + F-(aq)
03

Identify and eliminate the spectator ions.

In this reaction, there are no spectator ions. All ions are involved in the reaction.
04

Write the net ionic equation.

Since all ions are involved in the reaction, the net ionic equation is the same as the total ionic equation: NH3(aq) + HF(aq) → NH4+(aq) + F-(aq) (b) Perchloric acid and rubidium hydroxide
05

Write the balanced molecular equation.

The balanced molecular equation for the reaction between perchloric acid (HClO4) and rubidium hydroxide (RbOH) is: HClO4(aq) + RbOH(aq) → RbClO4(aq) + H2O(l)
06

Write the total ionic equation.

Break down the aqueous compounds into their respective ions: H+(aq) + ClO4-(aq) + Rb+(aq) + OH-(aq) → Rb+(aq) + ClO4-(aq) + H2O(l)
07

Identify and eliminate the spectator ions.

The spectator ions in this reaction are Rb+ and ClO4-. Eliminate them from the total ionic equation:
08

Write the net ionic equation.

The net ionic equation for this reaction is: H+(aq) + OH-(aq) → H2O(l) (c) Sodium sulfite and hydriodic acid
09

Write the balanced molecular equation.

The balanced molecular equation for the reaction between sodium sulfite (Na2SO3) and hydriodic acid (HI) is: Na2SO3(aq) + 2HI(aq) → 2NaI(aq) + H2O(l) + SO2(g)
10

Write the total ionic equation.

Break down the aqueous compounds into their respective ions: 2Na+(aq) + SO3^2-(aq) + 2H+(aq) + 2I-(aq) → 2Na+(aq) + 2I-(aq) + H2O(l) + SO2(g)
11

Identify and eliminate the spectator ions.

The spectator ions in this reaction are Na+ and I-. Eliminate them from the total ionic equation:
12

Write the net ionic equation.

The net ionic equation for this reaction is: SO3^2-(aq) + 2H+(aq) → H2O(l) + SO2(g) (d) Nitric acid and calcium hydroxide
13

Write the balanced molecular equation.

The balanced molecular equation for the reaction between nitric acid (HNO3) and calcium hydroxide (Ca(OH)2) is: 2HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2H2O(l)
14

Write the total ionic equation.

Break down the aqueous compounds into their respective ions: 2H+(aq) + 2NO3-(aq) + Ca^2+(aq) + 2OH-(aq) → Ca^2+(aq) + 2NO3-(aq) + 2H2O(l)
15

Identify and eliminate the spectator ions.

The spectator ions in this reaction are Ca^2+ and NO3-. Eliminate them from the total ionic equation:
16

Write the net ionic equation.

The net ionic equation for this reaction is: 2H+(aq) + 2OH-(aq) → 2H2O(l)

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Most popular questions from this chapter

A \(0.1375 \mathrm{M}\) solution of potassium hydroxide is used to titrate \(35.00 \mathrm{~mL}\) of \(0.257 M\) hydrobromic acid. (Assume that volumes are additive.) (a) Write a balanced net ionic equation for the reaction that takes place during titration. (b) What are the species present at the equivalence point? (c) What volume of potassium hydroxide is required to reach the equivalence point? (d) What is the \(\mathrm{pH}\) of the solution before any \(\mathrm{KOH}\) is added? (e) What is the \(\mathrm{pH}\) of the solution halfway to the equivalence point? (f) What is the \(\mathrm{pH}\) of the solution at the equivalence point?

There is a buffer system \(\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}-\mathrm{HPO}_{4}{ }^{2-}\right)\) in blood that helps keep the blood \(\mathrm{pH}\) at about \(7.40 .\left(\mathrm{K}_{\mathrm{a}} \mathrm{H}_{2} \mathrm{PO}_{4}^{-}=6.2 \times 10^{-8}\right)\). (a) Calculate the \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right] /\left[\mathrm{HPO}_{4}^{2-}\right]\) ratio at the normal \(\mathrm{pH}\) of blood. (b) What percentage of the \(\mathrm{HPO}_{4}{ }^{2-}\) ions are converted to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) when the \(\mathrm{pH}\) goes down to \(6.80\) ? (c) What percentage of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ions are converted to \(\mathrm{HPO}_{4}{ }^{2-}\) when the \(\mathrm{pH}\) goes up to \(7.80\) ?

For an aqueous solution of acetic acid to be called "distilled white vinegar" it must contain \(5.0 \%\) acetic acid by mass. A solution with a density of \(1.05 \mathrm{~g} / \mathrm{mL}\) has a \(\mathrm{pH}\) of \(2.95 .\) Can the solution be called "distilled white vinegar"?

Phenol red is an indicator with a \(\mathrm{pK}_{\mathrm{a}}\) of \(7.4\). It is yellow in acid solution and red in alkaline. (a) What is its \(K_{\mathrm{a}}\) ? (b) What is its \(\mathrm{pH}\) range? (c) What would its color be at \(\mathrm{pH}\) 7.4?

Write a balanced net ionic equation for the reaction of each of the following aqueous solutions with \(\mathrm{H}^{+}\) ions. (a) sodium fluoride (b) barium hydroxide (c) potassium dihydrogen phosphate \(\left(\mathrm{KH}_{2} \mathrm{PO}_{4}\right)\)
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