Chapter 4: Problem 33
Write a balanced net ionic equation for each of the following acidbase reactions in water. (a) nitrous acid and barium hydroxide (b) potassium hydroxide and hydrofluoric acid (c) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) and perchloric acid
Short Answer
Expert verified
Answer: The net ionic equations for the given acid-base reactions are:
(a) Nitrous acid and barium hydroxide: 2H^(+) + 2OH^(-) → 2H2O
(b) Potassium hydroxide and hydrofluoric acid: H^(+) + OH^(-) → H2O
(c) Aniline and perchloric acid: C6H5NH2 + H^(+) → C6H5NH3^(+)
Step by step solution
01
Write the balanced molecular equation
The reactants are nitrous acid (HNO2) and barium hydroxide (Ba(OH)2). Since this is an acid-base reaction, the products will be water (H2O) and a salt (barium nitrite, Ba(NO2)2). The balanced molecular equation is:
2HNO2 + Ba(OH)2 → 2H2O + Ba(NO2)2
02
Write the total ionic equation
Separate all soluble compounds into ions:
2H^(+) + 2NO2^(-) + Ba^(2+) + 2OH^(-) → 2H2O + Ba^(2+) + 2NO2^(-)
03
Write the net ionic equation
Eliminate the spectator ions (Ba^(2+) and 2NO2^(-)):
2H^(+) + 2OH^(-) → 2H2O
(b) potassium hydroxide and hydrofluoric acid
04
Write the balanced molecular equation
The reactants are potassium hydroxide (KOH) and hydrofluoric acid (HF). The products will be water (H2O) and a salt (potassium fluoride, KF). The balanced molecular equation is:
KOH + HF → H2O + KF
05
Write the total ionic equation
Separate all soluble compounds into ions:
K^(+) + OH^(-) + H^(+) + F^(-) → H2O + K^(+) + F^(-)
06
Write the net ionic equation
Eliminate the spectator ions (K^(+) and F^(-)):
H^(+) + OH^(-) → H2O
(c) aniline (C6H5NH2) and perchloric acid
07
Write the balanced molecular equation
The reactants are aniline (C6H5NH2) and perchloric acid (HClO4). The products will be a conjugate acid (anilinium ion, C6H5NH3^(+)) and a salt (perchlorate anion, ClO4^(-)). The balanced molecular equation is:
C6H5NH2 + HClO4 → C6H5NH3^(+) + ClO4^(-)
08
Write the total ionic equation
Since aniline is a weak base, it does not dissociate in water. Therefore, we will only separate perchloric acid into ions:
C6H5NH2 + H^(+) + ClO4^(-) → C6H5NH3^(+) + ClO4^(-)
09
Write the net ionic equation
Eliminate the spectator ions (ClO4^(-)):
C6H5NH2 + H^(+) → C6H5NH3^(+)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acid-Base Reactions
Acid-base reactions are pivotal in chemistry as they represent the processes where an acid donates a proton to a base. A simple and quintessential example of such a reaction is the neutralization process, which occurs when an acid reacts with a base to produce water and a salt. An acid is a substance that can donate a proton (hydrogen ion, represented as H+) to another substance, while a base is a substance that can accept a proton.
In the context of the exercise solution provided, such as when nitrous acid (HNO2) reacts with barium hydroxide (Ba(OH)2), it's a clear demonstration of an acid-base reaction. Here, nitrous acid donates protons to the hydroxide ions, creating water molecules and forming barium nitrite as the salt side-product.
Understanding the nuances of acid-base reactions involves not only writing the balanced equations but also recognizing the ions involved and the concept of spectator ions, which do not participate effectively in the reaction. This is where net ionic equations become particularly useful, stripping the reaction down to only the ions that undergo a chemical change.
In the context of the exercise solution provided, such as when nitrous acid (HNO2) reacts with barium hydroxide (Ba(OH)2), it's a clear demonstration of an acid-base reaction. Here, nitrous acid donates protons to the hydroxide ions, creating water molecules and forming barium nitrite as the salt side-product.
Understanding the nuances of acid-base reactions involves not only writing the balanced equations but also recognizing the ions involved and the concept of spectator ions, which do not participate effectively in the reaction. This is where net ionic equations become particularly useful, stripping the reaction down to only the ions that undergo a chemical change.
Chemical Equilibrium
When discussing chemical reactions, equilibrium plays a central role. It is the dynamic state where the rate of the forward reaction equals the rate of the reverse reaction. As a result, concentrations of reactants and products remain constant over time. This concept is crucial when examining acid-base reactions in aqueous solutions since reversible reactions are common.
However, in the exercises you're dealing with, the reactions tend to go to completion, meaning that a stable equilibrium point favoring the right-hand side of the equation (the products) is assumed. Therefore, it's not necessary to apply the principles of chemical equilibrium directly here. In other reactions, like the dissociation of weak acids or bases in water, equilibrium concepts become far more significant and require an understanding of equilibrium constants and their implications.
However, in the exercises you're dealing with, the reactions tend to go to completion, meaning that a stable equilibrium point favoring the right-hand side of the equation (the products) is assumed. Therefore, it's not necessary to apply the principles of chemical equilibrium directly here. In other reactions, like the dissociation of weak acids or bases in water, equilibrium concepts become far more significant and require an understanding of equilibrium constants and their implications.
Ionic Compounds in Water
Solubility and behavior of ionic compounds in water are also key concepts to grasp. Water, being a polar solvent, is particularly adept at dissolving ionic substances due to its ability to stabilize ions. When ionic compounds dissolve in water, they typically dissociate into their constituent ions.
This dissociation is clearly presented in the total ionic equations provided in the exercise, where soluble salts like Ba(NO2)2 and KF are separated into their ions. However, some compounds, such as the aniline (C6H5NH2) mentioned in the exercise, do not dissociate as they are either weak electrolytes or non-electrolytes. This distinction plays a critical role in writing net ionic equations and understanding the underlying chemistry of the substances involved. By focusing on the ions that directly participate in the reaction, students can gain a clearer picture of the process at hand.
This dissociation is clearly presented in the total ionic equations provided in the exercise, where soluble salts like Ba(NO2)2 and KF are separated into their ions. However, some compounds, such as the aniline (C6H5NH2) mentioned in the exercise, do not dissociate as they are either weak electrolytes or non-electrolytes. This distinction plays a critical role in writing net ionic equations and understanding the underlying chemistry of the substances involved. By focusing on the ions that directly participate in the reaction, students can gain a clearer picture of the process at hand.