Question

Below are indicated the formulas of some salts. Such salts could be formed by the reaction of the appropriate strong acid and strong base (with the other product of the reaction being, of course, water). For each salt, write an equation showing the formation of the salt from reaction of the appropriate strong acid and strong base. a. ${\mathrm{Na}}_2{\mathrm{SO}}_4$ b. ${\mathrm{RbNO}}_3$ c. ${\mathrm{KClO}}_4$ d. $\mathrm{KCl}$

Short answer

a. $2\mathrm{NaOH}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$ b. $\mathrm{RbOH}+{\mathrm{HNO}}_3\to {\mathrm{RbNO}}_3+{\mathrm{H}}_2\mathrm{O}$ c. $\mathrm{KOH}+{\mathrm{HClO}}_4\to {\mathrm{KClO}}_4+{\mathrm{H}}_2\mathrm{O}$ d. $\mathrm{KOH}+\mathrm{HCl}\to \mathrm{KCl}+{\mathrm{H}}_2\mathrm{O}$

Step-by-step solution

a. Formation of Na2SO4

To form ${\mathrm{Na}}_2{\mathrm{SO}}_4$, we need to have a Sodium ion (${\mathrm{Na}}^{+}$) and sulfate ion (${\mathrm{SO}}_4^{2-}$). We can find ${\mathrm{Na}}^{+}$ in a strong base like $\mathrm{NaOH}$ and ${\mathrm{SO}}_4^{2-}$ in a strong acid like ${\mathrm{H}}_2{\mathrm{SO}}_4$. The balanced chemical equation will therefore be: $2\mathrm{NaOH}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$

b. Formation of RbNO3

To form ${\mathrm{RbNO}}_3$, we need a Rubidium ion (${\mathrm{Rb}}^{+}$) and a nitrate ion (${\mathrm{NO}}_3^{-}$). We can find ${\mathrm{Rb}}^{+}$ in a strong base like $\mathrm{RbOH}$ and ${\mathrm{NO}}_3^{-}$ in a strong acid like ${\mathrm{HNO}}_3$. The balanced chemical equation will be: $\mathrm{RbOH}+{\mathrm{HNO}}_3\to {\mathrm{RbNO}}_3+{\mathrm{H}}_2\mathrm{O}$

c. Formation of KClO4

To form ${\mathrm{KClO}}_4$, we need a Potassium ion (${\mathrm{K}}^{+}$) and a perchlorate ion (${\mathrm{ClO}}_4^{-}$). We can find ${\mathrm{K}}^{+}$ in a strong base like $\mathrm{KOH}$ and ${\mathrm{ClO}}_4^{-}$ in a strong acid like ${\mathrm{HClO}}_4$. The balanced chemical equation will be: $\mathrm{KOH}+{\mathrm{HClO}}_4\to {\mathrm{KClO}}_4+{\mathrm{H}}_2\mathrm{O}$

d. Formation of KCl

To form $\mathrm{KCl}$, we need a Potassium ion (${\mathrm{K}}^{+}$) and a chloride ion (${\mathrm{Cl}}^{-}$). We can find ${\mathrm{K}}^{+}$ in a strong base like $\mathrm{KOH}$ and ${\mathrm{Cl}}^{-}$ in a strong acid like $\mathrm{HCl}$. The balanced chemical equation will be: $\mathrm{KOH}+\mathrm{HCl}\to \mathrm{KCl}+{\mathrm{H}}_2\mathrm{O}$

Key concepts

Acid-Base Reactions

Understanding acid-base reactions is foundational in chemistry, especially when it comes to the formation of salts. In such reactions, an acid and a base react to form a salt and water; this type of reaction is known as a neutralization reaction. Acids are substances that can donate a proton (H+ ion), and bases are substances that can accept a proton.

A simple way to remember what happens in an acid-base reaction is to think of it as an 'exchange of partners.' The hydrogen ions (H+) from the acid meet and bond with the hydroxide ions (OH-) from the base to form water (H2O), and the remaining ions from the acid and base come together to form a salt. The type of salt formed depends on the specific ions present in the reactants. For example, when sodium hydroxide (a base) reacts with sulfuric acid (an acid), they form sodium sulfate (a salt) and water.

Formation of Salts

The formation of salts is not only a fascinating chemical process but also hugely important for numerous applications, from industrial manufacturing to the pharmaceutical industry. Salts are ionic compounds consisting of positive and negative ions held together by the strong force of ionic bonds.

When an acid reacts with a base, the resulting formation of salts is through a process where the anions (negatively charged ions) from the acid bind with the cations (positively charged ions) from the base. In our textbook example, rubidium hydroxide (RbOH) and nitric acid (HNO3) react to form rubidium nitrate (RbNO3). It's important to notice that each acid and base pair in different reactions will result in distinct salts, reflecting the variety of acids and bases used.

Balancing Chemical Equations

Balancing chemical equations is akin to ensuring both sides of a scale are equal. It's a fundamental skill in chemistry, as it reflects the law of conservation of mass - matter is neither created nor destroyed. When you balance a chemical equation, you're ensuring that the same number of each type of atom appears on both sides of the equation.

For instance, the balanced equation for the formation of potassium perchlorate from potassium hydroxide and perchloric acid looks like this: $\mathrm{KOH}+{\mathrm{HClO}}_4\to {\mathrm{KClO}}_4+{\mathrm{H}}_2\mathrm{O}$. Each element's atoms are counted on both sides to confirm that they match, and coefficients are adjusted accordingly. This balancing act is a critical skill for anyone working in a field related to chemistry because it ensures that our chemical equations accurately reflect what happens in the laboratory or the real world.