Question

How do you complete and balance the following acid-base reactions?
-$HCl+Mn{{(OH)}_{2}}$
-$Ba{{(OH)}_{2}}+{{H}_{3}}P{{O}_{4}}$
-$C{{H}_{3}}COOH+L{{i}_{2}}C{{O}_{3}}$

Answer 1

Hint:
i) The acid-base reactions result in the formation of salt and water, as the proton from the acid is released which combines with the hydroxide which was released from the base, ultimately forming water.
ii) And the salt is formed by the combination of the left out conjugate base of the acid and conjugate acid of the base.

Complete step-by-step answer:
In order to balance a chemical reaction we need to complete them first, so we will follow a stepwise process for completion and balancing of the reactions one at a time. In the first reaction, hydrochloric acid reacts with the manganese hydroxide. At first we will write the complete chemical reaction which would take place,
$HCl+Mn{{(OH)}_{2}}\to MnC{{l}_{2}}~+~{{H}_{2}}O$

We can see that the formation of manganese di-chloride is formed along with the water molecule as a side product. Now, in order to balance this chemical equation we will see if the number of atoms on each element on both the sides are equal or not.

As we can see the number of hydrogen atoms is two on the product side of the reaction, and three on the reactant side. And the number of chlorine is two on the product side and one on the reactant side. Also the number of oxygen atoms are uneven. So, we will add a coefficient $2$ for the hydrochloric acid and the same coefficient for water. Now the equation becomes $2HCl+Mn{{(OH)}_{2}}\to MnC{{l}_{2}}~+~2{{H}_{2}}O$. Which is the balanced chemical equation.

In the next case the complete chemical equation can be written as,
\[Ba{{\left( OH \right)}_{2}}~+~{{H}_{3}}P{{O}_{4}}~\to ~BaHP{{O}_{4}}~+~{{H}_{2}}O\]
Here as we can see the number of hydrogen and oxygen are not balanced, and apart from these two atoms, the number of barium and phosphorus atoms are balanced. So, we will add a coefficient $2$ on the water which would make the number of hydrogen $5$ on both the sides and the number of oxygen $6$ on both the sides. So we get,
\[Ba{{\left( OH \right)}_{2}}~+~{{H}_{3}}P{{O}_{4}}~\to ~BaHP{{O}_{4}}~+~2{{H}_{2}}O\]

In the third scenario, the complete unbalanced chemical equation would be,
\[C{{H}_{3}}COOH~+~L{{i}_{2}}C{{O}_{3}}~\to ~C{{H}_{3}}COOLi~+~{{H}_{2}}O~+~C{{O}_{2}}\]
Now, if we notice carefully the number of lithium atoms is two, on the reactant side and it's one on the product side, so we will add a coefficient two on the lithium acetate, to make it even. And consequently, we would have to add a coefficient two, on the acetic acid to make the number of carbon hydrogen and oxygen equal on both the sides. So, we get,
\[2C{{H}_{3}}COOH~+~L{{i}_{2}}C{{O}_{3}}~\to ~2C{{H}_{3}}COOLi~+~{{H}_{2}}O~+~C{{O}_{2}}\]
Which is the required complete and balanced chemical equation.

Note: In order to balance the chemical equation we need to write the complete equation, and as it was mentioned it was acid-base reaction, so it was easy to predict the products, as the reaction of acid and base results in salt and water. In case of reaction of acid with metal carbonates, an extra side product which is carbon dioxide is released as an effervescence as in the case of the third reaction, where acetic acid reacted with lithium carbonate.