Q. 10.48 - An Introduction to General, Organic, and Biological Chemistry

Question

Balance each of the following neutralization equations:

a. ${\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right(s)⟶{\mathrm{H}}_2\mathrm{O}(l)+\mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)$

b. ${\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+\mathrm{Al}\left(\mathrm{OH}{)}_3\right(s)⟶{\mathrm{H}}_2\mathrm{O}(l)+{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)$

Step-by-Step Solution

Answer

The neutralization equation is balanced 

a. $2{\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right( \mathrm{S})\to \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

b.$3{\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{Al}\left(\mathrm{OH}{)}_3\right(aq)\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 1- Introduction

A neutralization is a chemical reaction in which acids and  bases are  quantitatively combined to form salts and water. It is  a powerful neutralizing methods that combines manner,$H+$ ions and $OH-$ ions to form water.  

Step 2- Explanation (part a)

a. Step 1 -Records by Reactant and Products

${\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right( \mathrm{S})\to \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)+{\mathrm{H}}_2\mathrm{O}(l)$

Step 2- Balance ${\mathrm{H}}^{+}$ in acid and base. Placing $2$ ahead of ${\mathrm{HNO}}_3$ makes $2{\mathrm{H}}^{+}$ from $2\mathrm{OH}¯$in $\mathrm{Ba}( OH{)}_2$.  

$2{\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right( \mathrm{S})\to \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 3- Balance ${\mathrm{H}}_2\mathrm{O}$ with ${\mathrm{H}}^{+}$

$2{\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right( \mathrm{S})\to \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 4- To the acid and base, add the salt of the remaining ion.

$2{\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right( \mathrm{S})\to \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

step 3- Explanation (part b)

b. Step 1 - Records by Reactant and merchandise

${\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+\mathrm{Al}\left(\mathrm{OH}{)}_3\right(aq)\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)+{\mathrm{H}}_2\mathrm{O}(l)$

Step 2 -Balance the  ${\mathrm{H}}^{+}$within the acid and within the base$3$ before of ${\mathrm{H}}_2{\mathrm{SO}}_4$ and $2$ before of $\mathrm{Al}(\mathrm{OH}{)}_2$ provides $6{\mathrm{H}}^{+}$for the $6{\mathrm{OH}}^{-}$in $Al\left(OH\right)₃.$

$3{\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{Al}\left(\mathrm{OH}{)}_3\right(aq)\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)+{\mathrm{H}}_2\mathrm{O}(l)$

Step 3-Balance the ${\mathrm{H}}_2\mathrm{O}$ with ${\mathrm{H}}^{+}$and

$3{\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{Al}\left(\mathrm{OH}{)}_3\right(aq)\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)+6{\mathrm{H}}_2\mathrm{O}(l)$

Step 4 -Within the acid and base, write the salt from the leftover ion.

$3{\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{Al}\left(\mathrm{OH}{)}_3\right(aq)\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$