Chapter 4, Problem 152AP

Interpretation Introduction

Interpretation:

The way in which ${\text{BaSO}}_4$ precipitates by neutralization and precipitation reactions by using different reactants is to be represented.

Concept introduction:

The reactions in which a salt is formed as a product in the form of precipitates are known as precipitation reactions. Precipitates are solid insoluble salts which separate from a solution.

Reactions between an acid and a base are known as neutralization reactions. When an acid reacts with a base, salt and water are formed as products. Salts are ionic compounds made up of cation from base and anion from acid.

Answer to Problem 152AP

Solution: The reactions are as follows:

$\text{Ba}{\left(\text{OH}\right)}_2\left(aq\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

${\text{BaCl}}_2\left(aq\right)+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2\text{NaCl}\left(l\right)$

${\text{BaCO}}_3\left(aq\right)+{\text{Na}}_2{\text{SO}}_4\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +{\text{Na}}_2{\text{CO}}_3\left(aq\right)$

$\text{Ba}{\left(\text{OH}\right)}_2\left(aq\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

${\text{K}}_2{\text{SO}}_4\left(aq\right)+{\text{BaCl}}_2\left(s\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2\text{KCl}\left(s\right)$

Explanation of Solution

${\text{BaSO}}_4$ precipitates show poor solubility in water. They are also insoluble in alcohols. It can be prepared by reacting different compounds.

$\text{Ba}{\left(\text{OH}\right)}_2$

: Adding an equivalent amount of ${\text{H}}_2{\text{SO}}_4$, ${\text{BaSO}}_4$ precipitates and water are formed as the products. This reaction is neutralization reaction as barium sulfate is a strong base and sulfuric acid is a strong acid.

The reaction is as follows:

$\text{Ba}{\left(\text{OH}\right)}_2\left(aq\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

${\text{BaCl}}_2$

: Adding an equivalent amount of ${\text{Na}}_2{\text{SO}}_4$ in it, ${\text{BaSO}}_4$ precipitates and NaCl are formed as the products. This is a precipitation reaction as here, the exchange of ions occurs between the salts.

The reaction is as follows:

${\text{BaCl}}_2\left(aq\right)+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2\text{NaCl}\left(l\right)$

${\text{BaCO}}_3$

: Adding an equivalent amount of ${\text{Na}}_2{\text{SO}}_4$ in it, ${\text{BaSO}}_4$ precipitates and ${\text{Na}}_2{\text{CO}}_3$

are formed as the products. This is also a precipitation reaction and it takes place by the exchange of ions between the salts.

The reaction is as follows:

${\text{BaCO}}_3\left(aq\right)+{\text{Na}}_2{\text{SO}}_4\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +{\text{Na}}_2{\text{CO}}_3\left(aq\right)$

${\text{H}}_2{\text{SO}}_4$

: Adding an equivalent amount of $\text{Ba}{\left(\text{OH}\right)}_2$

in it, ${\text{BaSO}}_4$ precipitates and water are formed as the products.

The reaction is as follows:

$\text{Ba}{\left(\text{OH}\right)}_2\left(aq\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

${\text{K}}_2{\text{SO}}_4$

: Adding an equivalent amount of ${\text{BaCl}}_2$ in it, ${\text{BaSO}}_4$ precipitates and KCl are formed as the products.

The reaction is as follows:

${\text{K}}_2{\text{SO}}_4\left(aq\right)+{\text{BaCl}}_2\left(s\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2\text{KCl}\left(s\right)$

Conclusion

The reactions for the formation of ${\text{BaSO}}_4$ precipitates by different reactants are as:

$\text{Ba}{\left(\text{OH}\right)}_2\left(aq\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

${\text{BaCl}}_2\left(aq\right)+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2\text{NaCl}\left(l\right)$

${\text{BaCO}}_3\left(aq\right)+{\text{Na}}_2{\text{SO}}_4\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +{\text{Na}}_2{\text{CO}}_3\left(aq\right)$

$\text{Ba}{\left(\text{OH}\right)}_2\left(aq\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

${\text{K}}_2{\text{SO}}_4\left(aq\right)+{\text{BaCl}}_2\left(s\right)\to {\text{BaSO}}_{\text{4}}\left(s\right)\downarrow +2\text{KCl}\left(s\right)$