Chapter 13, Problem 12E

(a)

To determine

Balance and complete the given reaction: ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+{\text{PbNO}}_{\text{3}}\left(\text{aq}\right)\to$

Answer to Problem 12E

The balanced chemical reaction is written as,

${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)\to 2{\text{KNO}}_{\text{3}}\left(\text{s}\right)+{\text{PbSO}}_{\text{4}}\left(\text{s}\right)$

Explanation of Solution

Given Info: The given chemical reaction is a double-replacement reaction.

Explanation:

A reaction in which positive ion of one reactant combines with negative ion of another reactant to form products is known as a double-replacement reaction.

In the given reaction, the positive ion ${\text{K}}^{\text{+}}$ combines with a negative ion ${\text{NO}}_{\text{3}}{}^{-}$ to form ${\text{KNO}}_{\text{3}}$. The positive ion ${\text{Pb}}^{\text{+2}}$ combines with a negative ion ${\text{SO}}_4{}^{-2}$ to form ${\text{PbSO}}_4$. Therefore, the given reaction is a double-replacement reaction.

The double-replacement reaction of ${\text{K}}_2{\text{SO}}_{\text{4}}$ and ${\text{PbNO}}_3$ is written as,

${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)\to {\text{KNO}}_{\text{3}}\left(\text{s}\right)+{\text{PbSO}}_{\text{4}}\left(\text{s}\right)$

In the above reaction, there are two potassium atoms and two ${\text{NO}}_{\text{3}}$ units on the reactant side and there is one potassium atom and one ${\text{NO}}_{\text{3}}$ unit on the product side. Therefore multiply ${\text{KNO}}_3$ with $2$. The balanced chemical reaction is written as,

${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)\to 2{\text{KNO}}_{\text{3}}\left(\text{s}\right)+{\text{PbSO}}_{\text{4}}\left(\text{s}\right)$

Conclusion:

Therefore, the balanced chemical reaction is written as,

${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)\to 2{\text{KNO}}_{\text{3}}\left(\text{s}\right)+{\text{PbSO}}_{\text{4}}\left(\text{s}\right)$

(b)

To determine

Balance and complete the given reaction: ${\text{K}}_3{\text{PO}}_4\left(\text{aq}\right)+{\text{CaBr}}_{\text{2}}\left(\text{aq}\right)\to$

Answer to Problem 12E

The balanced chemical reaction is written as,

${\text{2K}}_3{\text{PO}}_4\left(\text{aq}\right)+{\text{3CaBr}}_{\text{2}}\left(\text{aq}\right)\to 6\text{KBr(s})+{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)$

Explanation of Solution

Given Info: The given chemical reaction is a double-replacement reaction.

Explanation:

A reaction in which positive ion of one reactant combines with negative ion of another reactant to form products is known as a double-replacement reaction.

In the given reaction, the positive ion ${\text{K}}^{\text{+}}$ combines with a negative ion ${\text{Br}}^{-}$ to form $\text{KBr}$ The positive ion ${\text{Ca}}^{\text{+2}}$ combines with a negative ion ${\text{PO}}_{\text{4}}{}^{-3}$ to form ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$. Therefore, the given reaction is a double-replacement reaction.

The double-replacement reaction of ${\text{CaBr}}_{\text{2}}$ and ${\text{K}}_3{\text{PO}}_4$ is written as,

${\text{K}}_3{\text{PO}}_4\left(\text{aq}\right)+{\text{CaBr}}_{\text{2}}\left(\text{aq}\right)\to \text{KBr(s})+{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)$

In the above reaction, there is one ${\text{PO}}_{\text{4}}$ unit on the reactant side and two ${\text{PO}}_{\text{4}}$ units on the product side. Therefore, multiply ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ with $2$. The chemical equation is written as,

${\text{2K}}_3{\text{PO}}_4\left(\text{aq}\right)+{\text{CaBr}}_{\text{2}}\left(\text{aq}\right)\to \text{KBr(s})+{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)$

In the above reaction, there are six potassium atoms, one calcium atom and two bromine atoms on the reactant side. There are three calcium atoms, one potassium atom and one bromine atom on the product side. Therefore, multiply ${\text{CaBr}}_{\text{2}}$ with $3$ and $\text{KBr}$ with $6$.

The balanced chemical equation is written as,

${\text{2K}}_3{\text{PO}}_4\left(\text{aq}\right)+{\text{3CaBr}}_{\text{2}}\left(\text{aq}\right)\to 6\text{KBr(s})+{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)$

Conclusion:

Therefore, the balanced chemical reaction is written as,

${\text{2K}}_3{\text{PO}}_4\left(\text{aq}\right)+{\text{3CaBr}}_{\text{2}}\left(\text{aq}\right)\to 6\text{KBr(s})+{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)$