Q4.23P - Chemistry: Molecular Nature Of Matter And Change

Question

Water “softeners” remove metal ions such as  and  by replacing them with enough   ions to maintain the same number of positive charges in the solution. If 1.0 x 10³ L of “hard” water is 0.015 M  and 0.0010 M  how many moles of   are needed to replace these ions?

Step-by-Step Solution

Answer

0.33 mol of  ${\mathrm{Na}}^{+}$ are needed to replace these ions.

Step 1: Calculatethe number of moles of N a + required to remove both C a 2 + and F e 3 + metal ions

The no. of moles of  ${\mathit{C}\mathit{a}}^{\mathit{2}\mathit{+}}$ ion in the given sample is:

 $\begin{array}{rcl}Moles\mathit{ }of\mathit{ }{\mathit{C}\mathit{a}}^{\mathit{2}\mathit{+}}ion& =& concentrationxvolume\\ & =& \left(\mathrm{0}\mathrm{.}\mathrm{015}M\right)\left(\mathrm{1000}L\right)=15mol{\mathit{C}\mathit{a}}^{\mathit{2}\mathit{+}}\\ & & \end{array}$

Hence, to remove${\mathrm{Ca}}^{2+}$ ions we need a mole of  ${Na}^{+}$

Since the charge on $C{a}^{2+}$ is twice of${Na}^{+}$ 

Hence, to remove a${\mathrm{Ca}}^{2+}$ ion we need a 2 ${Na}^{+}$ ions

 $\begin{array}{rcl}2x moles of{Ca}^{2+}& =& 2x15 mol of {Na}^{+}\\ & =& 30 mol of {Na}^{+}\end{array}$

The no. of moles of ${Fe}^{+}$  ion in the given sample is:

 

Since the charge on ${Fe}^{3+}$  is thrice of ${Na}^{+}$ 

Hence, to remove a ${Fe}^{3+}$ ion we need 3 ${Na}^{+}$ ions

 

 $\begin{array}{rcl}3 x moles of F{e}^{3+}& =& 3x1mol of {Na}^{+}\\ & =& 3 mol of {Na}^{+}\end{array}$

Step 2: Calculate the total moles of N a + ion requires to replace of both metal ions C a 2 + , F e 3 + :

The total number of moles of ${Na}^{+}$ ions required is:

Moles of  ${Na}^{+}$=30 mol ${Na}^{+}$+3mol ${Na}^{+}$=33 mol ${Na}^{+}$