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?

Short answer

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

Step-by-step solution

Calculatethe number of moles of  Na+required to remove both  Ca2+and Fe3+ 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 onlocalid="1656759165424" $C{a}^{2+}$is twice of${Na}^{+}$

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

$\begin{array}{rcl}2xmolesof{Ca}^{2+}& =& 2x15molof{Na}^{+}\\ & =& 30molof{Na}^{+}\end{array}$

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

Since the charge on${Fe}^{3+}$is thrice oflocalid="1656760094429" ${Na}^{+}$

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

$\begin{array}{rcl}3xmolesofF{e}^{3+}& =& 3x1molof{Na}^{+}\\ & =& 3molof{Na}^{+}\end{array}$

Calculate the total moles of  Na+ ion requires to replace of both metal ions Ca2+,Fe3+ :

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

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