91Ó°ÊÓ

The ratio of moles of solute (in grams) to the volume of the solution is known as molarity (in litres).

The formula for calculating a solution's molarity is:

$\mathrm{Molarity}\left(\mathrm{M}\right)=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}\left(\mathrm{in}\mathrm{g}\right)}{\mathrm{Volume}\mathrm{of}\mathrm{solution}\left(\mathrm{in}\mathrm{L}\right)}$

Given,

$$\begin{gathered} 0.400{\mathrm{MCu}}^{+} \\ 0.0800\mathrm{MKSCN} \end{gathered}$$

The ratio of moles of copper ion to moles of thiocyanate ions is

$$\begin{gathered} 1:1 \\ {\mathrm{V}}_{\mathrm{e}}×0.400\mathrm{M}=0.05\mathrm{L}×0.0800\mathrm{M} \\ {\mathrm{V}}_{\mathrm{e}}=\frac{0.05\mathrm{L}×0.0800\mathrm{M}}{0.0400\mathrm{M}}=100\mathrm{mL} \end{gathered}$$

The molarity of thiocyanate is calculated as

By using the solubility constant

$$\begin{gathered} \left[C{u}^{+}\right]=\frac{K_{sp}}{SC{N}^{-}}=\frac{4.8×{10}^{-15}}{7.98×{10}^{-2}} \\ =6.0×{10}^{-14}MpC{u}^{+} \\ =-\log \left[C{u}^{+}\right] \\ =-\log \left[6.0×{10}^{-14}\right] \\ =13.22 \end{gathered}$$

The concentration of thiocyanate and copper ion are equal at equivalence point.

$$\begin{gathered} \left[{\mathrm{Cu}}^{+}\right]\left[{\mathrm{SCN}}^{-}\right]={\mathrm{x}}^2={\mathrm{K}}_{\mathrm{sp}} \\ \left[{\mathrm{Cu}}^{+}\right]\left[{\mathrm{SCN}}^{-}\right]={\mathrm{x}}^2=6.9×{10}^{-8} \\ \left[{\mathrm{Cu}}^{+}\right]=6.9×{10}^{-8}\mathrm{M} \end{gathered}$$

${\mathrm{pCu}}^{+}=7.16$Beyond equivalence point, there is only copper ion in excess

role="math" localid="1654842466204" $$\begin{gathered} \left[{\mathrm{Cu}}^{+}\right]=\left(0.0400\mathrm{M}\right)\left(\frac{101.0\mathrm{mL}-100\mathrm{mL}}{151.0\mathrm{mL}}\right) \\ =2.6×{10}^{-4}\mathrm{M} \\ {\mathrm{pCu}}^{+}=3.58 \end{gathered}$$

Thus, the volume of copper ion is${\mathrm{pCu}}^{+}=3.58$