Derive an expression analogous to Equation 7-12 for the titration of ${\mathbf{M}}^{\mathbf{+}}$(concentration, ${\mathbf{C}}_{\mathbf{M}}^{\mathbf{0}}$,volume=${\mathbf{V}}_{\mathbf{M}}^{\mathbf{0}}$) with ${\mathbf{X}}^{\mathbf{-}}$(titrant concentration=${\mathbf{c}}_{\mathbf{x}}^{\mathbf{0}}$). Your equation should allow you to compute the volume of titrant $\mathbf{\left(}{\mathbf{v}}_{\mathbf{x}}\mathbf{\right)}$as a function of$\mathbf{\left[}{\mathbf{x}}^{\mathbf{-}}\mathbf{\right]}$.

The titrant is a known-concentration solution that is introduced to another solution (titrated) to determine the concentration of a second chemical species.

Let consider the addition of ${\mathrm{v}}_{\mathrm{x}}$litre of titrant $x^{-}$(whose initial concentration is $c_x^6$) to $v_M^e$litre of solution containing anion $M^{+}$with a concentration $C_M^0$.

The total moles of added $X=(=C_x^0.V_x)$must equal the moles of ${\mathrm{M}}^{\mathrm{insolution}}\left(=\left[{\mathrm{M}}^{*}\right]({\mathrm{V}}_{\mathrm{M}}+{\mathrm{V}}_{\mathrm{x}}^0\right)$plus the moles of precipitated $\mathrm{MX}\left(\mathrm{s}\right)$. In a similar manner, we can write a mass balance for M.

role="math" localid="1654943713868" $$\begin{gathered} MassbalanceforX:C_x^0.V_x=\left[X^{-}\right]\left(V_M^e+V_x\right)+molofMX\left(s\right) \\ MassbalanceforM:C_M^0.V_M^{\theta }=\left[M*\right]\left(V_x+V_M^0\right)+molofMX\left(s\right) \\ NowequatemolMX\left(s\right)fromequation2withmolMX\left(s\right)fromequation1 \\ {C}_M^0{V}_M^0-\left[M^{*}\right]\left(V_M^0+V_x\right)=C_x^0{V}_x-\left[X^{-}\right]\left(V_M^0+V_x\right) \\ {C}_M^0{V}_M^0-\left[M^{+}\right]V_N^0-\left[M^{+}\right]V_x=C_x^0{V}_x-\left[X^{-}\right]V_M^0-\left[X^{-}\right]+V_x \\ {V}_M^0\left\{C_M^0-\left[M^{*}\right]+\left[X^{-}\right]\right\}=V_x\left\{C_x^0+\left[M^{*}\right]-\left[X^{-}\right]\right\} \\ Whichcanre-arrangedto \\ \Rightarrow V_x=\frac{V_u^e\left\{C_M^0-\left[M^{*}\right]+\left[X^{-}\right]\right\}}{\left\{C_x^{*}+\left[M^{*}\right]-\left[X^{-}\right]\right\}} \end{gathered}$$