Question

Derive the following equation for the titration of potassium hydrogen phthalate (K⁺HP^-) with NaOH:

$\mathit{\varphi }\mathbf{=}\frac{{\mathbf{C}}_{\mathbf{b}}{\mathbf{V}}_{\mathbf{b}}}{{\mathbf{C}}_{\mathbf{a}}{\mathbf{V}}_{\mathbf{a}}}\mathbf{=}\frac{{\mathbf{\alpha }}_{\mathbf{H}{\mathbf{P}}^{\mathbf{-}}}\mathbf{+}\mathbf{2}{\mathbf{\alpha }}_{{\mathbf{P}}^{\mathbf{2}\mathbf{-}}}\mathbf{-}\mathbf{1}\mathbf{-}{\displaystyle \frac{\left[H^{+}\right]\mathbf{-}\left[O{H}^{-}\right]}{{\mathbf{C}}_{\mathbf{a}}}}}{\mathbf{1}\mathbf{+}\frac{\left[H^{+}\right]\mathbf{-}\left[O{H}^{-}\right]}{{\mathbf{C}}_{\mathbf{b}}}}$

Short answer

It is proved thatfor the titration of potassium hydrogen phthalate (K⁺HP^-) with NaOH:

$\varphi =\frac{C_b{V}_b}{C_a{V}_a}=\frac{{\alpha }_{H{P}^{-}}+2{\alpha }_{P^{2-}}-1-{\displaystyle \frac{\left[H^{+}\right]-\left[O{H}^{-}\right]}{C_a}}}{1+\frac{\left[H^{+}\right]-\left[O{H}^{-}\right]}{C_b}}$

Step-by-step solution

Titration reaction

The titration reaction will be

$K^{+}H{P}^{-}+N{a}^{+}O{H}^{-}\to K^{+}N{a}^{+}{P}^{2-}+H_{2}O$

Let the initial amount of K⁺HP^- be C_aV_a and the amount of NaOH added will be C_bV_b

Fraction of titration

$\varphi =\frac{C_b{V}_b}{C_a{V}_a}$

Charge Balance Equation

Balancing the charges of titration reaction, we get the following equation

$\begin{array}{rcl}\left[K^{+}\right]+\left[N{a}^{+}\right]+\left[H^{+}\right]& =& \left[H{P}^{-}\right]+2\left[P^{2-}\right]+\left[O{H}^{-}\right]\\ \left[K^{+}\right]& =& \frac{C_a{V}_a}{V_a+V_b}\\ \left[N{a}^{+}\right]& =& \frac{C_b{V}_b}{V_a+V_b}\\ \left[H{P}^{-}\right]& =& {\alpha }_{H{P}^{-}}\frac{C_a{V}_a}{V_a+V_b}\\ \left[{P_2}^{-}\right]& =& {\alpha }_{{P_2}^{-}}\frac{C_a{V}_a}{V_a+V_b}\\ & & \end{array}$

Derivation

Plugging the respective terms in the balance equation, we get

$\begin{array}{rcl}\frac{C_a{V}_a}{V_a+V_b}+\frac{C_b{V}_b}{V_a+V_b}+\left[H^{+}\right]& =& {\alpha }_{H{P}^{-}}\frac{C_a{V}_a}{V_a+V_b}+2{\alpha }_{P^{2-}}\frac{C_a{V}_a}{V_a+V_b}+\left[O{H}^{-}\right]\\ C_a{V}_a+C_b{V}_b+\left[H^{+}\right]\left(V_a+V_b\right)& =& {\alpha }_{H{P}^{-}}{C}_a{V}_a+2{\alpha }_{P^{2-}}{C}_a{V}_a+\left[O{H}^{-}\right]\left(V_a+V_b\right)\\ C_a{V}_a+C_b{V}_b+V_a\left[H^{+}\right]+V_b\left[H^{+}\right]& =& {\alpha }_{H{P}^{-}}{C}_a{V}_a+2{\alpha }_{P^{2-}}{C}_a{V}_a+V_a\left[O{H}^{-}\right]+V_b\left[O{H}^{-}\right]\\ V_a\left(C_a+H^{+}-{\alpha }_{H{P}^{-}}{C}_a-2{\alpha }_{P^{2-}}{C}_a-\left[O{H}^{-}\right]\right)& =& V_b\left(\left[O{H}^{-}\right]-C_b-\left[H^{+}\right]\right)\\ \frac{V_b}{V_a}& =& \frac{\left(C_a+H^{+}-{\alpha }_{H{P}^{-}}{C}_a-2{\alpha }_{P^{2-}}{C}_a-\left[O{H}^{-}\right]\right)}{\left(\left[O{H}^{-}\right]-C_b-\left[H^{+}\right]\right)}\\ & & \end{array}$

Now multiply both side with $\frac{{\displaystyle \frac{1}{C_a}}}{{\displaystyle \frac{1}{C_b}}}$

width="385">CbVbCaVa=1CaCa+H+-αHP-Ca-2αP2-Ca-OH-1CbOH--Cb-H+CbVbCaVa=1-αHP--2αP2-+H+-OH-Ca-1+H+-OH-Caϕ=αHP-+2αP2--1-H+-OH-Ca-1+H+-OH-Ca