91Ó°ÊÓ

Similar to how an Arrhenius acid delivers a proton , a polyprotic acid also contributes protons. Contrarily, a polyprotic acid differs from a monoprotic acid in that it contains many acidics H + and can hence donate a large number of protons. It partially dissociates to form a mild polyprotic acid.Some examples of weak polyprotic acids are as follows:

• H3PO4 is a triphosphate acid.
• H2CO3 as a diprotic acid
• H2SO3 is a diprotic acid

a)

In task a), to calculate the equilibrium constant, we need to write two equations. The first one is the dissociation of the weak acid and the second one is the protonation of a weak base:

$$\begin{gathered} HA⇌H^{+}+A^{-} \\ {K}_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]} \\ B+H^{+}⇌B{H}^{+} \\ \frac{K_b}{K_w}=\frac{\left[B{H}^{+}\right]}{\left[B\right]\left[H^{+}\right]} \end{gathered}$$

By combining these two equations, we get the acid-base reaction

$HA+B⇌B{H}^{+}+A^{-}$

By putting everything in the acid-base equation for the equilibrium constant we get:

$K=\frac{\left[B{H}^{+}\right]\left[A^{-}\right]}{\left[HA\right]\left[B\right]}=\frac{K_aâ‹\dots K_b}{K_w}=\frac{({10}^{-2.86}â‹\dots {10}^{-3.36}}{{10}^{-14}}=6.03×{10}^7$

b)

The $p{K}_2$intersects the upper curve between the first and second equivalent point and there is a ${\mathrm{HA}}^{-}\mathrm{and}{\mathrm{A}}^{2-}$buffer in the solution. The pH is equal to ${\mathrm{pK}}_{2^{-}}$

The ${\mathrm{pK}}_2$intersects the lower curve after the equivalence point $\left(2V_e\right)$and there is a solution consisting of a base (B) and a conjugate acid $\left({\mathrm{BH}}^{+}\right)$The is equal to ${\mathrm{pK}}_{{\mathrm{BH}}^{+}}$.