Question

$K_{a_1},K_{a_2}$ and $K_{a_3}$ values of $H_{3}P{O}_4$ are ${10}^{-3},{10}^{-6}$ and ${10}^{-12}$ respectively. If $K_w\left(H_{2}O\right)={10}^{-14}$, then:
What is $K_b$ is $HP{O}_4{}^{-}$?

Answer 1

$H_{3}P{O}_4\to H^{+}+H_{2}P{O}_4^{-}$ $K{a}_1={10}^{-}$

$H_{2}P{O}_4^{-}\to H^{+}+HP{O}_4^{2-}$ $K{a}_1={10}^{-6}$

$HP{O}_4^{2-}\to H^{+}+P{O}_4^{3-}$ $K{a}_3={10}^{-12}$

$K_w=k_a{k}_b={10}^{-14}$

$k_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]$

$k_w=k_a{K}_b$

$k_a={10}^{-3}\times {10}^{-6}\times {10}^{-12}$

$={10}^{-21}$

$k_b=\frac{k_w}{k_a}=\frac{{10}^{-14}}{{10}^{-21}}$

$={10}^7$

$k_a=C{\alpha }^2$ $pH=-log(k_{a}C{)}^{\frac{1}{2}}$