what is kb of h₂po₄⁻?

1. Recall the relationship between (K_a), (K_b), and (K_w):
   • For a conjugate - acid - base pair, the product of the acid - dissociation constant ((K_a)) of the acid and the base - dissociation constant ((K_b)) of its conjugate base is equal to the ion - product constant of water ((K_w)). Mathematically, (K_a\times K_b = K_w), where (K_w=1.0\times10^{- 14}) at (25^{\circ}C).
   • (H_2PO_4^-) is an amphiprotic species. To find the (K_b) of (H_2PO_4^-), we need to consider its conjugate acid, which is (H_3PO_4). The relevant (K_a) value for (H_3PO_4) is (K_{a1}=7.5\times10^{-3}).
2. Calculate (K_b) using the (K_a - K_b) relationship:
   • Rearranging the equation (K_a\times K_b = K_w) for (K_b), we get (K_b=\frac{K_w}{K_a}).
   • Substituting (K_w = 1.0\times10^{-14}) and (K_a = 7.5\times10^{-3}) into the equation, we have (K_b=\frac{1.0\times10^{-14}}{7.5\times10^{-3}}).
   • Performing the division: (K_b = 1.33\times10^{-12}).

Explanation:

Step1: Identify the relationship

(K_a\times K_b = K_w)

Step2: Rearrange for (K_b)

(K_b=\frac{K_w}{K_a})

Step3: Substitute values

(K_b=\frac{1.0\times10^{-14}}{7.5\times10^{-3}})

Step4: Calculate (K_b)

(K_b = 1.33\times10^{-12})

Answer:

(1.33\times10^{-12})