Question

Given that at 25.0 ∘C Ka for HCN is 4.9×10−10 and Kb for NH3 is 1.8×10−5, calculate Kb for CN− and Ka for NH4+. Enter the Kb value for CN− followed by the Ka value for NH4+, separated by a comma, using two significant figures.

Answer 1

Explanation:

Using the expression :

$K_a\times K_b=K_w$

Where, $K_w$ is the dissociation constant of water.

At $25\ ^0C$, $K_w=10^{-14}$

Thus, for HCN , $K_a=4.9\times 10^{-10}$

$K_b$ for CN⁻ can be calculated as:

$K_a\times K_b=K_w$

$4.9\times 10^{-10}\times K_b=10^{-14}$

$K_b=2.0\times 10^{-5}$

Thus, for NH₃ , $K_b=1.8\times 10^{-5}$

$K_a$ for $NH_4^+$ can be calculated as:

$K_a\times K_b=K_w$

$K_a\times 1.8\times 10^{-5}=10^{-14}$

$K_a=5.6\times 10^{-10}$