The pH of 0.400 M sodium nitrite,NaNO₂,is 8.42.What is the K_b for this base?

Question

Quizplus · Accepted Answer

The $K_b$ for sodium nitrite can be calculated using the given pH and the concentration of the solution. Since the pH is 8.42, the $pOH$ is 14 - 8.42 = 5.58. The $OH^-$ concentration can be found using $pOH = -\log[OH^-]$, which gives $[OH^-] = 10^{-5.58}$. The $K_b$ is calculated using the formula $K_b = \frac{[OH^-]^2}{[Base] - [OH^-]}$. Given the concentration of the base is 0.400 M and assuming $[OH^-]$ is much smaller than the initial concentration of the base, the approximation $[Base] - [OH^-] \approx [Base]$ can be used. Plugging in the values, the $K_b$ calculation aligns with the value given in option A, $1.7 \times 10^{-11}$.

The PH of 0