The value of K, for nitrous acid is 4.50×10-4. What is the value of Kp, for its conjugate base, NO2"?

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Chapter1: Chemical Foundations
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**Question:**

The value of \(K_a\) for nitrous acid is \(4.50 \times 10^{-4}\).

What is the value of \(K_b\), for its conjugate base, \(NO_2^-\)? 

**Answer:** 

To find the value of \(K_b\) for the conjugate base \(NO_2^-\), we can use the relationship between the acid dissociation constant (\(K_a\)) and the base dissociation constant (\(K_b\)) for a conjugate acid-base pair, which is given by:

\[ K_a \times K_b = K_w \]

where \(K_w\) is the ion-product constant for water. At 25°C, \(K_w\) is \(1.0 \times 10^{-14}\).

Given: 
\[ K_a = 4.50 \times 10^{-4} \]

We need to find \(K_b\):

\[ K_b = \frac{K_w}{K_a} \]

\[ K_b = \frac{1.0 \times 10^{-14}}{4.50 \times 10^{-4}} \]

\[ K_b = 2.22 \times 10^{-11} \]

Thus, the value of \(K_b\) for the conjugate base \(NO_2^-\) is \(2.22 \times 10^{-11}\).
Transcribed Image Text:**Question:** The value of \(K_a\) for nitrous acid is \(4.50 \times 10^{-4}\). What is the value of \(K_b\), for its conjugate base, \(NO_2^-\)? **Answer:** To find the value of \(K_b\) for the conjugate base \(NO_2^-\), we can use the relationship between the acid dissociation constant (\(K_a\)) and the base dissociation constant (\(K_b\)) for a conjugate acid-base pair, which is given by: \[ K_a \times K_b = K_w \] where \(K_w\) is the ion-product constant for water. At 25°C, \(K_w\) is \(1.0 \times 10^{-14}\). Given: \[ K_a = 4.50 \times 10^{-4} \] We need to find \(K_b\): \[ K_b = \frac{K_w}{K_a} \] \[ K_b = \frac{1.0 \times 10^{-14}}{4.50 \times 10^{-4}} \] \[ K_b = 2.22 \times 10^{-11} \] Thus, the value of \(K_b\) for the conjugate base \(NO_2^-\) is \(2.22 \times 10^{-11}\).
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