Example 13.8: The Product \(K_{\mathrm{a}} \cdot K_{\mathrm{b}} = K_{\mathrm{w}}\)
Use the
\(K_{\mathrm{b}}\) for the nitrite ion,
\(\ce{NO2-}\), to calculate the
\(K_{\mathrm{a}}\) for its conjugate acid.
Solution
\(K_{\mathrm{b}}\) for
\(\ce{NO2-}\) is given in this section as
\(2.17\times 10^{-11}\).
\(K_{\mathrm{b,\ce{NO2-}}}\) \(= 2.17\times 10^{-11}\)
The conjugate acid of
\(\ce{NO2-}\) is
\(\ce{HNO2:}\)
\(\ce{NO2-}\)\(\ce{ + }\)\(\ce{H+}\)\(\ce{<=>}\)\(\ce{HNO2}\)\(\ce{ }\) \(K_{\mathrm{a}}\)
\(\ce{HNO2}\)\(\ce{ + }\)\(\ce{OH-}\)\(\ce{<=>}\)\(\ce{NO2-}\)\(\ce{ + }\)\(\ce{H2O}\)\(\ce{ }\) \(K_{\mathrm{b}}\)
for
\(\ce{HNO2}\) can be calculated using the relationship:
\(K_{\mathrm{a}} \cdot K_{\mathrm{b}} = K_{\mathrm{w}}\)
Solving for Ka, we get:
\(K_{\mathrm{a,\ce{NO2-}}}\) \(= \dfrac{K_{\mathrm{w}}}{K_{\mathrm{b,\ce{NO2-}}}}\)
\(\ \ \ =\dfrac{1.0\times 10^{-14}}{2.17\times 10^{-11}}\)
\(\ \ \ =4.6\times 10^{-4}\)
This answer can be verified by finding the
\(K_{\mathrm{a}}\) for
\(\ce{HNO2}\) in
Appendix H.