Question

Calculate the Ka for the following acids using the given information.
b) 0.0400 M solution of HClO2, pH = 1.80

Answer 1

you have to do an ICE table and consider the final concentration of H+ equal to the [ClO2- iqual to 10^-pH, and the final concentration of the no dissociated acid as the original concentration minus the concentration of H+. With these values you can calculate the Ka

Answer 2

\[pH=-\log_{}\left[ H ^{+} \right] ; \left[ H ^{+} \right] = 10^{-pH}\]

\[\left[ H ^{+} \right]= 10^{-1.8}= 0.0158 M\]
\[\left[ H ^{+} \right]=\left[ ClO _{2}^{-} \right] =0.0158 M\]
\[\left[ HClO \right]=0.0400 M - 0.0158 M =0.0242 M\]
\[Ka = \frac{ \left[ H ^{+} \right]\left[ ClO _{2}^{-} \right] }{ \left[ HClO _{2} \right] } = \frac{ \left( 0.0158 \right)^{2} }{ 0.0400-0.0158 }=0.0104\]

Answer 3

thank you