Answer to Question #208348 in Organic Chemistry for erin

Question

The percentage ionization of 0.12 mol/L nitrous acid is 

Answer 1

$HNO_2-->NO_2^-+H^+$

$PH=1.53$

$$H^+=10^{-1.53}= 2.9512092×10^{-3}$$

%ionization$=\frac{2.9512092×10^{-3}}{0.12}×100=2.46$

$\%ionization=\backslash frac\{2.9512092\times 10^\{-3\}\}\{0.12\}\times 100$