# Chapter 16 Principles of Chemical Reactivity: The Chemistry of Acids and Bases - Study Questions - Page 629d: 71

#### Work Step by Step

First equilibrium: $K_{a1}=[HSO_3^-][H_3O^+]/[H_2SO_3]$ $1.2\cdot10^{-2}=x^2/(0.45\ M-x)$ $x=0.068\ M$ Second equilibrium: $K_{a2}=[SO_3^{2-}][H_3O^+]/[HSO_3^-]$ $6.2\cdot10^{-8}=y(0.068+y)/(0.068-y)$ $y=6.2\cdot10^{-8}\ M$ a) $pH=-log(0.068\ M)=1.17$ b) $[SO_3^{2-}]=y=6.2\cdot10^{-8}\ M$

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