H2SO3(aq)⇌HSO−3(aq)+H+(aq)0.588 M=CCα1(Cα1+Cα1α2)
HSO−3(aq)⇌SO2−3(aq)+H+(aq)Cα1(1−α2)Cα1α2(Cα1+Cα1α2)
α1=√1.7×10−20.588=√17289×2
Therefore, α1<<1⇒(1−α1)≃1
Hence, α2<<1⇒(1−α2)≈1
∴[H+]=Cα1
=√Ka1×C=√17×10−3×0.588
=0.099
pH=−log100.099=1
Nearest integer =1