b
\(H _{2} SO _{3}\) [Dibasic acid]

\(c=0.588 \,M\)

\(\Rightarrow \quad pH\) of solution \(P\) due to First dissociation only

since \(K _{ a },>> Ka _{2}\)

\(\Rightarrow\) First dissociation of \(H _{2} SO _{3}\)

\(H _{2} SO _{3}( aq ) \rightleftharpoons H ^{\oplus}( aq )+ HSO _{3}^{-}( aq ): ka _{1}=1.7 \times 10^{-2}\)

\(t=0\) \(\;\;C\)

\(t\) \(\quad\;\;\;\;C - x\;\; \, x \;\;\, x\)

\(\Rightarrow \quad Ka _{1}=\frac{1.7}{100}=\frac{\left[ H ^{\oplus}\right]\left[ HSO _{3}^{-}\right]}{\left[ H _{2} SO _{3}\right]}\)

\(\Rightarrow \frac{1.7}{100}=\frac{ x ^{2}}{(0.58- x )}\)

\(\Rightarrow 1.7 \times 0.588-1.7 x =100 x ^{2}\)

\(\Rightarrow \quad 100 x^{2}+1.7 x-1=0\)

\(\Rightarrow \quad\left[ H ^{\oplus}\right]= x =\frac{-1.7+\sqrt{(1.7)^{2}+4 \times 100 \times 1}}{2 \times 100}=0.09186\)

Therefore \(pH\) of sol. is \(: pH =-\log \left[ H ^{\oplus}\right]\)

\(\Rightarrow \quad pH =-\log (0.09186)=1.036 \simeq 1\)