$0.1 M \quad\quad 0.2 M \quad0.1 M$
$BaSO _4 \rightleftharpoons Ba ^{+2}+ SO _4{ }^{2-}$
$a - S \quad S \quad S +0.1 \approx 0.1$
$K _{ SP }= S \times 10^{-1}$
$\Rightarrow 1 \times 10^{-10}= S \times 10^{-1}$
$\Rightarrow S =10^{-9} mol L^{-1 }$
So, $S =10^{-9} \times 233\,g\,L ^{-1}$
So, Answer : $233$
$i\,\, HCO_3$ $ii. \,H_3O^+$
$iii.\, HSO_4^-$ $iv.\, HSO_3F$
તો તેમની એસિડિક પ્રબળતાનો સાચો ક્રમ જણાવો.
[અહીં $K_b\,(NH_4OH) = 10^{-5}$ અને $log\,2 = 0.301$ ]