Concentration of H_2SO_4 and Na_2SO_4 in a so | vedclass

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(D) $H_2SO_4 \rightarrow H^{+} + HSO_4^{-}$$Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}$$HSO_4^{-} \rightleftharpoons H^{+} + SO_4^{2-}; \ K_{a2} = 1.2 \

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$6$

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(D) $H_2SO_4 ightarrow H^{+} + HSO_4^{-}$$Na_2SO_4 ightarrow 2Na^{+} + SO_4^{2-}$$HSO_4^{-} ightleftharpoons H^{+} + SO_4^{2-}; \ K_{a2} = 1.2 	imes 10^{-2} \ M$Let the concentration of $SO_4^{2-}$ be $[SO_4^{2-}] = 1.8 	imes 10^{-2} - x$,where $x$ is the amount of $SO_4^{2-}$ consumed by $H^{+}$ to form $HSO_4^{-}$.$K_{a2} = \frac{[H^{+}][SO_4^{2-}]}{[HSO_4^{-}]} = \frac{(1+x)(1.8 	imes 10^{-2} - x)}{(1-x)} = 1.2 	imes 10^{-2}$Assuming $x$ is small,$1+x \approx 1$ and $1-x \approx 1$,so $1.8 	imes 10^{-2} - x = 1.2 	imes 10^{-2} \Rightarrow x = 0.6 	imes 10^{-2} \ M$.Thus,$[SO_4^{2-}] = 1.8 	imes 10^{-2} - 0.6 	imes 10^{-2} = 1.2 	imes 10^{-2} \ M$.For $PbSO_4 ightleftharpoons Pb^{2+} + SO_4^{2-}$,$K_{sp} = [Pb^{2+}][SO_4^{2-}] = s(s + 1.2 	imes 10^{-2}) = 1.6 	imes 10^{-8}$.Since $s$ is very small,$s + 1.2 	imes 10^{-2} \approx 1.2 	imes 10^{-2}$.$s(1.2 	imes 10^{-2}) = 1.6 	imes 10^{-8} \Rightarrow s = \frac{1.6}{1.2} 	imes 10^{-6} = 1.33 	imes 10^{-6} \ M$.Comparing with $X 	imes 10^{-Y} \ M$,we get $Y = 6$.

Metal sulfide	Solubility product
$A$. $PbS$	$I$. $4.0 \times 10^{-53}$
$B$. $HgS$	$II$. $8.0 \times 10^{-28}$
$C$. $MnS$	$III$. $1.6 \times 10^{-24}$
$D$. $ZnS$	$IV$. $2.5 \times 10^{-13}$

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