Question

The Ksp of FeS = 4 x 10^19 at
298 K. The minimum
concentration of H+ ions
required to prevent the
precipitation of FeS from a
0.01 M solution Fe+2 salt by
passing H2S
(Given H2S ka1×ka2 = 10^21)​

Answer 1

Answer:

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Answer 2

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The Ksp of FeS=4×10⁻¹⁹ at 298 K. The minimum concentration of H+ ions required to prevent the precipitation of FeS from a 0.01 M solution Fe²⁺ salt by passing H₂S(0.1M) (Given H₂S ka₁×ka₁=10⁻²¹)?

Answer:

The minimum concentration of H⁺ ions needed to prevent the precipitation of FeS would be equal to 1.6×10⁻³M.

Explanation:

Given for FeS the value of $K_{sp}=4*10^{-19}$,

$[Fe^{2+}]=0.01M$ and $[H_{2}S]=0.1M$

$H_{2}S$  ⇄  $H^{+}+ HS^{-}$

$K_{a_{1}}}=\frac{[H^{+}][HS^{-}]}{[H_{2}S]}$

Further, $HS^{-}$  ⇄  $H^{+}+s^{2-}$

$K_{a_{2}}}=\frac{[H^{+}][S^{2-}]}{[HS^{-}]}$

Given  $K_{a_{1}}}*K_{a_{2}}}=10^{-21}$                    ................(1)

Put the values equilibrium constants in above eq.(1)

$\frac{[H^{+}][HS^{-}]}{[H_{2}S]} *\frac{[H^{+}][S^{2-}]}{[HS^{-}]}=10^{-21}$

$\frac{[H^{+}]^{2}[S^{2-}]}{[H_{2}S]} =10^{-21}$                                  .............(2)

To find concentration of $S^{2-}$:

$FeS$   ⇄   $F^{2+}+S^{2-}$

$K_{sp}=\frac{[Fe^{2+}][S^{2-}]}{[FeS]}$

$4*10^{-19}=\frac{[0.01][S^{2-}]}{(1)}$

$[S^{2-}]=4*10^{-17}$

Put the above value in eq. (2)

$\frac{[H^{+}]^{2}[4*10^{-17}]}{[0.1]} =10^{-21}\\{[H^{+}]^{2}}=\frac{10^{-5}}{4} \\{[H^{+}]}=1.6*10^{-3}M$