What is the concentration of ferrous ion?

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What is the concentration of ferrous ion?

A solution is prepared by mixing 75 ml of 0.030 M $F e S O_{4}$ $FeSO_4$ with 125 ml of 0.20 M KCN?The stability constant for the complex ${[F e {(C N)}_{6}]}_{6}^{- 4}$ $[Fe(CN)_6]^-4$ is $1 \cdot 10^{24}$ $1* 10^24$ .[Given: ${(0.0575)}^{6}$ $(0.0575)^6$ = $3.62 \cdot 10^{- 8}$ $3.62*10^-8$ ].

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Answer 1 · 2018-01-10T18:52:10

I got $[{Fe}^{2 +}] = 3.2 \times 10^{- 19} M$ $["Fe"^(2+)] = 3.2 xx 10^(-19) "M"$ .

Well, the reaction is:

${FeSO}_{4} (a q) + 6 KCN (a q) \to K_{4} [Fe {(CN)}_{6}] (a q) + K_{2} {SO}_{4} (a q)$ $"FeSO"_4(aq) + 6"KCN"(aq) -> "K"_4["Fe"("CN")_6] (aq) + "K"_2"SO"_4(aq)$

For simplicity, we include only the complexation participants:

${Fe}^{2 +} (a q) + 6 {CN}^{-} (a q) \to {[Fe {(CN)}_{6}]}_{6}^{4 -} (a q)$ $"Fe"^(2+)(aq) + 6"CN"^(-)(aq) -> ["Fe"("CN")_6]^(4-)(aq)$

The concentration of iron at the start of the reaction is

$0.030 M \times \frac{75 mL}{75 + 125 mL} = 0.0113 M$ $"0.030 M" xx "75 mL"/"75 + 125 mL" = "0.0113 M"$ ,

due to dilution by mixing.

The concentration of cyanide at the start of the reaction is

$0.20 M \times \frac{125 mL}{75 + 125 mL} = 0.125 M$ $"0.20 M" xx "125 mL"/"75 + 125 mL" = "0.125 M"$ ,

again due to dilution by mixing.

Therefore, we construct the ICE table using molar concentrations:

${Fe}^{2 +} (a q) + 6 {CN}^{-} (a q) \to {[Fe {(CN)}_{6}]}_{6}^{4 -} (a q)$ $"Fe"^(2+)(aq) " "+" " 6"CN"^(-)(aq) -> ["Fe"("CN")_6]^(4-)(aq)$

$I 0.0113 0.125 0$ $"I"" "0.0113" "" "" "" "0.125" "" "" "" "" "0$
$C - x - 6 x + x$ $"C"" "-x" "" "" "" "" "-6x" "" "" "" "+x$
$E 0.0113 - x 0.125 - 6 x x$ $"E"" "0.0113-x" "" "0.125-6x" "" "" "x$

The equilibrium formation of the complex is then given by:

$K_{f} = 1 \times 10^{24} = \frac{[{[Fe {(CN)}_{6}]}_{6}^{4 -}]}{[{Fe}^{2 +}] {[{CN}^{-}]}^{6}}$ $K_f = 1 xx 10^24 = ([["Fe"("CN")_6]^(4-)])/(["Fe"^(2+)]["CN"^(-)]^6)$

$= \frac{x}{(0.0113 - x) {(0.125 - 6 x)}^{6}}$ $= x/((0.0113 - x)(0.125 - 6x)^6)$

Since $K_{f}$ $K_f$ is so large, we approximate that $x \approx 0.0113$ $x ~~ 0.0113$ , and notice that iron is the limiting reactant. For the purposes of maximizing accuracy, consider the following solving method.

${ln K}_{f} = ln (10^{24}) = 24 ln 10$ $ln K_f = ln(10^24) = 24ln10$

$= ln [\frac{x}{(0.0113 - x) {(0.125 - 6 x)}^{6}}]$ $= ln[x/((0.0113 - x)(0.125 - 6x)^6)]$

$= ln x - ln (0.0113 - x) - 6 ln (0.125 - 6 x)$ $= lnx - ln(0.0113 - x) - 6ln(0.125 - 6x)$

Now, we can plug in $x = 0.0113$ $x = 0.0113$ EXCEPT for where we would get $ln \to - \infty$ $ln -> -oo$ . This gives us:

$24 ln 10 = ln (0.0113) - ln (0.0113 - x) - 6 ln (0.125 - 6 (0.0113))$ $24ln10 = ln(0.0113) - ln(0.0113 - x) - 6ln(0.125 - 6(0.0113))$

Evaluate the right-hand side to get:

$24 ln 10 = - 4.4830 - ln (0.0113 - x) - 6 ln (0.0572)$ $24ln10 = -4.4830 - ln(0.0113 - x) - 6ln(0.0572)$

If you notice, this would be where that hint came from. Without the $ln$ $ln$ , we would have gotten that ${(0.0575)}^{6} = 3.62 \times 10^{- 8}$ $(0.0575)^6 = 3.62 xx 10^(-8)$ , since ${ln a}^{c} = c ln a$ $lna^c = clna$ . We proceed to get:

$24 ln 10 = - 4.4830 - ln (0.0113 - x) + 17.1672$ $24ln 10 = -4.4830 - ln(0.0113 - x) + 17.1672$

$= 12.6842 - ln (0.0113 - x)$ $= 12.6842 - ln(0.0113 - x)$

Solving for $ln (0.0113 - x)$ $ln(0.0113 - x)$ , we obtain:

$ln (0.0113 - x) = 12.6842 - 24 ln 10$ $ln(0.0113 - x) = 12.6842 - 24ln10$

$= - 42.5778$ $= -42.5778$

As a result,

$[{Fe}^{2 +}] = 0.0113 - x$ $color(blue)(["Fe"^(2+)]) = 0.0113 - x$

$= e^{- 42.5778}$ $= e^(-42.5778)$

$= color(blue)ul(3.2 xx 10^(-19) "M")$ to two significant figures.

And just to verify that this is correct... $x = 0.0113 - e^(-42.5778)$ , and so we do indeed get $K_f$ back:

$K_f = (0.0113 - e^(-42.5778))/((0.0113 - 0.0113 + e^(-42.5778))(0.125 - 6(0.0113 - e^(-42.5778)))^6)$

$~~ (0.0113)/(e^(-42.5778)(0.125 - 6(0.0113))^6)$

$= ul(1.0_(0001349) xx 10^24) = 1 xx 10^24$ $color(blue)(sqrt"")$

where the subscripts indicate digits past the last significant figure.

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What is the concentration of ferrous ion?

A solution is prepared by mixing 75 ml of 0.030 M $F e S O_{4}$ $FeSO_4$ with 125 ml of 0.20 M KCN?The stability constant for the complex ${[F e {(C N)}_{6}]}_{6}^{- 4}$ $[Fe(CN)_6]^-4$ is $1 \cdot 10^{24}$ $1* 10^24$ .[Given: ${(0.0575)}^{6}$ $(0.0575)^6$ = $3.62 \cdot 10^{- 8}$ $3.62*10^-8$ ].

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A solution is prepared by mixing 75 ml of 0.030 M $F e S O_{4}$ $FeSO_4$ with 125 ml of 0.20 M KCN?The stability constant for the complex ${[F e {(C N)}_{6}]}_{6}^{- 4}$ $[Fe(CN)_6]^-4$ is $1 \cdot 10^{24}$ $1* 10^24$ .[Given: ${(0.0575)}^{6}$ $(0.0575)^6$ = $3.62 \cdot 10^{- 8}$ $3.62*10^-8$ ].