Trial #1:

$$\ce{Fe^3+(aq) + SCN^- (aq) -> FeSCN^2+(aq)}$$

\begin{array}{|r|c|c|c|}\hline \mathrm{Initial~}(M)&\phantom{-}6.00 &\phantom{-}10.00&\phantom{-}0.00\\\hline \mathrm{Change}~(M) & –4.00 & –4.00 &+4.00 \\\hline \mathrm{Equilibrium}~(M)&\phantom{-}2.00&\phantom{-}6.00&\phantom{-}4.00\\\hline \end{array} $$ K = 0.333$$

Consider the reaction represented by the equation: $$\ce{Fe^3+(aq) + SCN- (aq) \longrightarrow [Fe(SCN)]^2+(aq)}$$

Trial #2:

Initial: $10.0 ~M\ce{~Fe^3+(aq)}$ and $8.00~M\ce{ ~SCN- (aq)}$ (same temperature as Trial #1)

Equilibrium: $x~M\ce{~FeSCN^2+(aq)}$ 

I have tried putting $10.0$ and $8.00$ in an ice table, then when I have finished I start solving to get my equation as $$0=\frac{0.33x}{(10.0-x)(8.00-x)}$$ when I'm done I come out with $$0.33=\frac{x}{(x^2-18x+80)}$$ then I plug in for quadratic equation, $$x=\frac{-6.94\pm\sqrt{6.94^2-4\times0.33\times26.4}}{2\times0.33}$$ my answers came out to be $-5$ and $-16.03$. What am I doing wrong?

  • 2
    $\begingroup$ Welcome to Chemistry.SE. Take the tour to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that ‎we aren't doing your homework for you. $\endgroup$
    – user15489
    Commented Jun 22, 2015 at 7:24
  • 3
    $\begingroup$ Apart from this, please consider copying the content relating to your question of the linked power point presentation. This site should be self contained, but links can change and then this question would not help anyone any more. $\endgroup$ Commented Jun 22, 2015 at 7:34
  • $\begingroup$ This is getting better - could you please format your post according to meta.chemistry.stackexchange.com/questions/86/… $\endgroup$
    – user15489
    Commented Jun 22, 2015 at 10:54
  • $\begingroup$ Wow! That is a much improved edit - my retinas are saved! $\endgroup$
    – user15489
    Commented Jun 22, 2015 at 11:51
  • 1
    $\begingroup$ Thank you LordStryker & Martin, I understand it now that LordStryker have helped me edit it, and thank you Martin for further explanation for this problem, so I use fractions instead of decimals, correct? $\endgroup$
    – Kristi
    Commented Jun 22, 2015 at 16:44

1 Answer 1


You are just making a small sign error: \begin{align} ax^2 +bx +c &= 0 & x_{(1;2)} &= \frac{\color{\red}{\pmb{-}}b \pm\sqrt{b^2 - 4ac}}{2a} \end{align}

Taking $K=\frac13$: \begin{align} \frac13x^2 - \frac{21}{3}x +\frac{80}{3} &= 0 & x_{(1;2)} &= \frac{\color{\red}{\pmb{-}}(-\frac{21}{3})\pm\sqrt{\left(\frac{21}{3}\right)^2 - 4\cdot\frac13\frac{80}{3}}}{\frac{2}{3}}\\ && x_{(1;2)} &= \frac{7\pm\sqrt{49-\frac{320}{9}}}{\frac{2}{3}}\\ && &= \left(7\pm\frac{11}{3}\right)\cdot\frac{3}{2}\\ && x_{(1;2)} &= 16; 5 \end{align}

The first value obviously does not make any sense, hence $c(\ce{[Fe(SCN)]^2+})= 5~\mathrm{mol\, L^{-1}}$.


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