# Need help finding equilibrium constant for equilibrium system [closed]

The reaction between iron (III) ions and thiocyanate ions, SCN$$^-$$, is an equilibrium reaction:

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

When $$20$$cm$$^3$$ of $$1.0\times10^{-3}$$ mol dm$$^{-3}$$ Fe(NO$$_3$$)$$_3$$ (aq) is mixed with $$20$$cm$$^3$$ of $$1.0\times10^{-3}$$ mol dm$$^{-3}$$ KSCN (aq), the concentration of FeSCN$$^{2+}$$ (aq) at equilibrium is $$8.0 \times 10^{-3}$$ mol dm$$^{-3}$$. What is $$K_{c}$$ for the equilibrium system?

The question also includes a hint, saying that,

After mixing, the concentration of each reactant is half, because the volume of solution is doubled.

My attempts to solve this question are as follows:

The initial number of moles of Fe$$^{3+}$$ and SCN$$^- = 1\times 10^{-3}\times20\times10^{-3}=2\times10^{-5}$$mol

At equilibrium, the concentration of Fe$$^{3+}$$ and SCN$$^- = \displaystyle\frac{2\times10^{-5}}{20\times10^{-3}} - (8\times10^{-3})=1\times10^{-3} - (8\times10^{-3})$$mol dm$$^{-3}$$

I have obtained a negative value for the concentration Fe$$^{3+}$$ and SCN$$^-$$, so I think I have made some wrong calculations somewhere. I am confused about the volume and concentrations of Fe$$^{3+}$$ and SCN$$^-$$.

Could you tell me where I made a mistake?

• You don't not to write that cumbersome codes, your first equation is \ce{Fe^{3+}(aq) + SCN-(aq) <=> FeSCN^{2+}(aq)} – Zenix May 21 at 14:31
• I didn't see your calculations, but, concentration of $\ce{FeSCN^{2+}}$ is too high to be true. – Zenix May 21 at 14:39
• I think there's a typo: $\ce{[FeSCN^2+]}$ can't be higher than $\ce{[Fe^3+]}$ or $\ce{[SCN^-]}$. I think it should be $8 \times 10^{-5}$ – Aniruddha Deb May 21 at 15:22
• I find following your logic pretty hard since you are randomly omitting units. The measured quantity consists of a number and a unit, and you are not allowed to omit either of them. Also, your markup is indeed pretty cumbersome; please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. – andselisk May 21 at 15:30
• The $\ce{[FeSCN^2-]}$ at equilibrium should be $\lt 5.0 \times 10^{-4}$. – Mathew Mahindaratne May 21 at 15:53