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I have the $100$ microM of K$_3$Fe(CN)$_6$ is dissolved in water and the following data:

$$\begin{array}{c|c} \text{Experiment} & \ce{[Na_2SO_4]}/\pu{M} \\ \hline \mathrm{A} & 0 \\ \mathrm{B} & 1\cdot10^{-3} \\ \mathrm{C} & 3\cdot10^{-3} \\ \mathrm{D} & 8\cdot10^{-3} \\ \mathrm{E} & 1\cdot10^{-2} \\ \hline \end{array}$$

I want to calculate the ionic strength for each experiment and I know that I should use the equation:

$$I = \frac {1}{2} Σ c_i z_i^2$$

and that it for this case will look like:

$$I = \frac {1}{2} ((1)^2*[\ce{K+}]+(-3)^2\cdot[\ce{Fe(CN)6^3-}] + (1)^2*[\ce{Na+}]+(-2)^2\cdot[\ce{SO4^2-}] )$$

However, when I calculate this for the different experiments, I don't get the number which is in our answer key. There is no calculations in the answers and my answers differ quite much. This is the way I calculate the different ionic strengths:

$$I_{\mathrm{A}} = \frac {1}{2} ((1)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot0+(-2)^2\cdot0 ) = 5\cdot 10^{-4}$$

$$I_{\mathrm{B}} = \frac {1}{2} ((1)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(1\cdot10^{-3})+(-2)^2\cdot(1\cdot10^{-3}) ) = 3\cdot10^{-3}$$

$$I_{\mathrm{C}} = \frac {1}{2} ((1)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(3\cdot10^{-3})+(-2)^2\cdot(3\cdot10^{-3}) ) = 8\cdot10^{-3}$$

and so on.

In the answer key $I_{\mathrm{A}}=0.6\cdot10^{-3}$, $I_{\mathrm{B}}=3.6\cdot10^{-3}$ and $I_{\mathrm{C}}=9.6\cdot10^{-3}$. As you can see the difference increases and I don't understand what I am doing wrong. I would truly appreciate if someone could point out my mistakes. Thank you!

EDIT

I am also given this but I don't think they matter for the ionic strength: enter image description here

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  • $\begingroup$ When dissolving $\ce{K4Fe(CN)_6}$, it produces $4$ ions $\ce{K+}$ and $1$ ion $\ce{[Fe{CN}–6]^{4-}}$ . Why do you admit that it produces a negative ion containing some potassium like $\ce{[KFe(CN)_6]^{3-}}$ ? $\endgroup$
    – Maurice
    May 24 at 16:44
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    $\begingroup$ @confused - I have edited your title, removing MathJax (the use of which makes it harder to locate posts via search engines). Also note Maurice's comment. You have specified potassium ferricyanide as one of your salts, but you have the ferrocyanide anion later on. I might be off, but I encourage you to double-check. $\endgroup$
    – Todd Minehardt
    May 24 at 19:14
  • $\begingroup$ That is the way the problem was asked. It states that Fe(CN)$_6$$^{3-}$ reacts with an organic ion to form Fe(CN)$_6$$^{4-}$. $\endgroup$
    – confused
    May 25 at 10:26
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In aqueous solutions, $\ce{K3Fe(CN)6}$ and $\ce{Na2SO4}$ will completely dissociate:

$$\ce{K3Fe(CN)6 (aq) -> 3 K+ (aq) + Fe(CN)6^3- (aq)} \tag1$$ $$\ce{Na2SO4 (aq) -> 2 Na+ (aq) + SO4^2- (aq)} \tag2$$

Thus, from the equation $(1)$, $c_\ce{K+} = 3c_\ce{Fe(CN)6^3-} = 3c_\ce{K3Fe(CN)6}$ and from the equation $(2)$, $c_\ce{Na+} = 2c_\ce{SO4^2-} = 2c_\ce{Na2SO4}$

Hence for condition A $([\ce{Na2SO4}] = 0)$: $$I_{\mathrm{A}} = \frac {1}{2} \left((1)^2 \cdot (3\cdot10^{-4}) + (-3)^2\cdot(1\cdot10^{-4}) + (1)^2 \cdot 0 + (-2)^2\cdot 0 \right) = 6\times 10^{-4} = 0.6\times 10^{-3}$$

Similarly, for condition B $([\ce{Na2SO4}] = 1\cdot10^{-3})$: $$I_{\mathrm{B}} = \frac {1}{2} \left((1)^2\cdot(3\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(2\cdot10^{-3})+(-2)^2\cdot(1\cdot10^{-3}) \right) = 3.6 \times 10^{-3}$$

And, for condition C $([\ce{Na2SO4}] = 3\cdot10^{-3})$: $$I_{\mathrm{C}} = \frac {1}{2} \left((3)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(6\cdot10^{-3})+(-2)^2\cdot(3\cdot10^{-3}) \right) = 9.6 \times 10^{-3}$$

So on for other conditions. Note that for all calculations, ionic strength of $\ce{K3Fe(CN)6}$ is always $0.6\times 10^{-3}$ since $[\ce{K3Fe(CN)6}] = \pu{1.0 \times 10^{-4} M}$ for all conditions.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – confused
    May 25 at 10:37

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