Calculating the ionic strength for an aqueous solution containing potassium ferricyanide and sodium sulfate I have the $100$ microM of K$_3$Fe(CN)$_6$ is dissolved in water and the following data:
$$\begin{array}{c|c} 
\text{Experiment} &amp; \ce{[Na_2SO_4]}/\pu{M} \\ \hline
\mathrm{A} &amp; 0 \\
\mathrm{B} &amp; 1\cdot10^{-3} \\ 
\mathrm{C} &amp; 3\cdot10^{-3} \\ 
\mathrm{D} &amp; 8\cdot10^{-3} \\ 
\mathrm{E} &amp; 1\cdot10^{-2} \\
\hline
\end{array}$$
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:
In aqueous solutions, $\ce{K3Fe(CN)6}$ and $\ce{Na2SO4}$ will completely dissociate:
$$\ce{K3Fe(CN)6 (aq) -&gt; 3 K+ (aq) + Fe(CN)6^3- (aq)} \tag1$$
$$\ce{Na2SO4 (aq) -&gt; 2 Na+ (aq) + SO4^2- (aq)} \tag2$$
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.