How to prepare solutions of different concentrations of potassium from stock solutions with different concentartions?
If you mix two solutions of different concentrations with each other you can easily determine the resulting concentration from the general formula for it:
$$c = \frac{n}{V}\tag1\label{concentration-general}$$

You have one solution with $c_1 = \pu{100 mM}$ and one with $c_2 = \pu{5 mM}$, therefore with $\eqref{concentration-general}$ you can determine the amount of substance for each, and the total amount of substance:
$$n_\mathrm{tot} = c_1\cdot V_1 + c_2\cdot V_2\tag2\label{amount-of-substance-total}$$

Assuming there is not volume contraction, we can simply add them to know the total volume:
$$V_\mathrm{total} = V_1 + V_2 \tag3\label{volume-total}$$

The total concentration therefore is:
$$c_\mathrm{total} = \frac{n_\mathrm{total}}{V_\mathrm{total}}
  = \frac{c_1 V_1 + c_2 V_2}{V_1 + V_2}\tag4\label{concentration-total}$$

For your dilution series you will need to modify it, a $\eqref{volume-total}$ is fixed.
$$V_\mathrm{total} = V_1 + V_2 = \text{const.}$$

Therefore we modify $\eqref{concentration-total}$:
\begin{align}
  c_\mathrm{total} &amp;= 
    \frac{c_1 V_1 + c_2 (V_\mathrm{total} - V_1)}{V_\mathrm{total}}
    \tag{4a}\label{concentration-total-mod-a}\\
  c_\mathrm{total} &amp;= 
    \frac{c_1 (V_\mathrm{total} - V_2) + c_2 V_2}{V_\mathrm{total}}
    \tag{4b}\label{concentration-total-mod-b}\\
\end{align}

Now you have one remaining unknown, solve for it:
\begin{align}
  V_1 &amp;= \frac{c_\mathrm{total} - c_2}{c_1 -c_2} V_\mathrm{total}\\
  V_2 &amp;= \frac{c_\mathrm{total} - c_1}{c_2 -c_1} V_\mathrm{total}\\
\end{align}

Unless you have very accurate equipment, however, I do suggest using larger quantities than $\pu{4 mL}$.
\begin{array}{rrrrrr}
c_\mathrm{total}/\pu{mM} &amp;
V_\mathrm{total}/\pu{mL} &amp;
c_1/\pu{mM} &amp;
c_2/\pu{mM} &amp;
V_1/\pu{mL} &amp;
V_2/\pu{mL} \\\hline
10 &amp; 4 &amp; 100 &amp; 5 &amp; 0.21 &amp; 3.79 \\
20 &amp; 4 &amp; 100 &amp; 5 &amp; 0.63 &amp; 3.37 \\
30 &amp; 4 &amp; 100 &amp; 5 &amp; 1.05 &amp; 2.95 \\
40 &amp; 4 &amp; 100 &amp; 5 &amp; 1.47 &amp; 2.53 \\
50 &amp; 4 &amp; 100 &amp; 5 &amp; 1.89 &amp; 2.11 \\\hline
\end{array}