How to prepare solutions of different concentrations of potassium from stock solutions with different concentartions? I need to prepare different concentrations of $\ce{K+}$ ($10$, $20$, $30$, $40$, $\pu{50 mM}$) by diluting a solution of $\pu{100 mM}$ $\ce{K+}$ with an isotonic buffer of $\pu{5 mM}$ $\ce{K+}$. 
I need $\pu{4 mL}$ of each concentration.

I really don't remember how to calculate this. Because I need to consider the $\ce{K+}$ within the isotonic buffer right, because they will also add amount of $\ce{K+}$? If I would have diluted the $\pu{100 mM}$ $\ce{K+}$ with just water, then I understand how to do.
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}