# 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.

• The only thing you need to remember in chemistry is conservation of matter. "Just water" is a solution where $\ce{K^+}$ concentration is simply zero. Apr 26, 2016 at 9:57
• Thanks for answearing, but I still dont understand how to prepare the 4 concentrations :/ Apr 26, 2016 at 10:17

## 1 Answer

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} &= \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} &= \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 &= \frac{c_\mathrm{total} - c_2}{c_1 -c_2} V_\mathrm{total}\\ V_2 &= \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} & V_\mathrm{total}/\pu{mL} & c_1/\pu{mM} & c_2/\pu{mM} & V_1/\pu{mL} & V_2/\pu{mL} \\\hline 10 & 4 & 100 & 5 & 0.21 & 3.79 \\ 20 & 4 & 100 & 5 & 0.63 & 3.37 \\ 30 & 4 & 100 & 5 & 1.05 & 2.95 \\ 40 & 4 & 100 & 5 & 1.47 & 2.53 \\ 50 & 4 & 100 & 5 & 1.89 & 2.11 \\\hline \end{array}