# clock reaction dilutions for KI

Stock solution 0.1M KI solution 100mL

Volume of KI
30
10.5
5.25

Volume of water 0 19.5 14.75

concentration of KI 0.1 0.05 0.025

How did they calculate these numbers? I tried c1 v1 = c2 v2 For example for 0.05M: (0.1)(v1) = (0.05)(30/1000) v1= 0.015L = 15mL, what does this 15mL represent. How did they calculate the volume of KI and water.

I want to dilute KI and I want the final solution to be 30mL. And the concentrations that I need are shown above (0.1, 0.05, 0.025, 0.0125).

The formula

$$c_1\cdot V_1=c_2\cdot V_2$$

is correct.

Since your stock solution has a concentration of $c=0.1\ \mathrm{mol\ l^{-1}}$, the value for the initial concentration $c_1$ is always

$$c_1=0.1\ \mathrm{mol\ l^{-1}}$$

Since you want the final volume of each solution to be $V=30\ \mathrm{ml}$, the value for the final volume $V_2$ is always

$$V_2=30\ \mathrm{ml}$$

You want four individual solutions (labelled $\mathrm a$, $\mathrm b$, $\mathrm c$, and $\mathrm d$) with four different final concentrations ($0.1\ \mathrm{mol\ l^{-1}}$, $0.05\ \mathrm{mol\ l^{-1}}$, $0.025\ \mathrm{mol\ l^{-1}}$, $0.0125\ \mathrm{mol\ l^{-1}}$. These are the individual values for the final concentration $c_2$:

\begin{align} c_{2,\mathrm{a}}&=0.1\ \mathrm{mol\ l^{-1}}\\[6pt] c_{2,\mathrm{b}}&=0.05\ \mathrm{mol\ l^{-1}}\\[6pt] c_{2,\mathrm{c}}&=0.025\ \mathrm{mol\ l^{-1}}\\[6pt] c_{2,\mathrm{d}}&=0.0125\ \mathrm{mol\ l^{-1}} \end{align}

Now you can rearrange the above-mentioned equation to solve for the initial volume of the stock solution $V_1$:

\begin{alignat}{2} V_1&=\frac{c_2\cdot V_2}{c_1}\\[6pt] V_{1,\mathrm a}&=\frac{0.1\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=30\ \mathrm{ml}\\[6pt] V_{1,\mathrm b}&=\frac{0.05\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=15\ \mathrm{ml}\\[6pt] V_{1,\mathrm c}&=\frac{0.025\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=7.5\ \mathrm{ml}\\[6pt] V_{1,\mathrm d}&=\frac{0.0125\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=3.75\ \mathrm{ml} \end{alignat}

You measure these volumes of the stock solution and transfer them into four separate vessels.

Finally, you have to fill up each solution with pure water to the final volume of $V_2=30\ \mathrm{ml}$. The required additional volume of water $V_{\ce{H2O}}$ is approximately

\begin{alignat}{2} V_{\ce{H2O}}&\approx V_2-V_1\\[6pt] V_{\ce{H2O},\mathrm a}&\approx30\ \mathrm{ml}-30\ \mathrm{ml}&&=0\ \mathrm{ml}\\[6pt] V_{\ce{H2O},\mathrm b}&\approx30\ \mathrm{ml}-15\ \mathrm{ml}&&=15\ \mathrm{ml}\\[6pt] V_{\ce{H2O},\mathrm c}&\approx30\ \mathrm{ml}-7.5\ \mathrm{ml}&&=22.5\ \mathrm{ml}\\[6pt] V_{\ce{H2O},\mathrm d}&\approx30\ \mathrm{ml}-3.75\ \mathrm{ml}&&=26.25\ \mathrm{ml} \end{alignat}

• Are you sure? Because the person doing the experiment got different values. Oct 11, 2016 at 5:06
• Are you sure, because they got different results. For c2 =0.05M, they used 10.5mL Potassium iodide, and 19.5mL water. For c=3 = 0.025M, they used 5.25mL KI and 14.75mL water. please help. how did they get these exact values since its very different than the values you calculated Oct 11, 2016 at 5:11
• @Shahad You can check all values yourself using the given equation $c_1\cdot V_1=c_2\cdot V_2$. You can see that your first examples do dont give the concentrations that you want: $0.1\ \mathrm{mol\ l^{-1}}\times10.5\ \mathrm{ml}\neq0.05\ \mathrm{mol\ l^{-1}}\times\left(10.5\ \mathrm{ml}+19.5\ \mathrm{ml}\right)$ and $0.1\ \mathrm{mol\ l^{-1}}\times5.25\ \mathrm{ml}\neq0.025\ \mathrm{mol\ l^{-1}}\times\left(5.25\ \mathrm{ml}+14.75\ \mathrm{ml}\right)$. Also note that $5.25\ \mathrm{ml}+14.75\ \mathrm{ml}$ gives $20\ \mathrm{ml}$ of solution and you want $30\ \mathrm{ml}$.
– user7951
Oct 11, 2016 at 6:55