# How to effectively wash the precipitate or crystals on the filter and why?

My lab manual says that the precipitation on filter should be washed with a few small portions of the washing liquid. Why can't I just completely fill the filter funnel with the solvent instead of steadily add new tiny portions in it?

Also, how effective this washing process is? How many times should I wash the precipitation before it's completely free from impurities?

What filter funnel of what diameter of the paper filter should I use? Isn't the biggest one the is also the most effective?

Finally, why do I have to study limits when I want to become a chemist, not a mathematician?

Every time the precipitation on filter is washed, there is a certain constant volume $$V_\mathrm{ret}$$ of washing fluid retained on the filter. If the initial concentration of impurity is $$c_0$$, then initial amount of impurity on filter is

$$n_0 = c_0V_\mathrm{ret}$$

and after washing with a new portion of the fluid with the volume $$V_\mathrm{add}$$ concentration of impurity is reduced to

$$c_1 = \frac{V_\mathrm{ret}}{V_\mathrm{ret} + V_\mathrm{add}}c_0.$$

Analogously, after $$n$$-th washing act

$$c_n = \left(\frac{V_\mathrm{ret}}{V_\mathrm{ret} + V_\mathrm{add}}\right)^n c_0 \qquad\text{or}\qquad c_n = \left(\frac{1}{1 + \frac{V_\mathrm{add}}{V_\mathrm{ret}}}\right)^n c_0$$

and, assuming we have a total volume of washing liquid of $$V_\mathrm{tot} = nV_\mathrm{add}$$, the former formula becomes

$$c_n = \left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0$$

Now, lets assume we have a moderate amount of impurities (e.g. $$c_0 =\pu{0.1 mol L-1}$$) that we are trying to get rid of by washing with different amounts of the same volume of the same solvent (say, $$V_\mathrm{tot} = \pu{100 mL}$$). However, we use different filters, for instance:

• a large one (retains $$1/10$$ of the washing liquid, or $$V_\mathrm{ret} = \pu{10 mL}$$);
• a middle one (retains $$1/20$$ of the washing liquid, or $$V_\mathrm{ret} = \pu{5 mL}$$);
• a small one (retains $$1/100$$ of the washing liquid, or $$V_\mathrm{ret} = \pu{1 mL}$$),

and, of course, different portions of the liquid (various $$n$$). Lets plot the concentration of the impurity for each case as a function of washing frequency:

First and foremost, it's obvious that it is always better to wash with small portions many times, rather than two or three times with large quantities of washing liquid. Second, the smaller the filter, the less impurity will be left after the same number $$n$$ of washing acts.

Now, can we get rid of the impurity once and for all by washing with an extremely large number of small portions? In fact, lets use an infinite number and take the following limit

$$\lim_{n\to\infty}c_n = \lim_{n\to\infty}{\left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0},$$

which is, in fact, one of the notable special limits, sometimes called the second remarkable limit:

$$\lim_{x\to\infty} \left(\frac{1}{1+\frac{a}{x}}\right)^x = \frac{1}{e^a}$$

so that in our case we can conclude it's impossible to completely get rid of the impurity only by washing:

$$\lim_{n\to\infty}{\left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0} = c_0\exp{\left(-\frac{V_\mathrm{tot}}{V_\mathrm{ret}}\right)} \neq 0$$

### Conclusions

1. It is impossible to completely remove all impurities by washing the crystals/precipitate on the filter even by washing it infinite number of times.
2. It is more efficient to wash with small amounts of liquid many times than only a couple with large ones.
3. Use filter of the right size: not too big to retain the traces of impurities, but big enough to maintain efficient liquid transport through its surface. When possible, use pleated filter paper.

### Appendix

The figure have been plotted with pgflots. Feel free to play around with it or reuse for your lab manual. Source code:

\documentclass{standalone}
\usepackage{pgfplots}
\usepackage{siunitx}

\begin{document}

\begin{tikzpicture}
\begin{axis}[
title={Effectiveness of washing methods ($c_0 = \SI{0.1}{\mole\per\liter}$)},
xmin=0, xmax=20,
ymin=0, ymax=0.001,
xlabel = $n$,
ylabel = {$c_n$/\si{\mole\per\liter}},
minor tick num=4,
grid,
legend cell align={left},
]

domain=1:20,
samples=200,
solid,
thick,
]
{0.1*(1/(1 + 10/x))^x};
\addlegendentry{$$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.10$$}

domain=1:20,
samples=200,
dashed,
thick,
]
{0.1*(1/(1 + 20/x))^x};
\addlegendentry{$$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.05$$}

domain=1:20,
samples=200,
dotted,
thick,
]
{0.1*(1/(1 + 100/x))^x};
\addlegendentry{$$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.01$$}

\end{axis}
\end{tikzpicture}

\end{document}

• Good for what the explanation covers, but too much washing liquid tends to be bad since the desired compound is typically at least somewhat soluble in the washing liquid.
– MaxW
Jan 12, 2019 at 20:35
• @MaxW Hey, stop poking my spherical cow in a vacuum with a stick :D On a serious note, you are right, the desired product is going to be dissolved too, but who needs it if it's still contaminated? In fact, if plain old washing is the only available choice, then this math works out pretty good for the reasonable $n$, especially when the washing liquid is not just a solvent, but, say some electrolyte solution, containing either cations or anions of the recrystallized compound. Jan 12, 2019 at 20:48
• @andsekisk Thank you for the explanation. Now, assumingly when you say "$V_\rm{tot}$ remains constant across all series" you mean that it is independent of the choice of $i$, correct? In other words, for all $i=1, 2, 3, \dots$, the quantity $V_\rm{tot}$ is constant? Sep 23, 2020 at 9:20
• @andselisk No need at all to apologise, and were you to do so regardless, then apology accepted! I think this answer has a worthy effort. I also apologise because there are a few more comments to come :). (1/2) Sep 23, 2020 at 9:59
• @LinearChristmas Yep, I should think about how I can improve the writing. Thank you for pointing out this issue in logic. Sep 23, 2020 at 10:20