# Iterative mixing problem

This is a very math-heavy problem from the Chemistry Olympiad:

Beakers P and Q are placed side by side. Initially, P contains 100 cm³ of 1.00 M Na⁺ solution, while Q contains the same volume of 1.00 M K⁺ solution.
10.0 cm³ of P's contents are pipetted into Q which is then stirred. Then, 10.0 cm³ of Q's contents are pipetted back into P, which is then stirred. This constitutes one cycle of mixing.
(a) What is the concentration of Na⁺ in P after four cycles, in $$\pu{mol dm^{-3}}$$? [2 points] Leave your answer to 3 significant figures.
(b) How many cycles of mixing will it take, minimally, for the concentrations of Na⁺ and K⁺ in P to be within 0.0100 M of each other? [3 points]

I solved part (a) but I'm having trouble solving part (b). My solution for part (a):

Initial Setup:

Beaker P starts with 100 cm³ of 1.00 M Na⁺. Beaker Q starts with 100 cm³ of 1.00 M K⁺.

First Cycle:

10.0 cm³ of Na⁺ solution is transferred from P to Q. After mixing, Na⁺ concentration in Q is 0.0909 M. 10.0 cm³ is transferred back from Q to P. New Na⁺ concentration in P is 0.909 M.

Second Cycle:

10.0 cm³ of 0.909 M Na⁺ is transferred from P to Q. After mixing, Na⁺ concentration in Q is 0.1654 M. 10.0 cm³ is transferred back from Q to P. New Na⁺ concentration in P is 0.8345 M.

Third Cycle:

10.0 cm³ of 0.8345 M Na⁺ is transferred from P to Q. After mixing, Na⁺ concentration in Q is 0.2243 M. 10.0 cm³ is transferred back from Q to P. New Na⁺ concentration in P is 0.7745 M.

Fourth Cycle:

10.0 cm³ of 0.7745 M Na⁺ is transferred from P to Q. After mixing, Na⁺ concentration in Q is 0.2726 M. 10.0 cm³ is transferred back from Q to P. New Na⁺ concentration in P is 0.7242 M.

I need help with part (b). I do not even know how to start. Thank you!

• It is always good to summarize and/or complement the verbal description by describing the problem/task with symbolic algebraic expressions. And, keeping it that way until all is ready to just plug-in literal numbers with units. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger permanent value. Commented Aug 10 at 5:19
• The direct using of literal numbers can be advisable only when the procedure is simple and well known to the person, what is not the case. Commented Aug 11 at 10:03

Let $$x_n = c_{\text{Na,P,}n}$$ is the molar concentration of $$\ce{Na}$$ in the vessel $$\mathrm{P}$$ after the cycle $$n$$. $$x_0$$ is therefore the initial concentration.
Let $$y_n = c_{\text{K,Q,}n}$$ is the molar concentration of $$\ce{K}$$ in the vessel $$\mathrm{Q}$$ after the cycle $$n$$. $$y_0$$ is therefore the initial concentration.

$$c_{\text{Na,Q,}n} = x_0 - x_n$$ and $$c_{\text{K,P,}n} = y_0 - y_n$$ because of the conservation of total molar amounts.

$$n_\text{Na,tot} = c_\text{Na,P} \cdot V_\text{P} + c_\text{Na,Q} \cdot V_\text{Q} \\ =(c_\text{Na,P} + c_\text{Na,Q}) \cdot \pu{100 mL}=c_\text{Na,P,0} \cdot \pu{100 mL} \implies \\ \implies c_\text{Na,P} + c_\text{Na,Q} = c_\text{Na,P,0}$$

Note that

• The sum of concentrations of $$\ce{Na+}$$ (the same for $$\ce{K+}$$) in both vessels is after each cycle $$x_0$$ (= $$y_0$$).
• The sum of the concentrations of $$\ce{Na+}$$ and $$\ce{K+}$$ in each vessel is after each cycle $$x_0$$ (= $$y_0$$) as well.

The baseline volumes $$V=\pu{100 mL}$$. The exchanged volume $$V_1=\pu{10 mL}$$

After the start - the cycle $$n=0$$ :

$$\begin{array}{|c|c|c|} \hline \text{Element} & \text{mol. conc. in P} & \text{mol. conc. in Q} \\ \hline \ce{Na} & x_0 & 0 \\ \hline \ce{K} & 0 & y_0 \\ \hline \end{array}$$

After the cycle $$n$$ :

$$\begin{array}{|c|c|c|} \hline \text{Element} & \text{mol. conc. in P} & \text{mol. conc. in Q} \\ \hline \ce{Na} & x_n & x_0 - x_n \\ \hline \ce{K} & y_0 - y_n & y_n \\ \hline \end{array}$$

Transferring $$V_1$$ to $$\mathrm{Q}$$:

$$\begin{array}{|c|c|c|} \hline \text{Element} & \text{mol. conc. in P} & \text{mol. conc. in Q} \\ \hline \ce{Na} & x_n & \dfrac{V_1x_n + V(x_0 - x_n)}{V + V_1} \\ \hline \ce{K} & y_0 - y_n & \dfrac{V_1(y_0 - y_n) + V \cdot y_n}{V + V_1} \\ \hline \end{array}$$

Transferring $$V_1$$ back to $$\mathrm{P}$$ with $$V_\text{P}=V_\text{Q}= V$$. Here we use the fact that concentrations in P will become complementary to those in Q, with their sum being equal to the initial concentration, respectively:

$$\begin{array}{|c|c|c|} \hline \text{Element} & \text{mol. conc. in P} & \text{mol. conc. in Q} \\ \hline \ce{Na} & x_0 - \dfrac{V_1x_n + V(x_0 - x_n)}{V + V_1} & \dfrac{V_1x_n + V(x_0 - x_n)}{V + V_1} \\ \hline \ce{K} & y_0 - \dfrac{V_1(y_0-y_n) + V\cdot y_n}{V + V_1} & \dfrac{V_1(y_0-y_n) + V\cdot y_n}{V + V_1} \\ \hline \end{array}$$

Rearranging:

$$\begin{array}{|c|c|} \hline x_{n+1} = x_0 - \dfrac{V_1x_n + V(x_0 - x_n)}{V + V_1} & y_{n+1}=\dfrac{V_1(y_0-y_n) + V\cdot y_n}{V + V_1} \\ \hline x_{n+1} =\dfrac{x_0 (V + V_1) - V_1x_n - V(x_0 - x_n)}{V + V_1} & y_{n+1}=\left(\dfrac{V - V_1}{V + V_1}\right)y_n + \left(\dfrac{V_1}{V + V_1}\right)y_0 \\ \hline x_{n+1} =\left(\dfrac{V - V_1}{V + V_1}\right)x_n + \left(\dfrac{V_1}{V + V_1}\right)x_0 & y_{n+1}=\left(\dfrac{V - V_1}{V + V_1}\right)y_n + \left(\dfrac{V_1}{V + V_1}\right)y_0 \\ \hline x_{n+1} =\left(\dfrac{9}{11}\right)x_n + \left(\dfrac{1}{11}\right)x_0 & y_{n+1}=\left(\dfrac{9}{11}\right)y_n + \left(\dfrac{1}{11}\right)y_0 \\ \hline \end{array}$$

Here, we can notice that despite the mixing volume ratios are somewhat assymetrical (1 to 10, 1 to 9), the concentration shifts counterbalance that. As the result, the iterative dilution formulas are symmetrical for both $$\ce{Na+}$$ and $$\ce{K+}$$.

The last particular formulas has been used in online Excel (MS OneDrive) to calculate iterations of the values. Note that the values converge to the half of the initial concentrations, what is obvious, as each of elements is going to be distributed to the doubled volume. See the table at the botton of the post.

The answer for b) question is $$n=23$$.

More elegant is using the equation merging all iteration steps, tested in formula test column: $$x_n = \frac{1}{2} \cdot x_0 \left[\left(\frac{9}{11}\right)^n + 1\right]$$

We require:

$$x_n - c_{\text{P,K,}n} = x_n - (y_0 - y_n) = x_n - (x_0 - x_n) \le \dfrac{x_0}{100}$$ $$\frac{1}{2} \cdot x_0 \left[\left(\frac{9}{11}\right)^n + 1\right] - \left(x_0 - \frac{1}{2} \cdot x_0 \left[\left(\frac{9}{11}\right)^n + 1\right]\right) \le \dfrac{x_0}{100}$$

$$x_0 \left[\left(\frac{9}{11}\right)^n + 1\right] \le 1.01 \times x_0$$

$$\left(\frac{9}{11}\right)^n \le 0.01$$ $$n \cdot \log \left(\frac{9}{11}\right) \le -2$$ $$n \ge 22.94$$

what is in agreement with the previously determined $$n=23$$.

$$\begin{array}{|c|c|c|c|c|} \hline \text{Round} & c(\text{Na,P}) \, [\text{mol/L}] & c(\text{K,P}) \, [\text{mol/L}] & \text{diff} \, [\text{mol/L}] & \text{formula test} \\ \hline 0 & 1.000 & 0.000 & 1.000 & 1.000 \\ 1 & 0.909 & 0.091 & 0.818 & 0.909 \\ 2 & 0.835 & 0.165 & 0.669 & 0.835 \\ 3 & 0.774 & 0.226 & 0.548 & 0.774 \\ 4 & 0.724 & 0.276 & 0.448 & 0.724 \\ 5 & 0.683 & 0.317 & 0.367 & 0.683 \\ 6 & 0.650 & 0.350 & 0.300 & 0.650 \\ 7 & 0.623 & 0.377 & 0.245 & 0.623 \\ 8 & 0.600 & 0.400 & 0.201 & 0.600 \\ 9 & 0.582 & 0.418 & 0.164 & 0.582 \\ 10 & 0.567 & 0.433 & 0.134 & 0.567 \\ 11 & 0.555 & 0.445 & 0.110 & 0.555 \\ 12 & 0.545 & 0.455 & 0.090 & 0.545 \\ 13 & 0.537 & 0.463 & 0.074 & 0.537 \\ 14 & 0.530 & 0.470 & 0.060 & 0.530 \\ 15 & 0.525 & 0.475 & 0.049 & 0.525 \\ 16 & 0.520 & 0.480 & 0.040 & 0.520 \\ 17 & 0.516 & 0.484 & 0.033 & 0.516 \\ 18 & 0.513 & 0.487 & 0.027 & 0.513 \\ 19 & 0.511 & 0.489 & 0.022 & 0.511 \\ 20 & 0.509 & 0.491 & 0.018 & 0.509 \\ 21 & 0.507 & 0.493 & 0.015 & 0.507 \\ 22 & 0.506 & 0.494 & 0.012 & 0.506 \\ 23 & 0.505 & 0.495 & 0.0099 & 0.505 \\ 24 & 0.504 & 0.496 & 0.008 & 0.504 \\ \hline \end{array}$$

• Poutnik, Very nice answer which your rock solid skills, but I see a hidden 17th century French mathematician in you. They used to frighten readers by adding it is easy to show thay, it can be readily seen, it is trivial to show...this practice become so common in mathematics that there is a term now, called proof by intimidation :-)
– ACR
Commented Aug 10 at 12:05
• Just for testing, I asked ChatGPT-4o to solve the original question, it did the first cycle correctly and then went down the rabbit hole of incorrect answers. For students: don't rely on chatbots to solve chemistry problems. It is very good at the suggested items above but still bad at solving chem problems without any human guidance.
– ACR
Commented Aug 10 at 12:26
• Sure, I like ChatGPT but the suggestion was for new students only to use it carefully (in general). It is trivial to show that ChatGPT still needs to improve its chemistry problem solving skills.
– ACR
Commented Aug 10 at 12:33
• @ACT Feel free to make suggestions if you see space for more of simplification or clarification. We are often blind toward our own ways. Commented Aug 11 at 9:42