Iterative mixing problem

Question

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!

Answer 1

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 $n=23$.

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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 $n=23$.

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\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}