Question

You want to clean a 500.-ml flask that has been used to store a 0.900M solution. Each time the flask is emptied, 1.00 ml of solution adheres to the walls, and thus remains in the flask. For each rinse cycle, you pour 9.00 ml of solvent into the flask (to make 10.00 ml total), swirl to mix uniformly, and then empty it. What is the minimum number of such rinses necessary to reduce the residual concentration of 0.000100 M or below?

Answer 1

Answer:

At least four such rinses.

Explanation:

What's the concentration after each rinse?

Let $c_n$ denotes the concentration of the residue after the $n$th rinse. $n$ can be any non-negative integer (which includes zero.)

The initial concentration of the solution in the flask is $\rm 0.900\;M$. That is:

$c_0 = \rm 0.900\;M$.

$\rm 1.00\;mL$ of this $\rm 0.900\;M$ solution is left in the flask after it is emptied for the first time.

The $\rm 9.00\;mL$ solvent will increase the volume of the solution from $\rm 1.00\;mL$ to $\rm 10.00\;mL$. At this point the number of moles of solute in the flask stays unchanged. The concentration of the residue will drop to $1/10$ of the initial value. That is:

$\displaystyle c_1 = \rm \frac{1}{10}\;c_0 = \frac{1}{10}\times 0.900\;M$.

Repeat this process, and the concentration of the residue will drop by a factor of $1/10$ again. That is:

$\displaystyle c_2 = \rm \frac{1}{10}\;c_1 = \frac{1}{10}\times (\frac{1}{10}\times 0.900\;M) = {\left(\frac{1}{10}\right)}^{2}\times 0.900\;M$.

Summarize these values in a table:

$\begin{array}{l|cccc}\text{Number of Rinses} & 0 & 1 & 2 & \dots \\[0.5em]\displaystyle\text{Concentration}\atop\displaystyle\text{of the residue}& \rm 0.900\;M & \rm\dfrac{1}{10}\times 0.900\;M &\rm {\left(\dfrac{1}{10}\right)}^{2}\times 0.900\;M \dots \end{array}$.

The trend in $c_{n}$ is similar to that of a geometric series with

• The concentration of the residue before any rinse, $c_{0} = \rm 0.900\;M$, and
• Common ratio $\displaystyle r = \frac{1}{10}$.

Again, refer to the trend in $c_{n}$ as the value of $n$ increases. The general formula for $c_{n}$, the concentration after the $n$th rinse, will be:

$\displaystyle c_{n} = \rm {\left(\frac{1}{10}\right)}^{n}\; 0.900\;M$.

As a side note, $0 < 1/10 < 1$. As a result, the value of $c_{n}$ will decrease but stay positive as the value of $n$ increases. Increasing the number of rinses will indeed reduce the concentration of the residue.

How many such rinses are required?

In other words, what's the minimum value of $n$ for which $\c_{n} \le \rm 0.000100\;M$?

Recall that

$\displaystyle c_{n} = \rm {\left(\frac{1}{10}\right)}^{n}\; 0.900\;M$.

As a result, $n$ should satisfy the condition:

$\displaystyle \rm {\left(\frac{1}{10}\right)}^{n}\; 0.900\;M \le 0.000100\;M$.

Multiply both sides by

$\displaystyle \frac{1}{\rm 0.900\;M}$,

which is positive and will not change the direction of the inequality:

$\displaystyle \rm {\left(\frac{1}{10}\right)}^{n} \le \frac{0.000100}{0.900}$.

Take the natural logarithm $\ln$ of both sides of the inequality. The function $\ln(x)}$ is increasing as $x$ increases on the range $x > 0$. This function will not change the direction of the inequality, either.

$\displaystyle \rm \ln\left[{\left(\frac{1}{10}\right)}^{n}\right] \le \ln\left(\frac{0.000100}{0.900}\right)$.

Apply the power rule of logarithms: $n$ is an exponent inside the logarithm. That will be equivalent to an expression where $n$ is a coefficient in front of the logarithm operator:

$\displaystyle \rm \ln\left[{\left(\frac{1}{10}\right)}^{n}\right] = n\cdot \left[\ln{\left(\frac{1}{10}\right)}\right]$.

$\displaystyle n\cdot \left[\ln{\left(\frac{1}{10}\right)}\right]\le \ln\left(\frac{0.000100}{0.900}\right)$.

Multiply both sides by

$\displaystyle \frac{1}{\ln \left(\dfrac{1}{10}\right)}}$.

Keep in mind that logarithms of numbers between 0 and 1 (excluding 0 and 1) are negative. The reciprocal of a negative number is still negative. The direction of the inequality will change. That is:

$\displaystyle n\cdot \left[\ln{\left(\frac{1}{10}\right)}\right]\times \frac{1}{\ln \left(\dfrac{1}{10}\right)}}\ge \ln\left(\frac{0.000100}{0.900}\right)\times \frac{1}{\ln \left(\dfrac{1}{10}\right)}}$.

The right-hand side is approximately 3.95. However, $n$ has to be an integer. The smallest integer that is greater than 3.95 is 4. In other words, it takes at least four such rinses to reduce the residual concentration to under 0.000100 M (0.000090 M to be precise.) Three such rinses will give only 0.00900 M.