Solution Dilution#

Weed Killer#

I wanted to make some greener weed killer. I am not keen on using chemicals like Roundup. I thought I would try a greener solution. I found a random recipe on the internet:

  • 1 teaspoon dish soap per gallon

  • 2 cups of salt per gallon

  • strong vinegar

Note

The salt is only required

There are a couple of problems:

  • What is strong vinegar?

  • Recipe isn’t very quantitative.

  • I have a sprayer that has a capacity of about 10 litres.

The first problem is figuring out what concentration of vinegar to use. I purchased 473 ml of vinegar @ 45% concentration from Amazon. I want to keep the concentration at a safe level and diluted the vinegar to 10% concentration.

This article will be about deriving the dilution equation and diluting the vinegar to a safer concentration.

Dilution#

If I want a 10% solution, it stands to reason I have to add water to the concentrate. How much water is required and how much volume will that take?

(1)#\[\begin{equation} \frac{V_v}{V_c} = C \end{equation}\]

Where:

  • \(V_v\) - Volume of vinegar.

  • \(V_c\) - Volume of concentrate (i.e. volume of water + volume of vinegar).

  • \(C\) - The ratio of vinegar to concentrate volume. It can be expressed as a percentage.

We know:

  • \(V_c = 473\)

  • \(C = 0.45\)

\[ V_v = C \cdot V_c = 212.85\]

The volume of vinegar in the concentrate is 212.85 ml.

  • How does this help us determine how much water to add to dilute the solution?

  • Can we develop an equation that uses the information we have to determine the amount of water to add so we can dilute to the desired concentration?

Let’s express equation 1 in another form:

(2)#\[\begin{equation} \frac{V_v}{V_w + V_v} = C_f \end{equation}\]

  • \(C_f\) - is the final or target concentration.

We want to solve for V_w from equation 2:

\[ V_v = C_f \cdot (V_w + V_v) \]
\[ V_v = C_f \cdot V_w + C_f \cdot V_v \]
\[ V_v - C_f \cdot V_v = C_f \cdot V_w \]
\[ V_v \cdot (1 - C_f) = C_f \cdot V_w \]
\[ V_w = V_v \cdot \frac{(1 - C_f)}{C_f} \]
(3)#\[\begin{equation} V_w = \frac{(1 - C_f)}{C_f} \cdot V_v \end{equation}\]

Let’s say \(C_f = 0.1\) (10%):

\[ V_w = \frac{(1 - 0.1)}{0.1} \cdot 212.85 = 1915.65\]

\( \therefore \) we would need to add 1915.65 ml of water to dilute the solution to 10% vinegar.

We can verify using equation 1:

\[ \frac{212.85}{1915.65 + 212.85} = 0.1 \]

However, we know:

\[ V_v = C_I \cdot V_c \]

  • \(C_I\) - The initial concentration of vinegar in the solution.

\[ V_w = \frac{(1 - C_f)}{C_f} \cdot C_I V_c \]
\[ V_w = \frac{(C_I V_c - C_f C_I V_c)}{C_f} \]
\[ V_w = \frac{(C_I V_c)}{C_f} - C_I V_c \]
(4)#\[\begin{equation} V_w = V_c \cdot (\frac{C_I}{C_f} - C_I) \end{equation}\]

We can verify that is correct:

\[ V_w = 473 \cdot (\frac{0.45}{0.1} - 0.45) = 1915.65 \]

We end up with the same dilution amount. Using math and going from a practical use case, we can push further and create a general formula for dilution without having to calculate the intermediate steps.

The rest of the problem is ratio and proportions and a smidgen of unit conversion.

Note

The weed killer worked reasonably well. I ended up creating a 10% solution by adding 2.13 litres of water to the concentrated vinegar. I added 0.8 teaspoons of dish soup and 1.6 cups of salt. This was based on the fact that an American gallon is 3.785 litres.

It is also worth noting this isn’t rocket science, so the values can be off somewhat.

Preserves#

Another problem, I have a garden and have some jalapeño plants. I don’t go crazy with them. But I do have enough to pickle a couple of 500 ml jars (refrigerator pickles).

I purchased a pouch of pickle seasoning and the recipe on the back for one pouch was:

  • 3 1/3 cups vinegar @ 5%

  • 7 1/3 cups water

  • 1 pouch contained 184.3g of powder

Essentially, 1 pouch for 10 2/3 cups of fluids. I only need a litre, and I am going to use 7% vinegar solution.

The problem here is not so much the vinegar, I have vinegar at the correct concentration. I could use the methods developed above (and had intended too). What I need to determine is how much powder to use per millilitre.

I need to address approximately 1 litre of solution.

184.3 g/(10 2/3 cups) = 17.279 g/cup of powder.

That works out to be about 0.07g/ml

for 500 ml we would need 36.5g doubling that would be 73 g of powder.

Fairly simple calculations. I include this because I intended to dilute the vinegar concentration down to 5%.

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