Acidifying Water: Calculating the best bang-for-the-buck

Question

I have water that is at a pH of 8.0. I want to acidify the water to a pH of 4.0.

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I have two products that I can use to acidify the water:

1. Lemon Juice:
   • $12.47 * 13% tax = $14.09 per container
   • The volume of the container is 3.8 L
   • It takes 15 ml to acidify 1 L of water to 4.0 pH
2. Citric Acid:
   • $22.99 * 13% tax = $25.98 per bag
   • The weight of the bag is 2.27 kg. Note: according to my measurements, the dry granules equal a volume of 2.3 L (I did not add water to form a solution for the purpose this measurement).
   • It takes 1.25 ml of dry granules to acidify 1 L of water to 4.0 pH

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Which product is more cost effective at acidifying water?

Answer 1

I am going to make an assumption (always dangerous). You dissolve your citric acid into 2300 mL of water and then you use 1.25 mL of that solution to acidify the water. If that assumption is incorrect, then likely so is this answer. Of all the possible ways to convert kg of citric acid into volume, your way is the worst for trying to do calculations. However, if the assumption is correct, then this still works. For lemon juice, $$ \frac{14.09 dollars}{3800 mL}\times\;15 mL= .0556\;dollars.$$ For citric acid, $$\frac{25.98 dolllars}{2300\;mL}\times 1.25\;mL=0.0141\;dollars$$ Doing it this way, it takes more dollars worth of lemon juice.

In light of the comments, let me offer an amendment. The cost of citric acid is $$\frac{25.98\;dollars}{2270\;gm}=0.0114\;dollars/gm$$ Its bulk density is about $0.865\;gm/cm^3$ (googled it) and you used about $1.25 cm^3$ so $$\frac{0.01144\;dollars}{gm}\cdot \frac{0.865\;gm}{cm^3}\cdot 1.25\;cm^3=0.0124\; dollars$$