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

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

Which product is more cost effective at acidifying water?

User1974
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  • Wouldn't water have a pH of 7.0? – Hagen von Eitzen May 10 '18 at 17:45
  • No need for precise calculations: It takes more then ten times as much lemon juice than citric acid (in volume). The package sizes differ by less than a factor of two, the prices differ by less than a factor of two. Hence citric acid is really better (still by a factor greater than two) – Hagen von Eitzen May 10 '18 at 17:49
  • @HagenvonEitzen : My municipal tap water has a pH of 8.0. I'm not sure why, but it might be something to do with what the municipality treats it with. At any rate, pH is tricky; merely changing the temperature of water seems to change the pH reading. – User1974 May 10 '18 at 17:51
  • It would be good to see how you got 2.3L from 2.27 kg. Your calc. would make the citric concentration greater than 5M. – Narlin May 10 '18 at 17:58
  • @Narlin I poured the citric acid granules into an [empty] pitcher, marked how high it went, removed the citric acid, then poured water from a 1 L beaker into it. It took 2.3 litres of water from the beaker to fill the pitcher up to the marking. – User1974 May 10 '18 at 18:40

1 Answers1

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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 $$\frac{0.01144\;dollars}{gm}\cdot \frac{0.865\;gm}{cm^3}\cdot 1.25\;cm^3=0.0124\; dollars$$

Narlin
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  • Narlin, apologies, but your assumption was incorrect. The 2.27 kg of dry citric acid granules takes up 2.3 L of space in a pitcher. I did not "dissolve the citric acid into 2300 mL of water". – User1974 May 10 '18 at 18:45
  • What solution of citric acid composed the 1.25 mL? To get a correct answer, that needs to be known. We know how much Citric costs in the dry state, but it was added as a liquid solution. – Narlin May 10 '18 at 18:47
  • I see now how this was misleading (you'll have to forgive me, but I don't have a background in science, math, engineering, etc.). I measured out 1.25 ml (1/4 teaspoon) of dry citric acid granules using kitchen measuring spoons. I added it to 1 L of water and stirred the granules to dissolve them. Then I took a reading of the water using a pH gauge and determined that it lowered the pH from 8.0 to the desired 4.0 pH. I did not add the citric acid to the water as a liquid solution. Perhaps it was misleading (or even a bad practice) to measure dry granules in millimeter units? – User1974 May 10 '18 at 18:56
  • I updated the question to try and clear this up. – User1974 May 10 '18 at 19:01