[Plantologist]

I need some assistance with chemistry calculations.  In a research paper I found on Google entitled “Effects of salicylic acid on yield and quality characteristics of tomato fruit (Lycopersicum esculentum Mill.)”, the abstract states in part:

“…Foliar application of five concentrations of salicylic acid (0, 10-2, 10-4, 10-6, 10-8 M) were used. Results showed that application of salicylic acid affected tomato yield and quality characters of tomato fruits so that tomato plants treated with salicylic acid 10-6 M significantly had higher fruit yield (3059.5 g per bush) compared to non-treated plants (2220g per bush) due to an increase in the number of bunch per bush. Results also indicated that application of salicylic acid significantly improved the fruit quality of tomato. Application of salicylic acid increased the amount of vitamin C, lycopene, diameter of fruit skin and also increased rate of pressure tolerance of fruits. Fruit of tomato plants treated with salicylic acid 10-2M significantly had higher vitamin C (32.5 mg per 100 g of fruit fresh weight) compared to non-treated plants (24 mg per 100g fruit fresh weight). Salicylic acid concentration 10-2 M also increased the diameter of fruit skin (0.54 mm) more than two fold compared to control (0.26 mm). Fruit Brix index of tomato plants treated with salicylic acid 10-4 M significantly increased (9.3) compared to non-treated plants (5.9). These results suggest that foliar application of salicylic acid may improve quantity and quality of tomato fruits…”

I want to make 10 liters of a 10-4 M solution of salicylic acid in distilled water (0ppm).  I am using 99.5% pure dried crystalline salicylic acid.  It understand that salicylic acid of this purity does not readily dissolve into water but dissolves into alcohol, therefore, I intend to use the absolute minimum amount of 91% isopropyl alcohol necessary (maybe 5-10ml) to dissolve 100% of the salicylic acid crystals, and then introduce the isopropyl alcohol/salicylic acid solution into the 10 liters of distilled water.  I have used this same isopropyl alcohol method to dissolve gibberellic acid (GA3) for a foliar spray, and in my experience, the isopropyl alcohol was of such a small dilution that it evaporated off during atomization of the spray application and did not affect the plant health. My research has shown that salicylic acid has a molar mass (M) of 138.121 g/mol.

Can somebody please assist me in doing the chemistry math and calculating the dry weight (in grams) of 99.5% dried crystalline salicylic acid necessary to make 10 liters of a 10-4 M solution of salicylic acid in distilled water?  My last chemistry class was in high school many (and I mean many) years ago.

Thank you!

[Borek]

10 mL of ispopropyl alcohol added to 10 L of water yields a 0.012 M solution of the alcohol - 100 times higher concentration than the salicylic acid. While it definitely evaporates during spraying, I am not convinced its final concentration will be lower than that of the acid.

You need 10 L of the 10^-4 M solution. How many moles would that be? If one mole weights 138 grams, what is the weight of that number of moles?

[Plantologist]

Borek,

Please check my math:

10 liters x .0001 x 138 grams/liter = 0.138 grams

Does 0.138 grams look like the correct amount of salicylic acid to add to 10 liters of water to make a 10-4 M solution?

[Borek]

Does 0.138 grams look like the correct amount of salicylic acid to add to 10 liters of water to make a 10-4 M solution?

Yes.

[Plantologist]

Borek,

Thank you brother, I really appreciate your assistance  :-)

Plantologist