The pH of uncontaminated rain water is about 5.6. The pHof rain falling is Cleve

Question

The pH of uncontaminated rain water is about 5.6. The pHof rain falling is Cleveland, OH is 4.2 an that in Omaha, NE is5.5. a. Using a calculator, determine the number of liters ofrain water which would provide 1 mole of hydrogen ions for eachcity. b. What is the concentration of hydrogen ions in therainfall of each city? c. What is the concentration of hydroxide ions in therainfall of each city? I could not find ANYWHERE how to work these out...I'm takingEnvironmental Science, not chemistry! If someone could showme how to solve it without the answer that would be HUGE!!! Iam willing to learn not cheat! The pH of uncontaminated rain water is about 5.6. The pHof rain falling is Cleveland, OH is 4.2 an that in Omaha, NE is5.5. a. Using a calculator, determine the number of liters ofrain water which would provide 1 mole of hydrogen ions for eachcity. b. What is the concentration of hydrogen ions in therainfall of each city? c. What is the concentration of hydroxide ions in therainfall of each city? I could not find ANYWHERE how to work these out...I'm takingEnvironmental Science, not chemistry! If someone could showme how to solve it without the answer that would be HUGE!!! Iam willing to learn not cheat!

Explanation / Answer

hello! ok, so in (a) you found out how many moles of hydrogen ions(H+) in one liter. in chemistry, solution concentrations arewritten as M (which means moles/L). now that you know how manymoles of H+ you have, just divide that by one liter and then youhave your concentration of hydrogen ions in M. pOH is the measure of hydroxide ions in solution. if you takechemistry later, you'll learn that pOH = 14-pH. the problem giesyou the pH, so you can just solve for pOH. for example, since thepH in cleveland is 4.2, the pOH in cleveland is 14-4.2, which is9.8. pOH is the negative log of the concentration of hydroxideions, and to solve for concentration is exactly how we solved forthe concentration of H+ pOH = 9.8 -log[OH-] = 9.8 and then solve for [OH-] (which is another way of writing"concentration of OH- ions") let me know if you need anymore help!

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