Asked by jj
A titration of 15.0 cm3 of household ammonia, NH3, required 38.57 cm3 of 0.78M HCl. Calculate the molarity of the ammonia.
PLEASE SHOW ME HOW TO SOLVE THIS. Thank you very much!
PLEASE SHOW ME HOW TO SOLVE THIS. Thank you very much!
Answers
Answered by
drwls
Doesn't one mole of HCl neutralize one mole of NH3?
38.57 cm^3 = 0.03857 liters
There are (0.78)*(0.03857) = 0.0301 moles of HCl in the 0.78M titrating solution.
Assuming there are the same number of moles of NH3 in the 15.0 cm^3 of ammonia solution, calculate its molarity
38.57 cm^3 = 0.03857 liters
There are (0.78)*(0.03857) = 0.0301 moles of HCl in the 0.78M titrating solution.
Assuming there are the same number of moles of NH3 in the 15.0 cm^3 of ammonia solution, calculate its molarity
Answered by
Anonymous
Molarity = mols/liters.... so do .0301/.015.... and you get 2.01 M.
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