Asked by nick
How many grams of HCOOH are needed to make 3.7 liters of aqueous solution of pH=3.5?
Answers
Answered by
DrBob222
Here is a very similar problem.
http://www.jiskha.com/display.cgi?id=1286408005
http://www.jiskha.com/display.cgi?id=1286408005
Answered by
nick
I found that problem, however I'm having trouble finding the amount of gram. I found x to 3.16e-4 mol/L
Answered by
DrBob222
You want total moles of M x L.
Then moles = grams/molar mass. Solve for grams. Are you sure about 3.16 x 10^-4? Plug that back into Ka = x^2/(3.16e-4-x) and see if you get pH of 3.5. In fact, a H^+ of 3.16 x 10^-4 gives you a pH of 3.5 (what you want); however, the question doesn't ask for H^+, it asks for grams HCOOH.
Then moles = grams/molar mass. Solve for grams. Are you sure about 3.16 x 10^-4? Plug that back into Ka = x^2/(3.16e-4-x) and see if you get pH of 3.5. In fact, a H^+ of 3.16 x 10^-4 gives you a pH of 3.5 (what you want); however, the question doesn't ask for H^+, it asks for grams HCOOH.
Answered by
nick
x=3.6e-4 mol/L x 3.7L x 43.06g HCOOH/mol = 0.050g HCOOH.
my online homework counted this as incorrect. am I not going about it the right way?
my online homework counted this as incorrect. am I not going about it the right way?
Answered by
DrBob222
I don't believe either number you have quoted is correct. You can plug both 3.16 x 10^-4 or 3.6 x 10^-4 (you have each number in separate responses above) in to
Ka = (x^2)/(Y-x) where the Y is either of those numbers you have quoted above as HCOOH concn. If either number is right, x should give you a pH of 3.5 and I don't think it will.
Ka = (x^2)/(Y-x) where the Y is either of those numbers you have quoted above as HCOOH concn. If either number is right, x should give you a pH of 3.5 and I don't think it will.
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