Asked by Megan
Calculate the pH, to one decimal place, of the solution made by mixing 21.0 mL of 0.33 M HNO3 with 121.0 mL of 0.32 M formic acid (HCOOH)
Answers
Answered by
DrBob222
HNO3 is a strong acid.
HCOOH is a weak acid.
Therefore, the final H^+ concn will be that of HNO3 + that from HCOOH.
You know the H^+ from HNO3 since that is a strong acid. That H^+ will act as a common ion to HCOOH.
HCOOH ==> H^+ + HCOO^-
Ka = (H^+)(HCOO^-)/(HCOOH)
Set up an ICE chart for HCOOH and calculate the (HCOO^-) remembering that the concn of the H^+ to plug into that equation is 0.33M x (21 mL/(21mL + 121 mL) = ??
Then total H^+ is the sum of H^+ from HNO3 + (COO^-) from HCOOH.
Post your work if you get stuck.
HCOOH is a weak acid.
Therefore, the final H^+ concn will be that of HNO3 + that from HCOOH.
You know the H^+ from HNO3 since that is a strong acid. That H^+ will act as a common ion to HCOOH.
HCOOH ==> H^+ + HCOO^-
Ka = (H^+)(HCOO^-)/(HCOOH)
Set up an ICE chart for HCOOH and calculate the (HCOO^-) remembering that the concn of the H^+ to plug into that equation is 0.33M x (21 mL/(21mL + 121 mL) = ??
Then total H^+ is the sum of H^+ from HNO3 + (COO^-) from HCOOH.
Post your work if you get stuck.
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