Posted by Eddy on Wednesday, March 2, 2011 at 2:23pm.
40 mL x 0.25M HCOONa = 10 mmoles.
140 mL x 0.2M HCl = 28 mmoles.
..............HCOONa + HCl ==>HCOOH + NaCl
begin.........10.......0........0
change....... -10.....+28......+10...+10
final..........0......+18.......+10..+10
So your question is asking for the pH of 18 mmoles HCl + 10 mmoles HCOOH in (140mL + 40 mL = 180 mL). A good estimate is to ignore the HCOOH (much weaker than HCL) and you have 18 mmoles/180 mL = 0.1 M = pH of 1.0.
If you want to do it up brown, you can calculate the amount of H^+ contributed by HCOOH (M = mmoles/mL = 10/180 =0.0556M) as follows (remember that the HCl provides a common ion (H^+) and that will shift the HCOOH ionization to the left and DECREASE the production of H^+ from the HCOOH source.):
.............HCOOH ==> H^+ + HCOO^-.
begin.......0.0556.....0.1....0
change........-x.......+x.....+x
final......0.00556-x....x......x
(H^+)(HCOO^-)/(HCOOH) = Ka
Look up Ka in your text or notes and solve for x. I have estimated it at about 1E-6 and that added to 0.1M is inconsequential.
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