1.35% formic acid w/w means 1.35 g formic acid/100 g soln. 1.35g/molar mass = moles.
Use density to convert 100 g soln to volume, then mols formic acid/L soln = M formic acid.
Now go through the ICE chart and Ka and solve for pH. Is this enough to get you started?
This is probably easier then my making it but...
Calculate the ph of a forming acid solution that contains 1.35% forming acid by mass. Assume a density of 1.01g/ml for the solution.
3 answers
I think I made a mistake....
1.35g/100g solution=0.0135grams
0.0135g/1.01g=0.01337ml
Then...
1.35g/46.03g CH2O2= 0.0293mols
Then...
0.0293mol/ 1.337X10^-5 = 2.19X10^3 M
pH= -log[H+]
PH=-log(2.19X10^3)
But I get a negative ph.
1.35g/100g solution=0.0135grams
0.0135g/1.01g=0.01337ml
Then...
1.35g/46.03g CH2O2= 0.0293mols
Then...
0.0293mol/ 1.337X10^-5 = 2.19X10^3 M
pH= -log[H+]
PH=-log(2.19X10^3)
But I get a negative ph.
I think more than one. :-).
1.35 g/100g = 0.02933/100 g soln
mass = v x density
volume = m/d = 100/1.01 = 99 mL or 0.099 L
So the M of the soln is 0.02933/0.099 = 0.296 M.
You didn't set up an ICE chart. I am letting HX stand for formic acid.
.........HX ==> H^+ + X^-
initial..0.296...0......0
change...-y.......y.....y
equil...0.296-y...y.....y
Ka = (H^+)(X^-)/(HX)
LOOK UP Ka and substitute from the ICE chart above, solve for (H^+) then convert to pH. I would think the pH should be about 2 or so but I don't remember Ka for formic acid.
1.35 g/100g = 0.02933/100 g soln
mass = v x density
volume = m/d = 100/1.01 = 99 mL or 0.099 L
So the M of the soln is 0.02933/0.099 = 0.296 M.
You didn't set up an ICE chart. I am letting HX stand for formic acid.
.........HX ==> H^+ + X^-
initial..0.296...0......0
change...-y.......y.....y
equil...0.296-y...y.....y
Ka = (H^+)(X^-)/(HX)
LOOK UP Ka and substitute from the ICE chart above, solve for (H^+) then convert to pH. I would think the pH should be about 2 or so but I don't remember Ka for formic acid.