Beaker 2 solves for Ka.
...........X^- + HOH ==> HX + OH^-
initial...0.07...........0......0
change.....-y............y.....y
equil.....0.07-y.........y......y
Kb for X^- = (Kw/Ka for HX) = (y)(y)/(0.07-y)
Kw, of course, is 1E-14 and you solve for Ka for HX. y = OH and you can obtain that from the pH of 8.8.
Back to beaker #1.
6.00g/molar mass CH2O = 0.2M.
50 mL x 0.2M = 10 millimoles of HX
20 mL of 0.250 KOH (beaker 3) = 5 mmol.
Reaction of HX(beaker 1) and KOH (beaker 3).
.........HX + KOH ==> KX + H2O
initial..10.....0.......0....0
add............5.0...............
change...-5.0..-5.0....+5.0....+5.0
equil....5.0.....0.......5.0....5.0
Substitute from the above ICE chart into the Henderson-Hasselbalch equation and solve for pH.
pH = pKa + log [(base)/(acid)]
You know pKa from Ka you solved for from beaker 2.
Look up Ka in a table of Ka. I suspect that is formic acid and X^- is formate ion.
Post your work if you get stuck.
A student is given 3 beakers:
Beaker 1- 50.0 ml of a solution produced by dissolving 6.00 grams of a weak
monoprotic acid ,HX, in enough water to produce 1 liter of solution.
The empirical formula of HX is CH2O. The solution contains 3 drops
of phenolphthalein.
Beaker 2- A 0.07M solution of the salt NaX. It has a pH of 8.8
Beaker 3 – 50.0 ml of 0.250M KOH
The contents of beaker 3 is added drop-wise to beaker 1 until a pink color appears and remains for 30 seconds. This takes exactly 20.0 ml of the beaker 3 solution. Identify X and calculate the pH of beaker 1 after the addition of the 20 ml.
3 answers
So, if the X suppose to be HCH2O?
how do find the pH of beaker 1 after the addition of the 20 mL
1 answer