Posted by ashley on Sunday, February 12, 2012 at 3:17pm.
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.
So, if the X suppose to be HCH2O?
how do find the pH of beaker 1 after the addition of the 20 mL
Related Questions
chemistry - You have four beakers labeled A, B, C, and D. In beaker A, you place...
chemistry - (3 pts) The stockroom prepared a solution by dissolving 0.2300 g of ...
Chemistry - Two beakers are placed in a sealed container surrounded with air. In...
chemistry - 2. Imagine you have two beakers. Both beakers are filled with the ...
Chemistry - Please help me with the questions at the end of this chemistry lab....
Chemistry - I would really appreciate it if someone would please help me out ...
Chemistry - An empty beaker weighs 42.75 grams. When completely filled with ...
chemistry - If you were to dissolve 14.00 grams of salicylic acid in enough ...
CHEMISTRY HELP! - (1) A beaker with 165 mL of an acetic acid buffer with a pH of...
chemistry 101 - Given the following dissolving process: Fe2(SO4)3 (s) = 2Fe3+ (...
For Further Reading