methanoic acid = HM
millimols HM = 100 x 0.05 = 5.
...HM + OH^- ==> M^- + H2O
I...5....0.......0.....
add......x...........
C..-x...-x.......x
E..5-x...x.......x
Substitute the E line into the Henderson-Hasselbalch equation and solve for x = millimols NaOH.
Then M NaOH = mmols NaOH/mL NaOH. Substitute and solve for mL of 0.05 M NaOH. I've estimated the answer at about 25 mL of the .05 M NaOH. Post your work if you get stuck.
Calculate the volume of 0.05 M NaOH which must be added to 100 cm3 of 0.05 M methanoic acid to obtain a buffer of pH 4.23.�(Ka for methanoic acid = 1.77 x 10-4)
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