13. Challenge question (optional): An alternative way of making Buffer#2 pH 5.40
ID: 1017006 • Letter: 1
Question
13. Challenge question (optional): An alternative way of making Buffer#2 pH 5.40 would be to add a certain amount of NaOH to the acetic acid solution. Calculate how many mL of 0.50 M NaOH solution would need to be added to 20.0 mL of your concentration acetic acid solution to make a pH 5.40 buffer O1200 M (lr time permits, measure the actic acid solution into a beaker and slowly add the 0.50 M NaOH (If time permits, measure the acetic acid solution into a beaker and slowly add the 0.50 M NaoH with stirring while monitoring the pH. Stop when the pH is 5.40 and record the mL of NaOH used. Did it match the volume you calculated?) 13. (optional) mL of 0.50 M NaOH +20.0 mL of youro200 MCH COOH Calculation: Buffer #2 desired pH-5.40 Actual pH Tot (if time permits)Explanation / Answer
CH3COOH + OH ---> CH3COO- + H20
we know that
for buffers
pH = pKa + log [conjugate base / acid]
in this case
pH = pKa + log [CH3COO-/CH3COOH]
given
pH = 5.4
pKa for acetic acid is 4.76
so
5.4 = 4.76 + log [CH3COO- / CH3COOH]
[CH3COO- / CH3COOH] = 4.36516
now
let y ml of NaOH is added
we know that
moles = cocentration x volume (ml) / 1000
moles of CH3COOH = 0.2 x 20 / 1000 = 0.004
moles of NaOH added = 0.5 x y / 1000 = 0.0005y
now
consider the reaction
CH3COOH + OH- ---> CH3COO- + H20
we can see that
moles of CH3COOH reacted = moles of OH- added = 0.0005y
moles of CH3COO- formed = moles of OH- added = 0.0005y
now
finally
moles of CH3COOH remaining = 0.004 - 0.0005y
now
[CH3COO- / CH3COOH] = 4.36516
now
cocentration = moles / volume (L)
as the final volume is same for both CH3COOH and CH3COO-
so
ratio of concentrations = ratio of moles
so
0.0005y/ (0.004-0.0005y) = 4.36516
0.0005y = 4.36516 (0.004-0.0005y)
y = 6.51
so
6.51 ml of NaOH should be added
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