Question about buffer To \"buffer\" changes in pH effectively, the pK of the buf
ID: 564177 • Letter: Q
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Question about buffer
To "buffer" changes in pH effectively, the pK of the buffer should be near the desired pH. Under these conditions, the concentrations of the protonated and unprotonated forms are close to equal. One of the easiest ways to prepare a buffered solution is to add the right amount of each form to give the concentrations predicted by the Henderson Hassellbach equation. ACES is a buffer with a pka 6.9, and is frequently misrepresented with the structure shown below. OH O-S-o,NH2 Aces is never protonated on the sulfonic acid (as shown), since these have pKa 's of -6. Instead, it forms a "zwitterion" (from the German for hermaphrodite) by protonating the amino group shown with one hydrogen. The molar mass of ACES is 182.2 and the O-S NH NH2 molar mass of ACES.HCI is 218.66 To prepare an ACES buffer with a pH of 6.5, a volume of 2.00 L and a concentration of 0.050 M, how much of the acid form and how much of the base form would you need to add?Explanation / Answer
Let us call ACES as A and ACES.HCl as HA.
Now, according to Handerson Hasselbach equation,
pH = pKa + log [A]/[HA]
pH = 6.5
pKa = 6.9
So, 6.5 = 6.9 + log [A]/[HA]
log [A]/[HA] = -0.4
[A]/[HA] = 0.398
We can say that, [HA] = 1 and [A] = 0.398
Decimal fraction of each component:
HA = 1/(1+0.398) = 0.715
A = 0.398/(1+0.398) = 0.285
Desired concentration = 0.050 M
Molarities of each component:
[HA] = 0.050*0.715 = 0.036 M
[A] = 0.050*0.285 = 0.014
Moles = molarity*volume
Moles of HA needed = 0.036*2.00 = 0.072 mol
Moles of A needed = 0.014*2.00 = 0.028 mol
Mass of HA needed = moles*molar of ACES.HCl = 0.072*218.66 = 15.74 g
Mass of A needed = moles*molar mass of ACES = 0.028*182.2 = 5.10 g
Mass of acid form needed = 15.74 g
Mass of base form needed = 5.10 g
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