It doesn\'t take most student researchers very long to figure out that the actua

Question

It doesn't take most student researchers very long to figure out that the actual pH of the buffer they just carefully designed and prepared deviates somewhat from the target pH they were shooting for. It is very possible that you already had this experience in the Indicator pKa' lab exercise. Typically people just chalk up any error to impure reagents, sloppiness in technique, etc. In actual fact, knowing the real reason why this happens can be a big help in fixing things so it doesn't Problem Design a 0.01M buffer from chloroacetic acid and sodium chloroacetate having pH 2.00 Not a big deal. We've done this before. The equilibrium at hand is written: HOCIAcent +0CIAc" pK, = 2.865 21. Use the Henderson-Hasselbach equation (along with the specified formal buffer concentration) to calculate the concentrations of chloroacetic acid and sodium chloroacetate in the buffer. 22. Now use the systematic approach to the treatment of equilibrium to make a more careful estimate of what the pH of the solution will actually be, if the solution contains the HOCIA and OCIAc concentrations you calculated above. With judicious approximations, you will end up solving a quadratic equation.

Explanation / Answer

Q21 and 22:

the equation:

pH = pKa + log(OClAc-/HOClAc)

substitute data:

2 = 2.8655 + log([OClAc-]/[HOClAc])

10^(2-2.8655) = [OClAc-]/[HOClAc]

[OClAc-]/[HOClAc] = 0.1363

if we need

0.01 M in total , the

[OClAc-] + [HOClAc] = 0.01

[HOClAc] = 0.01 - [OClAc-]

[OClAc-]/[HOClAc] = 0.1363

[OClAc-] = (0.01 - [OClAc-] )*0.1363

[OClAc-] = (0.001363 -0.1363 [OClAc-]

[OClAc-] = (0.001363 ) / ( 1+0.1363 ) = 0.0012

[OClAc-] + [HOClAc] = 0.01

[HOClAc]  = 0.01-0.0012 = 0.0088 M

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