A. If 100.0 mL of 0.100 M solutions of BES HEPES or TRIS at pH 7.4 were each tre
ID: 193676 • Letter: A
Question
A. If 100.0 mL of 0.100 M solutions of BES HEPES or TRIS at pH 7.4 were each treated with 10.0 mL of 100.0 mM HCIO4, which would keep the pH closest to 7.4? Select one: a. BES b. HEPES c. TRIS B. Write the equation that relates pH and pKa to the log of the ratio of the concentrations of the acid and base components. If we add only the acid component, the concentration of added base is 0 M. If we use 0 M as a term in the equation it produces a mathematical conundrum. Show why this apparent mathematical impossibility does not, in reality, occur. C. 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
QUESTION A.
Answer b. HEPES buffer.
This is because the pka of HEPES buffer is 7.55. A buffer is most effective around its pka value. It is becasue outside this value, the concentration of the conjugate acid of the base is too small to resist effect of added H+ or OH- ions.
QUESTION B.
Answer. The pH and pka of acid and base components are related to each other by log of ratios of their concentrations. This relationship is called Handerson-Hassalbalch equation and can be written as,
pH = pka + log [Conjugate base] / [Acid]
or
pH = pka + log [A-] / [HA]
When we add acid component only, the equilibrium concentration of H+ and OH- changes in the added water. So we have an acidic solution not a buffer. A buffer is a combination of two components, an acid and its cojugate base. Hence the equation does not apply in this case.
Answer.C.
The relative concentration of acid and base can be determined from Handerson-Hassalbalch equation as follows.
The pH required is 6.5
pKa of ACES is 6.85
As per Handerson-Hassalbalch equation,
pH = pka + log [conjugate base] / [acid]
6.5 = 6.85 + log [B-]/ [BH]
-0.35 = log [B-]/ [BH]
Taking antilog
[B-] / [BH] = 0.7047
[B-] = 0.7047 [BH]
Since the molarity of buffer is 0.05M, so concentration of acid and base components when combined will be 0.05M.
[B-] + [BH] = 0.05M
Solving for [B-]gives, 0.014M and for [BH] yields 0.029M.
The amount in grams can be calculated after we know the molecular weight of acid and its conjugate base. Since we have to calculate for 2 L solution, the factor 2 have to be kept in mind.
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