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HEPES, is a commonly used component of physiological buffers. The PKa of the protonated amino group of HEPES is 7.55 at 25C. De...Asked by Angel
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HEPES, is a commonly used component of physiological buffers. The PKa of the protonated amino group of HEPES is 7.55 at 25C.
Describe the preparation of 1.00L of 0.100M HEPES buffer, pH 8.00, from the sodium salt of the basic form of HEPES (MW 260.3), 6.0M HCl and deionized water
HEPES, is a commonly used component of physiological buffers. The PKa of the protonated amino group of HEPES is 7.55 at 25C.
Describe the preparation of 1.00L of 0.100M HEPES buffer, pH 8.00, from the sodium salt of the basic form of HEPES (MW 260.3), 6.0M HCl and deionized water
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Answered by
DrBob222
B = base
A = acid = BH^+
B.Na + H^+ ==> A (BH^+)
pH = pKa + log[(base)/(acid)]
Plug in 8.00 for pH, 7.55 for pKa, and solve for ratio B/A.
You need another equation since there are two unknowns; and that one is A + B = 0.1M
Solve for A and B.
Then I would start with 0.1 mole B.Na which is 0.1 x 260.3 = ?? grams (I think about 26 grams). You know the concn needed for the acid and that MUST equal the H^+ needed. That will give you L of the 6M HCl needed and that should do it. Post your work if you get stuck. I arrived at about 26 grams of the sodium salt and about 4.5 mL of the acid (actually a little less), then make the solution to a final volume of 1.00 L in a volumetric flask.
A = acid = BH^+
B.Na + H^+ ==> A (BH^+)
pH = pKa + log[(base)/(acid)]
Plug in 8.00 for pH, 7.55 for pKa, and solve for ratio B/A.
You need another equation since there are two unknowns; and that one is A + B = 0.1M
Solve for A and B.
Then I would start with 0.1 mole B.Na which is 0.1 x 260.3 = ?? grams (I think about 26 grams). You know the concn needed for the acid and that MUST equal the H^+ needed. That will give you L of the 6M HCl needed and that should do it. Post your work if you get stuck. I arrived at about 26 grams of the sodium salt and about 4.5 mL of the acid (actually a little less), then make the solution to a final volume of 1.00 L in a volumetric flask.
Answered by
Needhelp
I am a little confused on the part after you solve for 26 grams. Can you explain a little more? Thanks
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