Assume you have successfully made the buffer from the experiment 6 post lab, suc

Question

Assume you have successfully made the buffer from the experiment 6 post lab, such that you have 0.250 M phosphate buffer at pH 7.2. You dissolve in this solution sufficient histidine (side chain pKa = 6.0), such that the final histidine concentration is 5 mM, all at 25°C. Therefore, the pH of the solution is controlled by the phosphate buffer. In this solution, what is the percent of histidine with a protonated side chain? What is the total charge on molecules in this fraction?

Lab 6 buffer : A common biological buffer is a “phosphate buffer,” which involves the equilibrium between H2PO4- and HPO42-, with a pKa at 25 oC of 6.86.

Explanation / Answer

Consider the side chain as an isolated weak acid RH with pKa= 6.0

The added histine doesn’t affect the buffer pH in a significant way (5mM << 250 mM).

At a given pH the R-/RH equilibrium is described by a buffer eq.:

pH = pKa + log ([R-]/[RH])

log ([R-]/[RH]) = pH-pKa = 7.2 – 6.0 = 1.2

[R-]/[RH] = 101.2 = 15.85 = 16

Thus [RH] = 5.0 /17 = 0.3mM and [R-] = 5.0 x 16/17 = 4.7 mM   

The percent of the side chain protonation is

    100 x 0.3mM / 5.0 mM = 6.0 %

At pH = 7.2

-- the acidic group – COOH is in the -COO- form ( -1 per molecule)

-the side chain (imidazole) is 94.0 % in R- form       ( -0.94 per molecule)

-the amino group is in - NH3+ form                             ( + 1 per molecule)

The total charge per molecule is - 0.94

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