Using Henderson Hasselbach equation: pH = pKa + log( conc of salt/conc of acid)

Question

Using Henderson Hasselbach equation:

pH = pKa + log( conc of salt/conc of acid)

Since the required pH value in this case is close to the pKa2 value, so we will use KH2PO4 as the acid and K2HPO4 as the salt in this buffer.

Putting values:

7.8 = 6.8 + log(conc of salt/conc of acid)

Solving we get:

(conc of salt/conc of acid) = 10

So,

conc of salt = 10*conc of acid

Also,

conc of salt + conc of acid = 0.1 M

Solving these two equations, we get:

conc of salt in buffer = 0.09 M

conc of acid in buffer = 0.009 M

Stock solutions conc for both acid and salt is 0.2 M

So,

Dilution factor for salt = 0.2/0.09 = 2.22

Volume of salt stock needed = 1000/2.22 = 450.45 mL

Dilution factor for acid = 0.2/0.009 = 22.22

Volume of acid stock needed = 1000/22.22 = 45 mL

So,

Mix 450.45 mL of K2HPO4 and 45 mL of KH2PO4 and make the final volume upto 1 L by adding distilled water, to make the reqd buffer.

Hope this helps !

Explanation / Answer

I

I need help with g

Chymotrypsin: Met-Arg-Ala-Tyr Asp-Met-Leu Gly-Asn Phe g) You need to prepare 1 L of a 0.1 M potassium phosphate buffer at pH 7.8 and you are given a supply of 0.2 M K2HPO4 and 0.2 M KH2PO4 solutions and no other reagents. Explain how you would prepare the buffer solution. The pKa values for dissociation of the three acid forms of phosphate are 2 (pKal), 6.8 (pKa2) and 9.8 (pKa3). h) Which of the following tripeptides would be closest to its pI (isoelectric point) in a solution at pH 5: A) HisLysArg; B) HisGluAla; C)TyrAspGluHis; D) GluAspGly. Show how you got your answer. Remember that His has an imidazole group whose pK is around 6.5.

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