Henderson-Hasselbach Equation (I think) 1) How was the equation rearranged to ge

Question

Henderson-Hasselbach Equation (I think)

1) How was the equation rearranged to get So = 1 x 10-6?

2)

What is the answer to that?

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What is the solubility of penicillin G (pKa 2.76) when only the non-dissociated formis present? (i.e. We need to find So, the solubility at low pH) The solubility at pH 8.o is o.174 moldm S-S pKa tlo 8.o 76 logl (o.174 So)/so) Hence by rearrangement So 1 x 10-6 mol dm 3 With so known we can obtain the solubility at any pH

Explanation / Answer

1. 8.0 = 2.76 + log((0.174-S°)/S°)

8.0-2.76 = log((0.174-S°/S°)

5.24 = log((0.174-S°)/S°)

antilog(5.24) = 0.174-S°/S°

1.74*105= 0.174-S°/S°

1.74*105S°= 0.174 - S°

1.74*105S° + S° = 0.174

1.74*105S° = 0.174

S° = 0.174/1.74*105

S° = 0.1*10-5 = 1*10-6mol/dm3

2. pH -pKa = log(s-s°/s°)

Keeping the given values-

6.4-4.1 = log(245-s°/s°)

2.3 = log(245-s°/s°)

antilog(2.3) =245-s°/s°

199.5 = 245-s°/s°

199.5s° = 245-s°

200.5s° = 245

s° = 245/200.5 = 1.22mg/L

4.0-4.1 = log(2.2-s°/s°)   

-0.1 = log(2.2-s°/s°)

antilog(-0.1) = 2.2-s°/s°

0.794 = 2.2 - s° / s°

0.794s° = 2.2 - s°

1.794s° = 2.2

s° = 2.2/1.794 = 1.22mg/L

So,from both the given values,we get S° = 1.22mg/L

Now we have to find S at pH = 7,

7 - 4.1 = log(S-1.22/1.22)

antilog(2.9) = (S-1.22/1.22)

794.3 = S-1.22/1.22

969 = S -1.22

S = 970.3mg/L

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