Question

In: Chemistry

A buffer was prepared by adding 0.1 mmoles of solid isoelectric histidine to 0.1 mmoles of...

A buffer was prepared by adding 0.1 mmoles of solid isoelectric histidine to 0.1
mmoles of solid histidine monohydrochloride and adjusting the final volume to
0.5 liters. The pKas for histidine are 1.8, 6.0, and 9.2. What is the pH of this
buffer?

Solutions

Expert Solution

The histidine is a molecule with two amino groups (mostly basic) and one carboxilic group (acid)

We know that the pKas for histidine are 1.8, 6.0 and 9.2; so, to realize at what pH this buffer works, it would be interesting to think a little about the carge pf the molecule at different pHs

pH < 1.8: NH+ , NH+, COOH (all groups are protonated, charge = +2)

1.8 < pH < 6.0: NH+ , NH+, COO- (amino groups are protonated but carboxile is desprotonated; charge= +1)

6.0 < pH < 9.2: NH+ , N , COO- ( one amino groups is protonated but carboxile and one amino group are desprotonated; charge= +0)

pH> 9.2 N, N, COO- (all groups are desprotonated; charge: -1)

The question says that we are mixing the histidine charged zero (solid isoelectric histidine) with histidine charged +1 (histidine monohydrochloride) , so from here we can realize we are trying to regulate pH around 6, considering that the acid is the histidine monohydrochloride and the conjugated "base" is the isoelectric histidine:

histidine monohydrochloride <---------> isoelectric histidine + H+   pKa = 6.0

(HA) (A-)

Now, using the Henderson- Hasselback equation we can calculate the pH of the buffer:

pH = 6.0 + Log (0.0001moles*0.5L/0.0001moles *0.5L)

pH = 6.0 As the concentration of both acid and base are the same, pH of the buffer is equal to pKa.

pH = 6.0


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