One liter of an aqueous solution contains 0.060 M valine at its isoelectric poin

Question

One liter of an aqueous solution contains 0.060 M valine at its isoelectric point. Given: pK1 = 2.3; pK2 = 9.6. If 0.020 moles of strong acid are added (at constant volume), the new pH will be One liter of an aqueous solution contains 0.060 M valine at its isoelectric point. Given: pK1 = 2.3; pK2 = 9.6. If 0.020 moles of strong acid are added (at constant volume), the new pH will be One liter of an aqueous solution contains 0.060 M valine at its isoelectric point. Given: pK1 = 2.3; pK2 = 9.6. If 0.020 moles of strong acid are added (at constant volume), the new pH will be

Explanation / Answer

The aminoa cid solutions aact as buffer soutions.

At the isoelectric point of valine the whole concentratio is as zwitterionic form.

That is RCH(NH3+)(COO-) form.

When 0.02 moles of strong acid is added , out of 0.060 moles of aminoacid , 0.020moles of it get converted to the form RCH(NH3+)(COOH) form.

Thus initially we have[ A-] = 0.060moles

After addition of acid [A-] = 0.040 moles and [HA] = 0.020 moles

The pH of this new buffer solution is calculated using Henderson equation as

pH = pKa1 + log [conjugate base] / [acid]

= 2.3 + log (0.040/0.020)

= 2.3 + 0.3010

= 2.6

The new pH of solution = 2.6

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