Question

(3) An acidic solute, HA, has a Ka of 1.00 x 10-5 and a ko between water and hexane of 3.00. (a) Calculate separately the extraction efficiencies if we extract a 50.00 mL sample of a 0.025 M aqueous solution of HA, buffered to a pH of 5.00 and 7.00, with 50.00 mL of hexane.

Answer 1

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سی ان کے T moles aw Pvesen st extrachen ;) pH ₂5.00 Catz 105.00 2 10x10s. (molestia & - [moles Sacs) Presentanter - not using [this formula) (sa solure) nital Vorg moles Soa) present aller Mechels - Vaa moles HA Corganic) Vorg males HA Car Vaq. using Arst eq. It calculate t 3-00 2 DUOD X. D 105.00 10 5.00 1.00x10 2 1.5 Toral moles [MA = 2 ler x moles Phase benzend then D 2 in aq phase invial 0.025m x 5o-oomL 1.25 mmoles extracted who organic) x modes So.-00ML [1.25-70 mnole 50.00ml los a 1-25 1-875 a 2-50 2) PC20-75 mmoles: extraction efficiency moles present Tabenzene mo moles present ninally inwater X Loot 2 0-75 mnoles To 25 meeles 100 4. - 607.

12 PM 27.00 3-00 107 - L 157 +1.00xio 2 0.0297. moles in volume organic phase of organic phase moles in volume aqueous phase of aq-phase De mmoles So ml o-0297 1-25 - x JomL : 0.037125 x 2 +0.0297x .036 mmoles. 0 -. efficiency 0.036 mmoles x Loot. 1-20 mnoles 2 2.88 7.