Question

Normality of Sodium Hydroxide: 0.200

Unknown Acid Letter: B

Utilized 50mL of Acid and 50mL of Diethyl Ether.

(a) Volume of NaOH to neutralize unextracted portion                                                33.28 mL

(b) Milliequivalents of base for unextracted portion                                                  ______meq

(c) Milliequivalents of acid initially present in unextracted portion ______meq

(d) Volume of NaOH to neutralize extracted portion 23.02 mL

(e) Milliequivalents of base for extracted portion                                                      ______meq

(f) Milliequivalents of acid remaining in water                                                          ______meq

(g) Milliequivalents of acid in ether                                                                           ______meq

(h) Distribution coefficient, K (show calculations):

K_D = concentration of acid in diethyl ether/concentration of acid in water

                   meq of acid in ether(g) / Volume of diethyl ether laver

       =     ----------------------------------------------------------------------------   

                   meq of acid in water(f) / Volume of water laver

       =      _______

i) % acid extracted = ((g)÷ (c)) ×100% = _______

Using the information given above, please show all answers with relevant formulas and all steps. Thanks!

Answer 1

(a) Volume of NaOH to neutralize unextracted portion = 33.28 mL

(b) milliequivalents of base for unextracted portion = volume in ml x normality

= 33.28 ml x 0.2N = 6.656 meq

(c) let Milliequivalents of acid initially present in unextracted portion = 50 ml X x ( x is concentration of acid in N)

[ ACID CONCENTRATION NEED NOT TO BE CALCULATED ]

total volume of NaOH used for untracted and extracted acids = 33.28 ml + 23.02 mL = 55.3 ml

so, acid equivalents = base equivalents

50 x = 55.3 x 0.2

milliequivalents of acid intially in unextracted portion = 11.06 meq [== 55.3 x 0.2]

(d) Volume of NaOH to neutralize extracted portion = 23.02 mL

(e) Milliequivalents of base for extracted portion = 23.02 x 0.2 = 4.604 meq

(f) Milliequivalents of acid remaining in water = 11.06- 4.604 =   6.456 meq

(g) Milliequivalents of acid in ether =     4.604 meq   

(h) Distribution coefficient, K (show calculations):

KD = concentration of acid in diethyl ether/concentration of acid in water

                   meq of acid in ether(g) / Volume of diethyl ether laver

       =     ----------------------------------------------------------------------------   

                   meq of acid in water(f) / Volume of water laver

4.604 /50

= ----------------------------- = 0.7131

6.456/50

i) % acid extracted = ((g)÷ (c)) ×100% =    4.604 x100 /11.06

= 41.62 %