Q: 1: The distribution constant for iodine between an organic solvent and H;O is 85. Find the concentration of I; remaining in the aqueous layer after extraction of 50.0ml. of 1.00 x 10 M I with the following quantities of the organic solvent: (a) 50.0 ml.; (b) two 25.0 ml. portions; (c) five 10.0 ml. portions.
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- UEid- base tit ration the Consentrat ion of ammonia Solution excess HCL was titrated ith o.07om Na, cO3 Solutian. The volume of Na, Coz Solution reguired was 15.50 ml using the following eguations: 2HCLem> + Na, COgcaps CO2 Cays + HzO1) t NaCL Calculate the concentration (molarity and ppm) of ammonia in the Original Solution.(MM of NHzis 17.031.9/4) 10.0 was by of o.10OMI HCE to Then10. A 0.514 gram sample of NazCO, (106.0 g/mol) was dissolved in distilled water in a 100.0 mi volumetric flask. The molar concentration of Na,CO, in solution is: 0.0485 M 0.0370 M b. 0.4849 M 0.0340 M 0.0330 M d. е.Q: 1: The distribution constant for iodine between an organic solvent and H;O is 85. Find the concentration of I; remaining in the aqueous layer after extraction of 50.0ml. of 1.00 x 103 M I; with the following quantities of the organic solvent: (a) 50.0 ml.; (b) two 25.0 ml. portions; (c) five 10.0 ml. portions. Q: 2: An acidic solute, HA, has an acid dissociation constant of 1.00 x10, and a partition coefficient between water and benzene of 3.00. Calculate the extraction efficiency when 50.00 mL of a 0.025 M aqueous solution of HA buffered to a pH of 3.00, is extracted with 50.00 mL of benzene. Repeat for cases in which the pH of the aqucous solution is buffered to 5.00 and 7.00.
- 3. Complete neutralization of 10 ml of phosphoric acid solution by NaOH 0.I N in the presence of phenol phthalein until the appearance of purple color (pH, = 9) 11 ml. of NaOH is consumed. (a) What is the concentration of phosphoric acid? (b) Calculate the indicator error. pk2.1 pK. - 7.2 pK - 124A solution is prepared in which [Pb2+]= 0.0150 M and [Br 0.00350 M. Given that PbBr2 has a Kgo = 2.1 x 106, which of the following is true? %3D Select one: O a. Qsp Kg. and the solution is supersaturated %3D O d. Q.,< Kgp and the solution is supersaturatedThe distribution constant for solute Z. between Water is 3.7. Calculate the remaining consentration of 2. remaining in a solution when 0.036 M and 31 ml. aparous solution of A is extracted with 4 portions of 8 ml. hexane AY 7.28 x 10M B ) 7.37 x 10M C) 148 x 10M D) 2.61 x 10M E ) 1.87 x 10M
- Preparation of sodium standard solutions with concentrations of 10, 20, 40, 60, 80, 100 (ppm) in a volume of 100 ml for each solution. [using a 100 ppm reference solution prepared from a weight of sodium chloride in a volume of 1 liter]Calc. the alkaline strength of a sample of impure K2CO3 in terms of percent K2O from the ff. data: wt. of sample = 1.000g; HCL used = 55.90 ml;NaOH used = 0.42ml; 1.000mlNaOH 0.008473 g KHC2O4 .H2C2O4 . 2H2O; 1.000ml HCL 2.500 ml NaOH.Ans. 17.15 mg. 54) A 1.000-g sample of a mixture which contains only NaCl and KCl gare a precipitate of AgCl which weighed 2.000 g. What are the percentages of Na and K in the mixture? ins 10 g Ans. 44.77% K; 5.75% Na. 1:J 55. I- can be sernarated from othor
- Q: 1: The distribution constant for iodine between an organic solvent and HO is 85. Find the concentration of I, remaining in the aqueous layer after extraction of 50.0mL of 1.00 X 103 M I2 with the following quantities of the organic solvent: (a) 50.0 mL; (b) two 25.0 mL portions; (c) five 10.0 mL portions.If K = 2.35 & K, 1.32 x 1010, Suppose 50ml of 0.010M aqueous amine is extracted with 100ml of benzene. What % will remain in aqueous sol" at (a) pH - 9 and (b) pH = 11.0 OA a. 51.1% b. 18.9% В. a. 29.2% b. 14.9% a. 48.6% b. 24.2% O D. a. 64.6%ng resourc.. (27) Could've Been - [References] Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. 100.0 mL of 0.30 M HC7H502 (Ka = 6.4 x 10-5) titrated by 0.10 M NaOH pH at the halfway point = pH at the equivalence point = b. 100.0 mL of 0.40 M C2H;NH2 (K½ = 5.6 × 10¬4) titrated by 0.10 M HNO3 pH at the halfway point = pH at the equivalence point = c. 100.0 mL of 0.70 M HC1 titrated by 0.15 M NaOH pH at the halfway point = pH at the equivalence point =