close
標題:
發問:
A 1.127 g sample of commercial potassium hydroxide [KOH] contaminated by potassium carbonate [K2CO3] was dissolved in water, and the resulting solution was diluted to 500. 0 mL in a graduated flask. A 50. 0mL aliquot of this solution was treated with 40. 0 mL of 0.05304 M hydrochloric acid [HCl] and... 顯示更多 A 1.127 g sample of commercial potassium hydroxide [KOH] contaminated by potassium carbonate [K2CO3] was dissolved in water, and the resulting solution was diluted to 500. 0 mL in a graduated flask. A 50. 0mL aliquot of this solution was treated with 40. 0 mL of 0.05304 M hydrochloric acid [HCl] and boiled to remove CO2. The excess acid consumed 4.74 mL of 0.04983 M sodium hydroxide solution [NaOH] (phenolphthalein indicator). An excess of neutral barium chloride solution [BaCl2] was added to another 50. 0mL aliquot to precipitate the carbonate as BaCO3 and filtered, the solution was then titrated with 28.56 mL of the acid to phenolphthalein end point. Calculate the percentage of potassium hydroxide, potassium carbonate, and water in the sample, assuming that these are the only compound present. (Given: Na = 23; Cl = 35.5; K = 39; H = 1; C = 12; O = 16) Please show the steps clearly. ThX~~~~~~~~~ 更新: I am sorry I forgot to show the final answer are KOH = 75.29% K2CO3 = 22.68% H2O = 2.04%~~~~~~
最佳解答:
Mass of the sample in 500 mL of the solution = 1.127 g Mass of the sample in 50 mL of the solution = 1.127 x (50/500) = 0.1127 g ========== Consider the second titration: The excess BaCl2 solution followed by filtration removes all the CO32- ions as BaCO3 solid. Therefore, the titration only involves the reaction between HCl and KOH. HCl + KOH → KCl + H2O Mole ratio HCl : KOH = 1 : 1 No. of moles of HCl used = MV = 0.05304 x (28.56/1000) = 0.001515 mol No. of moles of KOH used = 0.001515 mol Molar mass of KOH = 39 + 16 + 1 = 56 g mol-1 Mass of KOH in 50 mL of solution = mol x (molar mass) = 0.001515 x 56 = 0.08484 g ========== Consider the first titration (back titration): Part of the HCl in 40 mL of the solution is used to react with KOH and K2-CO3 in 50 mL of the solution, and then the excess is titrated with NaOH solution. HCl + KOH → KOH + H2O ...... (*) 2HCl + K2CO3 → 2KCl + CO2 + H2O ...... (**) Mole ratio HCl : K2CO3 = 2 : 1 HCl + NaOH → NaCl + H2O ...... (***) Mole ratio HCl : NaOH = 1 : 1 No. of moles of NaOH used in (***) = MV = 0.04983 x (4.74/1000) = 0.0002362 mol No. of moles of HCl used in (***) = 0.000236 mol As calculated above, no. of moles of HCl used in (*) = 0.001515 mol Total no. of moles of HCl used = MV = 0.05304 x (40/1000) = 0.002122 mol No. of moles of HCl used in (**) = 0.002122 - (0.001515 + 0.0002362) = 0.0003708 mol No. of moles of K2CO3 used in (**) = 0.0003708 x (1/2) = 0.0001854 mol Molar mass of K2CO3 = 39x2 + 12 + 16x3 = 138 g mol-1 Mass of K2CO3 in 50 mL of solution = mass x (molar mass) = 0.0001854 x 138 = 0.02559 g ========== In 50 mL of the solution: Total mass of the sample = 0.1127 g Mass of KOH = 0.08484 g Mass of K2CO3 = 0.02559 g % by mass of KOH = (0.08484/0.1127) x 100% = 75.28% % by mass of K2CO3 = (0.02559/0.1127) x 100% = 22.71% % by mass of H2O = 100% - (75.28% + 22.71%) = 2.01%
其他解答:
此文章來自奇摩知識+如有不便請留言告知
Analysis of Titration (20 points!!!)發問:
A 1.127 g sample of commercial potassium hydroxide [KOH] contaminated by potassium carbonate [K2CO3] was dissolved in water, and the resulting solution was diluted to 500. 0 mL in a graduated flask. A 50. 0mL aliquot of this solution was treated with 40. 0 mL of 0.05304 M hydrochloric acid [HCl] and... 顯示更多 A 1.127 g sample of commercial potassium hydroxide [KOH] contaminated by potassium carbonate [K2CO3] was dissolved in water, and the resulting solution was diluted to 500. 0 mL in a graduated flask. A 50. 0mL aliquot of this solution was treated with 40. 0 mL of 0.05304 M hydrochloric acid [HCl] and boiled to remove CO2. The excess acid consumed 4.74 mL of 0.04983 M sodium hydroxide solution [NaOH] (phenolphthalein indicator). An excess of neutral barium chloride solution [BaCl2] was added to another 50. 0mL aliquot to precipitate the carbonate as BaCO3 and filtered, the solution was then titrated with 28.56 mL of the acid to phenolphthalein end point. Calculate the percentage of potassium hydroxide, potassium carbonate, and water in the sample, assuming that these are the only compound present. (Given: Na = 23; Cl = 35.5; K = 39; H = 1; C = 12; O = 16) Please show the steps clearly. ThX~~~~~~~~~ 更新: I am sorry I forgot to show the final answer are KOH = 75.29% K2CO3 = 22.68% H2O = 2.04%~~~~~~
最佳解答:
Mass of the sample in 500 mL of the solution = 1.127 g Mass of the sample in 50 mL of the solution = 1.127 x (50/500) = 0.1127 g ========== Consider the second titration: The excess BaCl2 solution followed by filtration removes all the CO32- ions as BaCO3 solid. Therefore, the titration only involves the reaction between HCl and KOH. HCl + KOH → KCl + H2O Mole ratio HCl : KOH = 1 : 1 No. of moles of HCl used = MV = 0.05304 x (28.56/1000) = 0.001515 mol No. of moles of KOH used = 0.001515 mol Molar mass of KOH = 39 + 16 + 1 = 56 g mol-1 Mass of KOH in 50 mL of solution = mol x (molar mass) = 0.001515 x 56 = 0.08484 g ========== Consider the first titration (back titration): Part of the HCl in 40 mL of the solution is used to react with KOH and K2-CO3 in 50 mL of the solution, and then the excess is titrated with NaOH solution. HCl + KOH → KOH + H2O ...... (*) 2HCl + K2CO3 → 2KCl + CO2 + H2O ...... (**) Mole ratio HCl : K2CO3 = 2 : 1 HCl + NaOH → NaCl + H2O ...... (***) Mole ratio HCl : NaOH = 1 : 1 No. of moles of NaOH used in (***) = MV = 0.04983 x (4.74/1000) = 0.0002362 mol No. of moles of HCl used in (***) = 0.000236 mol As calculated above, no. of moles of HCl used in (*) = 0.001515 mol Total no. of moles of HCl used = MV = 0.05304 x (40/1000) = 0.002122 mol No. of moles of HCl used in (**) = 0.002122 - (0.001515 + 0.0002362) = 0.0003708 mol No. of moles of K2CO3 used in (**) = 0.0003708 x (1/2) = 0.0001854 mol Molar mass of K2CO3 = 39x2 + 12 + 16x3 = 138 g mol-1 Mass of K2CO3 in 50 mL of solution = mass x (molar mass) = 0.0001854 x 138 = 0.02559 g ========== In 50 mL of the solution: Total mass of the sample = 0.1127 g Mass of KOH = 0.08484 g Mass of K2CO3 = 0.02559 g % by mass of KOH = (0.08484/0.1127) x 100% = 75.28% % by mass of K2CO3 = (0.02559/0.1127) x 100% = 22.71% % by mass of H2O = 100% - (75.28% + 22.71%) = 2.01%
其他解答:
文章標籤
全站熱搜
留言列表
