31. A 0.4584 g sample of a pure soluble bromide compound is dissolved in water,
ID: 541949 • Letter: 3
Question
31. A 0.4584 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 0.8018 g. What is the mass percentage of bromine in the original compound? 32. A student determines the aluminum content of a solution by first precipitating it as aluminum hydroxide, and then decomposing the hydroxide to aluminum oxide by heating. How many grams of aluminum oxide should the student obtain if his solution contains 32.0 mL of 0.557 M aluminum nitrate? 33. The iodide ion concentration in a solution may be determined by the precipitation of silver iodide. Ag(aq) + Iraq) AgI(s) A student finds that 15.69 mL of 0.5430 M silver nitrate is needed to precipitate all fo the iodide ion in a 50.00-ml sample of an unknown. What is the molarity of the iodide ion in the student's unknown. 34. W hat volume of a 0.180 M barium hydroxide solution is required to neutralize 15.8 mL of a 0.330 M hydrobromic acid solution? 35. What volume of a 0.116 M nitric acid solution is required to neutralize 25.7 mL of a 0.146 M barium hydroxide solution? of a 0.287 M perchloric acid solution? solution of hydrobromic acid. If 15.8 mL of base are required to neutralize 25.3 mL of 36. What volume of a 0.124 M calcium hydroxide solution is required to neutralize 29.1 mL 37. An aqueous solution of barium hydroxide is standardized by titration with a 0.180 M the acid, what is the molarity of the barium hydroxide solution? 38. An aqueous solution of nitric acid is standardized by titration with a 0.146 M solution of barium hydroxide. If 25.7 mL of base are required to neutralize 11.1 mL of the acid, what is the molarity of the nitric acid solution?Explanation / Answer
32) Al(NO3)3(aq) + 3NaoH(aq) ---> Al(OH)3(s) + 3 NaNO3(aq)
no of mol of Al(NO3)3 reacted = 32*0.557/1000 = 0.0178 mol
no of mol of Al(OH)3 = 0.0178 mol
2Al(OH)3(s) ----> Al2O3 + 3H2O
no of mol of Al2O3 formed = 0.0178 mol
amount of Al2O3 formed = 0.0178*102 = 1.8156 g
33) ksp of AgI = 8.5*10^-17
[Ag+] = 15.69*0.543/65 = 0.131 M
[I-] = ?
(8.5*10^-17 ) = x*0.131
x = [I-] = 6.5*10^-16 M
34) Ba(OH)2 + 2HBr ---> BaBr2 + 2H2O
M1V1/n1 = M2V2/n2
(0.18*V1/1) = (0.33*15.8/2)
V1 = Vol of Ba(OH)2 required =14.483 ml
35) Ba(OH)2 + 2HNO3 ---> Ba(NO3)2 + 2H2O
M1V1/n1 = M2V2/n2
(25.7*0.146/1) = (0.116*V2/2)
V2 = Volume of HNO3 required = 64.7 ml
36) Ca(OH)2 + HClO4 ---> Ca(ClO4)2 + 2 H2O
(0.124*V1/1) = (0.287*29.1/2)
V1 = VOLUME OF Ca(OH)2 required = 33.67 ml
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