Question

Suppose 0.459 g of iron(II) bromide is dissolved in 150. mL of a 18.0 m Maqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) bromide is dissolved in it. Round your answer to 3 significant digits. OM x 6 ?

Answer 1

No. of moles of febra = mass of Febry -3 0.459 g -= 2.13 x10 moe Mw of Feborg 215.65 g mor ImL No. of moles of AgNO3 = 1800 x10 ml x 150 ml x 0.0012 2. to xio 3 more Fe Porz (92) + 2 AgNO3 (26) ► 20g br (5) + Fe (NO3) (22) No. of moles of Ri= 2x (2:13 x10 moe) = 4.26 x 103 moe No. of moles of Agf = 2, to xio 3 moe Az AND Boo Refer in til mole RATO to FORM Aglor PRECIPITATE. So 9.70 x103 mot Ast Resers with 2.70 x10 more (out of 40 26x10" moe) for Ion to form Agfor. Excess Por = 4026 x103 d. to xio? = 150 x10 moe molarity of Bi Ion = 1.56 x10 mol = 10.01049 1 XI 11 150 ml x 0.001L Ime