What is the minimum volume in mL of 0.3867 M potassium fluoride KF solution required to precipitate as insoluble calcium fluoride CaF2 all the calcium ions from 232 mL of a 0.6199 M solution of CaCl2 solution? The balanced equation should be written to guide this calculation.
2007-05-21 20:25:32 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
CaCl2+2KF--->CaF2+2KCl
you see you need twice as much KF Than CaCl2
232mL =0.232L
as the solution is 0.6199 it contains 0.6199*0.232= 0.1438mole of CaCl2
so you need 0.1438*2 =0.2876 mole of KF
1liter of your solution of KF contains 0.3867 mole
to have 0.2876 mole , you need 0.2876/0.3867=0744L
and in mL 744 mL
2007-05-21 22:59:21 · answer #1 · answered by maussy 7 · 0⤊ 0⤋
CaCl2 + 2KF ----> CaF2 + 2KCl
use this formula: M1V1/M2V2 = n1/n2 where 1= KF and 2= CaCl2
0.3867 x V1/(0.6199 x 232) = 2/1
V1 = 743.82 mL
2007-05-22 07:19:40 · answer #2 · answered by 1-man-show 3 · 0⤊ 0⤋