The solubility constant of agbr is 5.0 x 10-13. the formation constant of the silver(i) ammonia complex, ag(nh3)2, is 1.6 x 107 calculate the solubility, in moles/l, of agbr in 1.00 mammonia solution.

S AgBr = 2.82 E-3 mol/L

Explanation:

∴ Ksp = [ Ag+ ] * [ Br- ] = 5.0 E-13

∴ Kf = 1.6 E7 = α[Ag(NH3)2]+ / ( αAg+ )*( αNH3 )²

∴ C NH3(sln) = 1.00 M

from (1) + (2):

                             1 M                     0                       0

                            1 - 2x                   x                        x

∴ K = Ksp*Kf = ( 5.0 E-13 )*( 1.6 E7 ) = 8.0 E-6

⇒ K = ( [ Br- ] * [ Ag(NH3)2+] ) / [ NH3 ]² = 8.0 E-6

⇒ K = (( x )*( x )) / ( 1 - 2x ) = 8.0 E-6

⇒ x² = 8.0 E-6*( 1- 2x )

⇒ x² + 1.6 E-5x - 8.0 E-6 = 0

⇒ x = 2.82 E-3 M

⇒ [ Br- ] = [ AgBr ] = 2.82 E-3 M

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.