Question

For the aqueous [A1F,] complex K,=4.0 x 10" at 25 °C. Suppose equal volumes of 0.0090 M Al(NO3), solution and 0.62 M KF solution are mixed. Calculate the equilibrium molarity of aqueous A1°* ion. Round your answer to 2 significant digits. Щм xs ?

Answer 1

equilibrium molarity Al³⁺ = 2.2 x 10^-19 M

Explanation

let us take 1 L of each solution

concentration Al(NO₃)₃ = 0.0090 M

moles Al³⁺ = (concentration Al(NO₃)₃) * (volume of solution in Liter)

moles Al³⁺ = (0.0090 M) * (1.0 L)

moles Al³⁺ = 0.0090 mol

Similarly, moles F^- = 0.62 mol

Al³⁺ (aq) + 6 F^- (aq) $\rightleftharpoons$ AlF₆^3- (aq) : K_f = 4.0 x 10¹⁹

moles AlF₆^3- formed = moles Al³⁺ = 0.0090 mol

concentration AlF₆^3- = (moles AlF₆^3-) / (total volume)

concentration AlF₆^3- = (0.0090 mol) / (1 L + 1 L)

concentration AlF₆^3- = 0.0045 M

moles F^- remaining = (initial moles F^-) - (moles F^- consumed)

moles F^- remaining = (0.62 mol) - (6 * 0.0090 mol)

moles F^- remaining = 0.566 mol

concentration F^- = (moles F^- remaining) / (total volume)

concentration F^- = (0.566 mol) / (1 L + 1 L)

concentration F^- = 0.283 M

K = 1/K_f

K = 1/(4.0 x 10¹⁹)

K = 2.5 x 10^-20

ICE table	AlF63- (aq)	$\rightleftharpoons$	Al3+ (aq)	6 F- (aq)
Initial conc.	0.0045 M		0	0.283 M
Change	-x		+x	+6x
Equilibrium conc.	0.0045 M - x		+x	0.283 M + 6x

K = [Al³⁺]_eq[F^-]_eq⁶ / [AlF₆^3-]_eq

2.5 x 10^-20 = [(x) * (0.283 M + 6x)⁶] / (0.0045 M - x)

Solving for x, x = 2.2 x 10^-19 M

[Al³⁺] = x = 2.2 x 10^-19 M