Question

What is the value of the equilibrium constant (three sig figs even though the answer is good to only two) for the reaction of hydrofluoric acid and ammonia? K = Fill in the following reaction table with millimoles (to the nearest 0.01 mmol) for the reaction occurring when 32.7 mL of 0.118-M ammonia are mixed with 56.0 mL of 0.264-M hydrofluoric acid.

HF + NH3 F1- + NH41+ initial mmol delta mmol final mmol (fill in ICE table)

What is the final concentration of ammonium ion to three significant figures? [NH41+] = M How many mmols (three sig figs) of the limiting reactant are actually present at equilibrium?

Note that this value is not based on the final amount shown in the reaction table because the reaction table assumes complete reaction (not an equilibrium) of the limiting reactant. mmol

Answer 1

K_a(HF) = 6.9×10^-4 and K_b(NH₃) = 1.8×10^-5

HF + NH₃ NH₄⁺ + F^-

K_eq = K_a×K_b/K_w = 1.8×10^-5×6.9×10^-4/10^-14 = 1.242×10⁶

Number of millimole of ammonia = 32.7ml×0.118M = 3.8586mmol

Number of millimole of HF = 56.0ml×0.264M = 14.784mmol

	NH3	HF		NH4+	F-
Initail	3.8586	14.784		-	-
Change	-3.8586	- 3.8586		+3.8586	3.8586
Equilibrium	~0	10.9254		3.8586	3.8586

K_eq = 1.242×10⁶ = 3.8586 × 3.8586/(10.9254 ×[NH₃])

[NH₃] = 1.097×10^-6 mmol

Millimole.of limiting reagent ( NH₃) = 1.10×10^-6 mol

[NH₄⁺] = 3.8586mmol/(88.7ml) = 0.0435 M