Finding equilibrium concentration and equilibrium constant. An experiment was ca

Question

Finding equilibrium concentration and equilibrium constant.

An experiment was carried out to determine the value of the equilibrium constant K_c for the reaction. Total moles of Ag* present = 3.6 times 10^-3 moles Total moles of NH_3 present = 6.9 times lO^-3 moles Measured concentration of (Ag(NH_3)_2]* at equilibrium = 3.4 times 10^-2 M Total solution volume = 100 mL b. Calculate the equilibrium concentration of Ag* (uncomplexed or free Ag*) c. Calculate the equilibrium concentration of NH_3 (uncomplexed or free NH_3) d. Calculate the value of the equilibrium constant.

Explanation / Answer

b)

we knoow that

concentration = moles x 1000 / volume (L)

so

[Ag+] = 3.6 x 10-3 x 1000 / 100

[Ag+] = 0.036 M

c)

now

[NH3] = 6.9 x 10-3 x 1000 / 100

[NH3] = 0.069

d)

the equation is

Ag+ + 2NH3 ---> Ag(NH3)2+

the equilibrium constant is given by

Keq = [Ag(NH3)2+] / [Ag+] [NH3]^2

so

Keq = [ 3.4 x 10-2] / [0.036] [0.069]^2

Keq = 198.37

so

the value of equilibrium constant is 198.37

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