Question

The following reaction is conducted, starting with 0.60 M BF3 and 0.60 M NH3.

BF3 (aq) + NH3 (aq) ⇌ BF3NH3 (aq)

Plot the value of Q for this reaction against the concentration of BF3NH3 on the axes provided, including adding value markers to the axes. To do this, calculate the value of Q at the following BF3NH3 concentrations: 0.0 M, 0.1 M, 0.2 M, 0.3 M, 0.4 M, and 0.5 M. If the value of Kc for this reaction is 22.8, use your plot to determine the equilibrium concentration of BF3NH3. This problem is trickier than you think! Be sure you are correctly calculating all relevant concentrations at each point!

Answer 1

Given data:

BF3 (aq) + NH3 (aq) ⇌ BF3NH3 (aq)

BF3 = 0.60M ; NH3=0.60M ; BF3NH3=0.1M, 0.2M, 0.3M, 0.4M, 0.5M

Kc = 22.8

                      BF3 (aq) + NH3 (aq)     ⇌ BF3NH3 (aq)

Initial:             0.60             0.60                    0.1

Change:          -x                -x                        +x

Equilibrium:   0.60-X           0.60-X                  0.1+X

Equilibrium constant K (Concentration) is Kc = [BF3NH3]/[BF3][NH3] ----------- equ(1)

As per above equation (1) substitute the given values.

22.8=[0.1+X]/[0.60-X][0.60-X]

22.8=[0.2+X]/[0.60-X][0.60-X]

22.8=[0.3+X]/[0.60-X][0.60-X]

22.8=[0.4+X]/[0.60-X][0.60-X]

22.8=[0.5+X]/[0.60-X][0.60-X]