An aqueous mixture of hydrocyanic acid and ammonia has initial concentrations of 0.100 M HCN(aq) and 0.140 M NH3(aq). At equilibrium, the CN–(aq) concentration is 0.055 M. Calculate Kc for the reaction

Kc = 0.791

Explanation:

Step 1: Data given

Hydrocyanic acid = HCN

ammonia = NH3

initial concentrations of HCN = 0.100 M

initial concentrations of NH3 =  0.140 M

At equilibrium, the CN–(aq) concentration is 0.055 M.

Step 2: The balanced equation

HCN(aq) + NH3(aq) ↔ CN-(aq) + NH4+(aq)

Step 3: The concentrations at the equilibrium

For 1 mol HCN we need 1 mol NH3 to produce 1 mol CN- and 1 mol NH4+

[CN-] = [NH4+] = 0.055 M

[HCN] = 0.100 - 0.055 = 0.045 M

[NH3] = 0.140 - 0.055 = 0.085 M

Step 4: Calculate Kc

Kc = [CN-][NH4+] / [HCN][NH3] = (0.055 * 0.055) / (0.045 * 0.085) =0.791

Kc = 0.791

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons