Question

HNO2(aq)+NH3(aq)⇄NH4+(aq)+NO2−(aq)
Kc=1×10^6
Nitrous acid reacts with ammonia according to the balanced chemical equation shown above. If 50.mL of 0.20MHNO2(aq) and 50.mL of 0.20MNH3(aq) are mixed and allowed to reach equilibrium at 25°C , what is the approximate [NH3] at equilibrium?

Answer 1

I have looked at this two ways and came up with close values.
1. The solution is 0.1 M in NH4NO2 or 0.2 x (50 mL/100 mL) = 0.1 M in each.
................NH4^+ + NO2^- hydrolysis = NH3 + HNO2
I................0.1..........0.1............................0..........0
C...............-x.............-x.............................x...........x
E...........0.1 - x.........0.1- x.........................x..........x
This is the reverse of the Kc given in the problem so Kc for this reaction is 1E-6
Kc = (x)(x)/(0.1-x)(0.1-x) = 1E-6
Take sqrt both sides
(x)/(0.1-x) = 0.001
x = (0.1-x)*0.001
Solve for x = 9.99E-5
or
2. Calculate (H^+) from (H^+) = sqrt (KwKa/Kb)
Ka for HNO2 = 7.2E-4 from the web.
Kb for NH3 = 1.8E-5 from the web.
I get (H^+) = 6.3E-7 so (OH^-) = Kw/H^+ = 1.58E-8
Then NH3 + HOH --> NH4 + OH^-
Kb = 1.8E-5 = (NH4+)(OH^-)/(NH3)
1.8E-5 = (0.1)(1.58E-8)/(NH3)
(NH3) = 8.7E-5 M which is close to that from above.
You should check the math. Check my reasoning. This is not the usual question from a problem of this kind.

Answer 2

After mixing, the initial [NH3]=0.1M
. The approximate [NH3]
at equilibrium is calculated by using the following mathematical expression, where x=[NH3]
at equilibrium, and is based on the assumption that [NH3]=[HNO2]
and the equilibrium [NH3]
is very small relative to the initial [NH3]
. Kc=1×106=[NH4+][NO2−][HNO2][NH3]=(0.10−x)(0.10−x)(x)(x)≈0.102x2
, which gives x=0.00010M
.