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?
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
.
Well, I'm not really good at chemistry, but I'll give it a shot! Why did the acid go to the gym? To become a buffer, of course! 😄
Now, let's approach your question. We're given the equilibrium constant, Kc, which is 1×10^6. Kc relates the concentrations of the products and reactants at equilibrium.
In this case, we'll assume that the initial concentrations of NH4+ and NO2- are negligible compared to the initial concentrations of HNO2 and NH3.
Now, to find the equilibrium concentration of NH3, we need to set up an ICE table.
Initial:
[HNO2] = 0.20 M
[NH3] = 0.20 M
[NH4+] = 0 M
[NO2-] = 0 M
Change:
[HNO2] decreases by x
[NH3] decreases by x
[NH4+] increases by x
[NO2-] increases by x
Equilibrium:
[HNO2] = 0.20 - x M
[NH3] = 0.20 - x M
[NH4+] = x M
[NO2-] = x M
Now, using the given equilibrium constant expression, Kc = [NH4+][NO2-]/[HNO2][NH3], we can substitute the equilibrium concentrations into it to get:
Kc = (x*x) / ((0.20 - x)(0.20 - x))
Now, since Kc = 1×10^6, we can make some approximations. Since x is expected to be small compared to 0.20, we can ignore it in the denominator:
Kc = x*x / (0.20*0.20)
Simplifying, we have:
1×10^6 = x^2 / 0.04
Rearranging and solving for x, we find:
x^2 = (1×10^6)(0.04)
x^2 = 40,000
x ≈ 200
Now, since we're looking for the approximate concentration of NH3 at equilibrium, we have:
[NH3] ≈ 0.20 - x ≈ 0.20 - 0.20
[NH3] ≈ 0 M
So, the approximate concentration of NH3 at equilibrium is 0 M.
Always remember, chemistry jokes are always funny because all the good ones are argon! Good luck with your chemistry studies! 😄
To solve this question, we can use the stoichiometry of the balanced chemical equation and the given equilibrium constant (Kc). The balanced chemical equation is:
HNO2(aq) + NH3(aq) ⇄ NH4+(aq) + NO2-(aq)
Given:
Initial volume of HNO2 solution = 50 mL
Initial concentration of HNO2 solution = 0.20 M
Initial volume of NH3 solution = 50 mL
Initial concentration of NH3 solution = 0.20 M
To find the equilibrium concentration of NH3, we need to know the concentration of HNO2 at equilibrium. Let's assume the change in concentration of HNO2 is "x".
Now, let's set up an ice table to keep track of the changes in concentration:
[Species] HNO2(aq) NH3(aq) NH4+(aq) NO2-(aq)
Initial 0.20 M 0.20 M 0 M 0 M
Change -x -x +x +x
Equilibrium 0.20-x 0.20-x x x
Using the equilibrium concentrations, we can calculate the equilibrium constant (Kc):
Kc = [NH4+][NO2-] / [HNO2][NH3]
Substituting the given Kc value:
1 × 10^6 = x*x / (0.20 - x)(0.20 - x)
Since the value of Kc is very large, we can assume that "x" is small compared to 0.20 in the denominator.
Thus, we can approximate (0.20 - x)(0.20 - x) as (0.20)^2.
1 × 10^6 = x*x / (0.20)^2
Simplifying the equation:
x^2 = (1 × 10^6)(0.20)^2
Taking the square root of both sides:
x = sqrt[(1 × 10^6)(0.20)^2]
x ≈ 200
Therefore, the equilibrium concentration of NH3 is approximately (0.20 - x):
[NH3] at equilibrium ≈ 0.20 M - 200 ≈ 0.20 M - 0.20 M ≈ 0 M
So, the approximate concentration of NH3 at equilibrium is 0 M.
To solve this problem, we need to apply the concept of equilibrium and the equilibrium constant (Kc).
First, let's write down the balanced chemical equation and the expression for the equilibrium constant (Kc):
HNO2(aq) + NH3(aq) ⇄ NH4+(aq) + NO2-(aq)
Kc = [NH4+][NO2-] / [HNO2][NH3]
Next, we'll use the given information to determine the equilibrium concentration of NH3:
1. Convert the volumes into liters:
50 mL = 50/1000 L = 0.05 L
2. Calculate the initial number of moles for HNO2 and NH3:
moles HNO2 = Molarity × Volume
moles HNO2 = 0.20 M × 0.05 L = 0.01 moles
moles NH3 = Molarity × Volume
moles NH3 = 0.20 M × 0.05 L = 0.01 moles
3. Since the initial concentration of NH3 is the same as its equilibrium concentration, we can assume it to be "x" (in moles/liter).
4. The change in concentration of HNO2 will be -x (due to the stoichiometry of the reaction), and for NH4+ and NO2-, it will be +x.
5. Now, let's express the equilibrium concentrations:
[NH3] = x
[HNO2] = 0.01 - x
[NH4+] = x
[NO2-] = x
6. Plug these concentrations into the equilibrium expression for Kc and solve for x:
Kc = ([NH4+][NO2-]) / ([HNO2][NH3])
1 × 10^6 = (x * x) / ((0.01 - x) * x)
Rearrange the equation: x^2 / (0.01 - x) = 1 × 10^6
Using this equation, we can solve for x using numerical methods or approximation techniques. However, since the value of Kc is very large (indicating that the reaction significantly favors the product's formation), we can make the assumption that x will be small compared to 0.01. Therefore, we can ignore -x in the denominator and simplify the equation as follows:
x^2 / 0.01 = 1 × 10^6
x^2 = (1 × 10^6) * 0.01
x^2 = 1 × 10^4
x ≈ 100
Therefore, the approximate equilibrium concentration of NH3, [NH3], is approximately 100 moles per liter (M).
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.