What is the concentration of ammonium ion in a solution of #NH_3(aq)# that is nominally #0.157*mol*L^-1#?

Question

Peyton Adams · Accepted Answer

#[NH_4^+]=1.67xx10^-3*mol*L^-1#

We scrutinize the balance:

#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + ""^(-)OH#,

for which #K_b=1.8xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])#

Initially, #[NH_3]=0.157*mol*L^-1#, and we assume some ammonia accepts a proton from water to give stoichiometric #NH_4^+# and #HO^-#. If we let the amount of protonation be #x#, then we write:

#K_b=1.8xx10^-5=(x^2)/(0.157-x)#

This is a quadratic in #x#, which we could solve exactly, but we make that approximation that #0.157-x~=0.157#, and thus........

#x_1~=sqrt(1.8xx10^-5xx0.157)=1.68xx10^-3#

This is indeed small compared to #0.157#, but we could plug it in again, now that we have an approx. for #x#.

#x_2=1.67xx10^-3#, and since the values have converged, I am willing to accept this value for our answer.

Now #x=[NH_4^+]=[HO^-]# by our definition. What is the #pH# of this solution?

I hope this clarifies your model response. If you need something rephrased or clarified, just ask, and someone will assist you.

Natalie Anton · Answer

undefined To find the concentration of ammonium ion ($\text{NH}_4^+$) in a solution of ammonia ($\text{NH}_3(aq)$), we need to consider the equilibrium that is established when ammonia dissolves in water:

$$\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)$$

This equilibrium can be described by the equilibrium constant expression for the reaction, known as the $K_b$ for ammonia ($\text{NH}_3$):

$$K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$$

The $K_b$ for ammonia at 25°C is $1.76 \times 10^{-5}$.

Given that the initial concentration of $\text{NH}_3$ is 0.157 M, we can use the $K_b$ value to solve for the concentration of $\text{NH}_4^+$ at equilibrium. Assuming $x$ is the change in concentration of $\text{NH}_3$ that reacts to form $\text{NH}_4^+$ and $\text{OH}^-$, the equilibrium concentrations can be represented as:

$$[\text{NH}_3] = 0.157 - x$$
$$[\text{NH}_4^+] = x$$
$$[\text{OH}^-] = x$$

Substituting these into the $K_b$ expression gives:

$$1.76 \times 10^{-5} = \frac{x \cdot x}{0.157 - x}$$

Because $K_b$ is small and the initial concentration of $\text{NH}_3$ is relatively large, we can make the approximation that $x$ is much smaller than 0.157, allowing us to simplify the denominator to just 0.157. This gives:

$$1.76 \times 10^{-5} = \frac{x^2}{0.157}$$

Solving for $x$ will give us the concentration of $\text{NH}_4^+$. Let's calculate $x$.The concentration of ammonium ion ($\text{NH}_4^+$) in a solution of ammonia ($\text{NH}_3$) that is nominally $0.157 \, \text{M}$ is approximately $0.00166 \, \text{M}$.