A 0.10 M NH3 solution has a degree of dissociation 1.3% at temperature T. Calculate Kb for NH3 at this temperature?

Question

Naomi Aronson · Accepted Answer

#K_b = 1.7 * 10^(-5)#

To tackle this problem, two things are necessary for you to understand.

As you know, ammonia, #"NH"_3#, is a weak base, which means that it does not dissociate completely in aqueous solution to form ammonium ions, #"NH"_4^(+)#, its conjugate acid, and hydroxide anions, #"OH"^(-)#.

Instead, only a fraction of the number of molecules of ammonia will actually dissociate. That fraction depends on ammonia's base dissociation constant , #K_b#.

You are told that your ammonia solution has a degree of dissociation of #1.3%# at a given temperature #T#. What that means is that for every #100# molecules of ammonia present in solution, only #1.3# will dissociate.

Alternatively, you can think about it like this - for every #1000# molecules of ammonia, only #13# will ionize to form the aforementioned ions. The rest will remain as molecules of ammonia.

This is how the balanced chemical equation for this reaction appears.

#"NH"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "NH"_text(4(aq])^(+) + "OH"_text((aq])^(-)#

The base dissociation constant by definition will be equal to

#K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3])#

You are aware of that

#["NH"_3] = "0.10 M"#

Now, use the degree of dissociation to find the concentrations of the two ions. You have #1:1# mole ratios between all chemical species, so you can say that

#["NH"_4^(+)] = ["OH"^(-)] = 1.3/100 * ["NH"_3]#

#["NH"_4^(+)] = ["OH"^(-)] = 1.3 * 10^(-2) * "0.10 M" = 1.3 * 10^(-3)"M"#

This implies that you receive

#K_b = (1.3 * 10^(-3) * 1.3 * 10^(-3))/0.10 = color(green)(1.7 * 10^(-5))#

The number of sig figs you have for the initial ammonia concentration is the answer, which is rounded to two sig figs.

Cooper Anderson · Answer

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

Given that NH₃ dissociates as follows:

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

The degree of dissociation ($\alpha$) is given as 1.3%, which means $\alpha = 0.013$.

Since NH₃ is a weak base, the concentration of NH₃ after dissociation is approximately equal to its initial concentration minus the concentration of NH₄⁺ formed. Therefore, $[NH_3] \approx [NH_3]_0 - [\text{NH}_4^+]$.

Using the equilibrium expression:

$$K_b = \dfrac{(x)(x)}{(0.10 - x)}$$

Given that $\alpha = \frac{x}{0.10}$, we can express $x$ in terms of $\alpha$.

Solving for $x$:

$$x = \sqrt{K_b(0.10 - x)}$$

$$x = \sqrt{K_b(0.10 - \alpha \times 0.10)}$$

$$x = \sqrt{K_b(0.10 - 0.013 \times 0.10)}$$

$$x = \sqrt{K_b(0.10 - 0.0013)}$$

$$x ≈ \sqrt{K_b \times 0.0987}$$

$$x ≈ 0.3137 \sqrt{K_b}$$

Now, substituting the expression for $x$ into the equation for $\alpha$:

$$0.013 = \frac{0.3137 \sqrt{K_b}}{0.10}$$

$$K_b ≈ \frac{(0.013 \times 0.10)^2}{0.3137^2}$$

$$K_b ≈ \frac{(0.0013 \times 0.10)^2}{0.3137^2}$$

$$K_b ≈ \frac{0.000169}{0.098187}$$

$$K_b ≈ 1.72 \times 10^{-3}$$

So, $K_b$ for NH₃ at this temperature is approximately $1.72 \times 10^{-3}$ M.