Calculating equilibrium constant for a general acid base reaction

Question

I am trying to find the equilibrium value for different acid-base reactions. I have understood that when an acid and base are in an aqueous solution, many different reactions will occur. By combining these reactions, and multiplying their equilibrium constants, we can find the equilibrium value for an acid-base reaction.

Example 1: $NH_3$ and $HCl$

To illustrate my question I will take an example: if $NH_3$ and $HCl$ are in an aqueous solution, what reactions will occur? I believe the following reactions are those that will occur:

$HCl + H_2O ⇌ Cl^- + H_3O^+$, $K_1 = K_a(HCl) = 1.3 * 10^6$

$Cl^- + H_2O ⇌ HCl + OH^-$, $K_2 = K_b(Cl^-) = K_w/K_a(HCl)$

$NH_3 + H_2O ⇌ NH_4^+ + OH^-$, $K_3 = K_b(NH_3) = 1.8 * 10^{-5}$

$NH_4^+ + H_2O ⇌ NH_3 + H_3O^+$, $K_4 = K_a(NH_4^+) = K_w/K_b(NH_3)$

$2H_2O ⇌ H_3O^+ + OH^-$, $K_5 = K_w = 10^{-14}$ (the temperature is 25 degrees Celsius)

These are all the reactions that will occur in this solution, right? I noticed that adding all these will be of no use since the combined reaction will be the autoprotolysis of water. Instead, we only add reactions 1 and 3. That gives:

$HCl + NH_3 + 2H_2O ⇌ Cl^- + NH_4^+ + H_3O^+ + OH^-$, $K_6 = K_1 * K_3 = K_a(HCl) * K_b(NH_3)$

We can subtract the autoprotolysis of water from this reaction to get:

$HCl + NH_3 ⇌ Cl^- + NH_4^+$, $K_7 = K_6 / K_w = K_a(HCl) * K_b(NH_3) / K_w = 1.3 * 10^6 * 1.8 * 10^{-5} * 10^{14} = 2.34 * 10^{15}$

So the equilibrium constant for the reaction between $HCl$ and $NH_3$ is $2.34 * 10^{15}$; in other words, it essentially goes to completion. Is this correct?

Example 2: $HCl$ and $NaOH

Another example is the neutralization reaction between $HCl$ and $NaOH$. How should one deal with $NaOH$? Here is my attempt. The reactions that occur are:

$ HCl + H_2O ⇌ Cl^- + H_3O^+ $, $K_8 = 1.3 * 10^6$

$ NaOH → Na^+ + OH^- $, $K_9$, what should $K_9$ be?

In addition to these, $Cl^- + H_2O$ and the autoprotolysis of water will occur. Similarly to the first example, these reactions will be of no use to find the equilibrium constant for the neutralization reaction.

Adding reactions 8 and 9 gives:

$HCl + NaOH + H_2O ⇌ Cl^- + OH^- + Na^+ + H_3O^+$, $K_{10} = K_8 * K_9$

Can we here subtract the autoprotolysis of water as in the first example (since there is no "2" in front of "H_2O" on the left side of the reaction)? If we can do that then the reaction becomes:

$HCl + NaOH ⇌ Cl^- + Na^+$, $K_{11} = K_8 * K_9 / K_w$

I know that reaction 11 is unbalanced. $K_{11}$ is very large since $K_a(HCl)$ is very large and $K_9$ also is very large. So the reaction will essentially go to completion. However, it feels like something is wrong because reaction 11 is not balanced. What have I done wrong?

Summary

My goal is to understand the underlying theory of acid-base reactions and equilibrium. I have here presented in these two examples what I understand. What am I misunderstanding about how to calculate the equilibrium value for general acid-base reactions?

Thank you in advance!

Answer 1

The rule No 0: Learn to enumerate chemical reactions. The prerequisite is the knowledge of arithmetic of small natural numbers. Total charges and total atom counts must be equal on both reaction sites. No atom is created nor destroyed during any chemical reaction.

The rule No 1: Simplify everything where simplification brings negligible error, otherwise you often end with too complicated math. Considering strong acids as HCl or strong bases as NaOH as completely dissociated brings very negligible error ( with much bigger ignored error due activity versus concentration).

The rule No 2: After the calculation, verify if conditions for the simplification are met. If they are not, you have to reject simplification.

E.g., we want to derive an equation of pH of diluted solution of weak acid. The dissociation constant for a weak acid is:

$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$

If we assume $\ce{H+}$ from water autodissociation is negligible, compared to the acid dissociation ($[\ce{H+}] \gg [\ce{OH-}]$), we can simplify it using $[\ce{H+}] \approx [\ce{A-}]$ to:

$$K_\mathrm{a} = \frac{[\ce{H+}]^2}{[\ce{HA}]}$$

If we then also assume near all acid is not dissociated ($c \approx [\ce{HA}]$), we can simplify it to:

$$K_\mathrm{a} = \frac{[\ce{H+}]^2}{c}$$

leading to

$$c \gg [\ce{H+}] \gg [\ce{OH-}] \implies \mathrm{pH} = \frac 12(\mathrm{p}K_\mathrm{a} - \log{c})$$

The rule No 3: Any respective conjugate pair acid/base (like $\ce{NH4+}$/$\ce{NH3}$) is in water solutions in first place in equilibrium with $\ce{H2O}$/$\ce{H+(aq)}$/$\ce{OH-(aq)}$ respectively. Any equilibrium with other acid/base system is indirect, dependent and redundant.

The rule No 4: Any acid-base reaction in water is generally reaction of some acid passing a proton to some base, the former becoming a base and vice versa.

$$\ce{\mathrm{acid}_1 + \mathrm{base}_2 <=> \mathrm{base}_1 + \mathrm{acid}_2}$$

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How to calculate the equilibrium constant for the reaction between $\ce{H2O}$ and $\ce{NH3}$ (in my example nr 1) and between $\ce{HCl}$ and $\ce{NaOH}$ (in my example nr 2). Could you please show me how you would do that using your method (if my method is wrong?).

• Example No 1: The equilibrium constant is experimentally determined for NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq). You have already listed it in your question as $K_\mathrm{b} = \pu{1.8E-5}$.
• Example No 2: It is equilibrium $\ce{H2O <=> H+(aq) + OH-(aq)}$ with $K_\mathrm{w} = \pu{e-14}$ at $\pu{25 ^\circ C}$, as HCl and NaOH are in water (with negligible error) fully dissociated. $\ce{Na+}$ and $\ce{Cl-}$ ions are often called spectator/bystander ions that do not take part in the reaction.