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Suppose we want to calculate pH of an $\ce{NH4HCO3}$ solution of known concentration. Now, $\ce{NH4+}$ will be hydrolised to give $\ce{H+}$ ions. Some $\ce{HCO3-}$ ions will be converted to (i) $\ce{H2CO3}$ and $\ce{OH-}$ and some will give (ii) $\ce{H+}$ and $\ce{CO3^2-}$

\begin{align} \ce{HCO3- + H2O &<=> H2CO3 + OH− }& 1/K_\mathrm{a1}\\ \ce{HCO3- + H2O &<=> CO3^2- + H3O+ }& K_\mathrm{a2}\\ \ce{NH4+ + H2O &<=> NH3 (aq) + H3O+ }& 1/K_\mathrm{b}\\ \end{align}

  1. I am not sure how to take account of both $\ce{H3O+}$ and $\ce{OH-}$ in the calculation of pH.

  2. Will the third equilibium costant be $1/K_\mathrm{b}$?

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  • $\begingroup$ Can you use the formula given in my answer and check if if it gives you the correct answer? $\endgroup$ – Aditya Dev Nov 12 '15 at 11:33
  • $\begingroup$ 3rd equilibrium is written terribly - it's simply NH4+ +H2o = NH3 +H3o+ (proton exchange) $\endgroup$ – Mithoron Nov 12 '15 at 11:53
  • $\begingroup$ @Mithoron: That's the reaction given in my textbook. User intends hydrolysis of cation. BTW, did the formula work? $\endgroup$ – Aditya Dev Nov 12 '15 at 13:05
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    $\begingroup$ @AdityaDev You should get new textbook ;) "NH4OH" is only in old ones - there's no such molecule, there's NH3(aq) written instead now $\endgroup$ – Mithoron Nov 12 '15 at 17:02
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    $\begingroup$ My supervisor spent about five minutes grinning and laughing about a practical joke a TA did in an undergrad lab course to a fellow student, sending her away to get ‘solid $\ce{NH4OH}$’ while telling us the story in the post-serious drinking session during a symposium. $\endgroup$ – Jan Nov 12 '15 at 17:43
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The main reactions which take place are

1)Ionization of $\ce{HA-}$: $\ce{HA- -> H+ +A^2-}$

2)Hydrolysis of $\ce{HA-}$: $\ce{HA- + H+ -> H2A}$

3)Hydrolysis of cation: $\ce{B+ + H2O -> BOH + H+}$

for reaction $1$, $$k_{a2}=\frac{[H^+][A^{2-}]}{[HA^-]}$$

for reaction $2$, $$k_{a1}=\frac{[HA^-][H^+]}{[H_2A]}$$

for reaction $3$, $$k_h=\frac{k_w}{k_b}=\frac{[BOH][H^+]}{[B^+]}$$

At any moment, $$\ce{[H^+]=[BOH] +[A^{2-}]-[H2A]}$$ $$[H^+]=\frac{[NH^{4+}].k_w}{[H^+]k_b}+\frac{k_{a2}.[HCO_3^-]}{[H^+]}-\frac{[HCO_3^-][H^+]}{k_{a1}}$$ Rearranging the terms to get $\ce{[H+]}$, $$[H^+]=\sqrt{\frac{\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]}{1+\frac{[HCO_3^-]}{k_{a1}}}}$$ $$[H^+]=\sqrt{\frac{k_{a1}\cdot \big(\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]\big)}{k_{a1}+[HCO_3^-]}}$$

Now we will have to make 2 assumptions. First lets assume that $k_{a1}<<[HCO_3^-]$ so that $k_{a1}+[HCO_3^-]=[HCO_3^-]$

$$[H^+]=\sqrt{\frac{k_{a1}\cdot \big(\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]\big)}{[HCO_3^-]}}$$

If you assume that neither hydrolysis nor the dissociation goes too far, you can assume the concentration of bicarbonate ion to be equal to the initial concentration (say C). Since Ammonium bicarbonate dissociates to give two products in 1:1 ratio, concentration of ammonium produced is equal to the concentration of bicarbonate ion produced.

I saw this formula for calculation pH of such salts in my textbook but no derivation was given: $$[H^+]=\sqrt{k_{a1}\cdot\big(\frac{k_w}{k_b}+k_{a2}\big)}$$ The above formula can be obtained if you consider the last assumption. (Clarification needed)

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  • $\begingroup$ I don't understand why $\ce {[H+]=[B+] + [A2−] − [H2A]}$ It should be $\ce [BOH] + [A2−] − }$ [how to take acount of this? here OH- was produced] $\endgroup$ – Archisman Panigrahi Nov 12 '15 at 17:36
  • $\begingroup$ I have edited that part. It's not B+. It's BOH. Reason is, H+ is produced by B+ and equal amount of H+ and BOH are formed. So [H+] from B+ is same as the amount of BOH formed. $\endgroup$ – Aditya Dev Nov 12 '15 at 17:40
  • $\begingroup$ And, how do you take account of the last part? (why $\ce {−[H2A]}$) I mean, how to substitute the [OH-]? $\endgroup$ – Archisman Panigrahi Nov 12 '15 at 17:42
  • $\begingroup$ Amount of H+ used up and amount of H2A formed are same. $\endgroup$ – Aditya Dev Nov 12 '15 at 17:44
  • $\begingroup$ If you have any numerical type question, see if the formula works. $\endgroup$ – Aditya Dev Nov 12 '15 at 17:45
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$$\ce{Lets assume a general salt BHA (a salt of weak base and weak acid)}$$ $$\ce{(Here B^+= NH_{4}^+ and HA^-= HCO_{3}^-)}$$

$\ce{BHA}$ dissociates in water as follows: $$\ce{BHA => B+ + HA-}$$

The following three main reactions take place:- $$\ce{B+ + H2O <=> BOH + H+}$$ $$\ce{HA- + H2O <=> H2A + OH- (HA- acting as base)}$$ $$\ce{HA- <=> H+ + A^2- (HA- acting as acid)}$$

for reaction $1$, $$K_{h1}=\frac{[BOH][H^+]}{[B^+]}=\frac{K_w}{K_b}$$ for reaction $2$, $$K_{h2}=\frac{[H_{2}A][OH^-]}{[HA^-]}=\frac{K_w}{K_{a1}}$$ for reaction $3$, $$K_{h3}=\frac{[H^+][A^{2-}]}{[HA^-]}=K_{a2}$$

Now, conserving moles of $B$ $$n(B^+)_{t=0}=n(B^+)_{t=eq}+n(BOH)_{t=eq}$$ $$\frac{n(B^+)_{t=0}}{V}=\frac{n(B^+)_{t=eq}}{V}+\frac{n(BOH)_{t=eq}}{V}$$ $$[B^+]_{t=0}=[B^+]_{t=eq}+[BOH]_{t=eq}$$ $$[B^+]_{t=0}=[B^+]_{t=eq}+\frac{K_{h1}*[B^+]_{t=eq}}{[H^+]}$$ $$[B^+]_{t=eq}=\frac{[B^+]_{t=0}}{1+\frac{K_{h1}}{[H^+]}}=\frac{C}{1+\frac{K_{h1}}{[H^+]}}$$

Similarly, $$[HA^-]_{t=0}=[A^{2-}]_{t=eq}+[HA^-]_{t=eq}+[H_{2}A]_{t=eq}$$ $$[HA^-]_{t=0}=\frac{K_{h3}*[HA^-]_{t=eq}}{[H^+]}+[HA^-]_{t=eq}+\frac{K_{h2}*[HA^-]_{t=eq}}{[OH^-]}$$ $$[HA^-]_{t=eq}=\frac{[HA^-]_{t=0}}{\frac{K_{h3}}{[H^+]}+1+\frac{K_{h2}}{[OH^-]}}=\frac{C}{\frac{K_{h3}}{[H^+]}+1+\frac{K_{h2}}{[OH^-]}}$$

Now, $$[B^+]_{t=eq} \approx[HA^-]_{t=eq}$$ $$( [HA^-]_{t=0}=[B^+]_{t=0}=C )$$ $$( And\ there\ is\ negligible\ change\ in\ their\ respective\ concentrations\ since\ hydrolysis\ is\ a\ weak\ phenomena\ and\ HA^-\ is\ a\ weak\ acid)$$ $$\frac{C}{1+\frac{K_{h1}}{[H^+]}}\approx\frac{C}{\frac{K_{h3}}{[H^+]}+1+\frac{K_{h2}}{[OH^-]}}$$ $$1+\frac{K_{h1}}{[H^+]}\approx\frac{K_{h3}}{[H^+]}+1+\frac{K_{h2}}{[OH^-]}$$ $$\frac{K_{h1}}{[H^+]}\approx\frac{K_{h3}}{[H^+]}+\frac{K_{h2}}{[OH^-]}$$ $$\frac{K_{w}}{K_{b}*[H^+]}\approx\frac{K_{a2}}{[H^+]}+\frac{K_{w}}{K_{a1}*[OH^-]}$$ $$\frac{K_{w}}{K_{b}*[H^+]}\approx\frac{K_{a2}}{[H^+]}+\frac{[H^+]}{K_{a1}}$$

Solving which, $$[H^+]\approx\sqrt{\frac{K_{w}*K_{a1}}{K_{b}}-K_{a1}*K_{a2}}$$

$$For\ NH_{4}HCO_{3}\ solution,\ pH \approx 5.06$$

Cheers!

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