We need to calculate the $\ce{[H+]}$ of 0.1M $\ce{(NH4+)(HCO3^-)}$ given the $\ce{k_b(NH4OH), k_{a_1}(H2CO3), k_{a_2}(HCO3^-)}$
The hint tells me to directly use the result $\ce{[H+]=\sqrt{k_{a_1}(\frac{k_w}{k_b}-k_{a_2})}}$
I'm trying to reach this result by systematic equilibrium analysis.
I considered the following simultaneous equilibrium:
$\ce{NH3 +H2O<=>NH4+ +OH^-:k_b = \frac{[NH4+][OH-]}{[NH3]}}$
$\ce{H2CO3<=>H+ +HCO3-:k_{a_1}=\frac{[H+][HCO3-]}{[H2CO3]}}$
$\ce{HCO3^-<=>H+ +CO3^{2-}:k_{a_2}=\frac{[H+][CO3^{2-}]}{[HCO3^-]}}$
$\ce{H2O(l)<=>H+ +OH-:k_w=[H+][OH-]}$
I'm not sure how I shall proceed further. I also tried to write the following charge balance equation:
$\ce{[NH4+] +[H+]=[HCO3-] +2[CO3^{2-}] +[OH-]}$
But in this case, wouldn't the $\ce{[NH4+],[HCO3-]}$ cancel out as they both are equal to 0.1M? Upon solving the charge balance I don't reach the given result. Where am I going wrong? How do I derive the formula?
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$\begingroup$ It reduces to the generally applicable substitution method of solving the set of n nonlinear equations for n variables, with justified applying simplifications based of strong inequalities. The starting points are all equilibrium equations, the ion charge balance equation and total substance concentration equations. I highly recommend to perform formal symbol substitution by single-letter variables. BTW there is no NH4OH in water, it is decades old myth. $\endgroup$– PoutnikCommented May 11, 2023 at 14:27
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$\begingroup$ wouldn't the [NH4+],[HCO3−] cancel out as they both are equal to 0.1M? They will hydrolyze and not in the same extent. [NH3] + [NH4+] = [H2CO3] + [HCO3-] + [CO3^2-] $\endgroup$– PoutnikCommented May 11, 2023 at 14:33
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$\begingroup$ @Poutnik I understand that it comes down to solving a system of equations, but I think either my method or the result given in the problem has an issue. If I can get the chemistry right, then the equations won't be a problem. About the $\ce{NH4OH}$, so should I edit that equilibrium to $\ce{NH3 +H2O<=>NH4+ +OH-}$? Would that be correct? But then how would I use the $\ce{k_b(NH4OH)}$? $\endgroup$– SolusCommented May 11, 2023 at 14:35
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$\begingroup$ the Kb, or pKb (about 4.75) remains it is just NH4OH(aq) versus NH3(aq). N atom is unable to form 5 bonds, having only one 2s and three 2p orbitals. // [NH3] + [NH4+] = 0.1 M AND [H2CO3] + [HCO3-] + [CO3^2-] = 0.1 M $\endgroup$– PoutnikCommented May 11, 2023 at 14:41
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$\begingroup$ @Poutnik Okay, got it! What I did was I put the values in the given result to verify if it was correct and in the end I ended up getting that it'll only be true if $\ce{[HCO3-]=[NH4+]}$. Is the result wrong? $\endgroup$– SolusCommented May 11, 2023 at 17:35
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