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Which formula can be used to calculate the exact hydronium concentration present in sodium hydrogen carbonate solution ?

  1. $$\ce{[H3O+]}=\sqrt{K_\mathrm{a1}K_\mathrm{a2}}$$
  2. $$[\ce{H3O+}]=\sqrt{\frac{K_\mathrm{a1}K_\mathrm{w}}{[\ce{NaHCO3}]}}$$
  3. $$[\ce{H3O+}] = \sqrt{\frac{K_\mathrm{a1}K_\mathrm{a2}\times[\ce{NaHCO3}]+K_\mathrm{a1}K_\mathrm{w}}{[\ce{NaHCO3}]+K_\mathrm{a1}}}$$

I tried to order the series of reactions once sodium hydrogen carbonate is dissociated according the following sequence : series of reactions

  1. Taking into consideration equation two and equation four, and taking the assumption: Degree of ionization=degree of hydrolysis, an approximate calculation hydronium concentration can be done by the formula one.

  2. Recognizing equation four and equation two , that for every one mole of carbonate- formed, one mole of hydronium was formed, and for every mole of carbonic acid formed, one mole of hydronium was used up and when comparing K value between both of these equation, I can ignore the contribution of equation four. And because of the low concentration of carbonic acid in equation three and low carbonate ion concentration in equation five , I can ignore the contribution of both equation. So I take into my consideration equation two to derive formula two.

  3. I tried to derive formula three but I don't know the assumptions beyond this formula and how to derive it.

  4. I need to know how to calculate concentrations of all species are present in sodium hydrogen carbonate solution ? I appreciate any help.

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    $\begingroup$ Well none of the formulas calculate the exact pH. One will calculate an approximate pH, assuming $\ce{[HCO3^-] > 10^{-6}}$ so autoionization of water is irrelevant. Also assume that $\ce{[CO3^{2-}] \approx [H2CO3]}$ $\endgroup$ – MaxW Jul 15 '18 at 6:41
  • $\begingroup$ @MaxW 2According to the two assumption mentioned above in the comment : pH is independent of salt concentration ,so we choose formula one , but how to calculate pH without the second assumption? $\endgroup$ – Adnan AL-Amleh Jul 15 '18 at 9:26
  • $\begingroup$ It is a mess. See Section 9 starting on page 42 of Acid-base Equilibria and Calculations. In a nut shell the complete solution considers the autodissociation of water, and the concentrations of species CO2(g), H2CO3, HCO3-, and CO3(-2). Take a look at figure 16. $\endgroup$ – MaxW Jul 15 '18 at 14:03
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    $\begingroup$ Note that by assuming $\ce{[CO3^{2-}] = [H2CO3] }$ the pH is nailed to 8.31. The more realistic assumption is that $\ce{[HCO3-] \gg [CO3^{2-}] + [H2CO3] }$ which means that pH will vary with changes in initial concentration of $\ce{[HCO3-]}$. (This still neglects autodissociation of water.) $\endgroup$ – MaxW Jul 15 '18 at 14:45
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    $\begingroup$ @AdnanALAmleh - I'd be real careful when mixing Ka and Kb equations for a problem like this. Personally I like to use either the two Ka formulas or the two Kb formulas depending on if the solution is going to be acidic or basic and then the explicitly use the Kw expression if needed. The chemical equations in 2-4 just contain three different formulas not four. Three because you can get $K_w$ from a Ka/Kb pair. $\endgroup$ – MaxW Jul 16 '18 at 20:59
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Wording of the Question...

The question asks: Which formula can be used to calculate the exact hydronium concentration present in sodium hydrogen carbonate solution?

What does "exact" mean?

The word exact gives me pause. In general there isn't usually an exact answer, but rather only an answer that is within the required experimental error. The values of $K_\mathrm{a1}$, $K_\mathrm{b1}$, and $K_\mathrm{b2}$ are given to three significant figures. The value for $K_\mathrm{a2}$ is only given to two significant figures, but

$$K_\mathrm{a2} = \dfrac{K_\mathrm{w}}{K_\mathrm{b2}} = \dfrac{1.00\times10^{-14}}{2.13\times10^{-4}} = 4.69483\times10^{-11}\tag{SF1}$$

So it seems reasonable to assume that $K_\mathrm{a2} = 4.70\times10^{-11}$ and that exact means to within 1 part per thousand (ppt).

Since the pH of pure water is taken to be 7.000, this means that for any pH between 4.000 and 10.000 the autoionization of water will have to be taken into account to get results within 1 ppt. (Remember that the characteristic of the pH doesn't count in significant figures, only the mantissa does.)

Realistically that is terribly precise. At 25 $^\circ \mathrm{C}$ $K_w \approx 13.995$ and the change is about 0.033 per $^\circ \mathrm{C}$. So a 1 ppt change is only a 0.03 $^\circ \mathrm{C}$ change in temperature. For such calculations I'd probably expect the typical error would be on the order of ± 5% of reality. But the problem poses 1 ppt precision and so that is how the calculations "should" be done. If the problem poser wanted less precision them the equilibrium constants should have been specified with fewer significant figures.

Setup

Assumptions

  1. Ignoring relationship between $\ce{CO2}$ gas phase and dissolved $\ce{CO2}$ in the liquid phase. Mass balance would require relative volumes of gas and liquid phases.

  2. Ignoring relationship between dissolved $\ce{CO2}$ and carbonic acid. Accept that $K_{\mathrm{a1}}$ is the "combined" equilibrium constant which takes that relationship into account.

  3. The equilibriums will be calculated using concentrations instead of the more appropriate activities. So above 1 millimolar concentration of sodium bicarbonate the calculated results are suspect.

Starting with the basic relationships.

$$K_{\mathrm{a1}} = \dfrac{\ce{[H+][HCO3^-]}}{\ce{[H2CO3]}}\tag{S1}$$

$$K_{\mathrm{a2}} = \dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^-]}}\tag{S2}$$

$$K_\mathrm{w} = \ce{[H^-][OH^-]}\tag{S3}$$

Letting $\mathrm{[C_T]}$ be the total concentration of all the various carbonate species (which we know from is nominal molarity of the sodium bicarbonate).

$$\mathrm{[C_T]} = \ce{[H2CO3] + [HCO3^-] + [CO3^{2-}]}\tag{S4}$$

But this leaves us with five unknowns ($\ce{[H+], [OH^-], [H2CO3], [HCO3^-]}$ and $\ce{[CO3^{2-}]}$) but only four equations (S1), (S2), (S3) and (S4). So another equation is needed relating the five unknowns to completely numerically solve the system.

Fraction of Species

The fraction of each species can be calculated for a particular pH value using equations (F1), (F2), and (F3). Basically you substitute values into (S4) as a function of $\ce{[H+]}$ and whatever species ($\ce{[H2CO3], [HCO3^-], [CO3^{2-}]}$) that you're interest in. The equations for the fractions of the species are exact.

$$\dfrac{\ce{[H_2CO_3]}}{\ce{[C_T]}} = \dfrac{\ce{[H^+]^2}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F1}$$

$$\dfrac{\ce{[HCO3^-]}}{\ce{[C_T]}} = \dfrac{\ce{K_\mathrm{a1}[H^+]}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F2}$$

$$\dfrac{\ce{[CO3^{2-}]}}{\ce{[C_T]}} = \dfrac{\ce{K_\mathrm{a1}K_\mathrm{a2}}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F3}$$

Think of dissolving $\ce{H2CO3}$ ($\ce{C_T}$) and then adding between 0 and 2 equivalents of $\ce{NaOH}$. The equations relate the three variables pH, $\ce{[C_T]}$ and the $x$ equivalents of $\ce{NaOH}$. Knowing any two fixes the third.

Another interesting fact is found by differentiating equation (F2) for $\ce{[H^+]}$ which shows that the maximum of $\ce{[HCO3^-]/[C_T]}$ (or any diprotic acid...) occurs when

$$\ce{[H^+] = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}}\tag{F4}$$

It is the pH as which the bicarbonate solution has its maximum buffering capacity. Also at that pH $\ce{[H2CO3] = [CO3^{2-}]}$.

For the bicarbonate system the maximum concentration of $\ce{HCO3^-}$ is 97.99% which occurs at a pH of 8.340. At that pH $\ce{[H2CO3] = [CO3^{2-}] = 1.01\%}$. So for an accuracy of 1 ppt the hydrolysis of water and the multiple species cannot be neglected.

If the pH is above 7.648, then the concentration of the bicarbonate species will be at least 95%.

\begin{array}{|c|c|c|c|c|}\hline \ce{[H+]} & \mathrm{pH} & \ce{F(H2CO3)} & \ce{F(HCO3^-)}&\ce{F(CO3^{2-})}\\ \hline 1.00E+00 & 0.000E+00 & 1.000E+00 & 4.450E-07 & 2.091E-17 \\ \hline 8.00E-01 & 9.691E-02 & 1.000E+00 & 5.562E-07 & 3.268E-17 \\ \hline 6.00E-01 & 2.218E-01 & 1.000E+00 & 7.417E-07 & 5.810E-17 \\ \hline 4.00E-01 & 3.979E-01 & 1.000E+00 & 1.112E-06 & 1.307E-16 \\ \hline 2.00E-01 & 6.990E-01 & 1.000E+00 & 2.225E-06 & 5.229E-16 \\ \hline 1.00E-01 & 1.000E+00 & 1.000E+00 & 4.450E-06 & 2.091E-15 \\ \hline 8.00E-02 & 1.097E+00 & 1.000E+00 & 5.562E-06 & 3.268E-15 \\ \hline 6.00E-02 & 1.222E+00 & 1.000E+00 & 7.417E-06 & 5.810E-15 \\ \hline 4.00E-02 & 1.398E+00 & 1.000E+00 & 1.112E-05 & 1.307E-14 \\ \hline 2.00E-02 & 1.699E+00 & 1.000E+00 & 2.225E-05 & 5.229E-14 \\ \hline 1.00E-02 & 2.000E+00 & 1.000E+00 & 4.450E-05 & 2.091E-13 \\ \hline 8.00E-03 & 2.097E+00 & 9.999E-01 & 5.562E-05 & 3.268E-13 \\ \hline 6.00E-03 & 2.222E+00 & 9.999E-01 & 7.416E-05 & 5.809E-13 \\ \hline 4.00E-03 & 2.398E+00 & 9.999E-01 & 1.112E-04 & 1.307E-12 \\ \hline 2.00E-03 & 2.699E+00 & 9.998E-01 & 2.225E-04 & 5.228E-12 \\ \hline 1.00E-03 & 3.000E+00 & 9.996E-01 & 4.448E-04 & 2.091E-11 \\ \hline 8.00E-04 & 3.097E+00 & 9.994E-01 & 5.559E-04 & 3.266E-11 \\ \hline 6.00E-04 & 3.222E+00 & 9.993E-01 & 7.411E-04 & 5.805E-11 \\ \hline 4.00E-04 & 3.398E+00 & 9.989E-01 & 1.111E-03 & 1.306E-10 \\ \hline 2.00E-04 & 3.699E+00 & 9.978E-01 & 2.220E-03 & 5.217E-10 \\ \hline 1.00E-04 & 4.000E+00 & 9.956E-01 & 4.430E-03 & 2.082E-09 \\ \hline 8.00E-05 & 4.097E+00 & 9.945E-01 & 5.532E-03 & 3.250E-09 \\ \hline 6.00E-05 & 4.222E+00 & 9.926E-01 & 7.362E-03 & 5.767E-09 \\ \hline 4.00E-05 & 4.398E+00 & 9.890E-01 & 1.100E-02 & 1.293E-08 \\ \hline 2.00E-05 & 4.699E+00 & 9.782E-01 & 2.177E-02 & 5.115E-08 \\ \hline 1.00E-05 & 5.000E+00 & 9.574E-01 & 4.260E-02 & 2.002E-07 \\ \hline 8.00E-06 & 5.097E+00 & 9.473E-01 & 5.269E-02 & 3.096E-07 \\ \hline 6.00E-06 & 5.222E+00 & 9.310E-01 & 6.905E-02 & 5.409E-07 \\ \hline 4.00E-06 & 5.398E+00 & 8.999E-01 & 1.001E-01 & 1.176E-06 \\ \hline 2.00E-06 & 5.699E+00 & 8.180E-01 & 1.820E-01 & 4.277E-06 \\ \hline 1.00E-06 & 6.000E+00 & 6.920E-01 & 3.080E-01 & 1.447E-05 \\ \hline 8.00E-07 & 6.097E+00 & 6.426E-01 & 3.574E-01 & 2.100E-05 \\ \hline 6.00E-07 & 6.222E+00 & 5.741E-01 & 4.258E-01 & 3.336E-05 \\ \hline 4.00E-07 & 6.398E+00 & 4.733E-01 & 5.266E-01 & 6.187E-05 \\ \hline 2.00E-07 & 6.699E+00 & 3.100E-01 & 6.898E-01 & 1.621E-04 \\ \hline 1.00E-07 & 7.000E+00 & 1.834E-01 & 8.162E-01 & 3.836E-04 \\ \hline 8.00E-08 & 7.097E+00 & 1.523E-01 & 8.472E-01 & 4.977E-04 \\ \hline 6.00E-08 & 7.222E+00 & 1.187E-01 & 8.806E-01 & 6.898E-04 \\ \hline 4.00E-08 & 7.398E+00 & 8.239E-02 & 9.165E-01 & 1.077E-03 \\ \hline 2.25E-08 & 7.648E+00 & 4.803E-02 & 9.500E-01 & 1.984E-03 \\ \hline 2.00E-08 & 7.699E+00 & 4.291E-02 & 9.548E-01 & 2.244E-03 \\ \hline 1.00E-08 & 8.000E+00 & 2.188E-02 & 9.735E-01 & 4.576E-03 \\ \hline 8.00E-09 & 8.097E+00 & 1.756E-02 & 9.767E-01 & 5.738E-03 \\ \hline 6.00E-09 & 8.222E+00 & 1.320E-02 & 9.791E-01 & 7.670E-03 \\ \hline 4.57E-09 & 8.340E+00 & 1.006E-02 & 9.799E-01 & 1.008E-02 \\ \hline 4.00E-09 & 8.398E+00 & 8.806E-03 & 9.797E-01 & 1.151E-02 \\ \hline 2.00E-09 & 8.699E+00 & 4.372E-03 & 9.728E-01 & 2.286E-02 \\ \hline 1.00E-09 & 9.000E+00 & 2.142E-03 & 9.531E-01 & 4.479E-02 \\ \hline 8.00E-10 & 9.097E+00 & 1.695E-03 & 9.429E-01 & 5.540E-02 \\ \hline 6.00E-10 & 9.222E+00 & 1.249E-03 & 9.262E-01 & 7.255E-02 \\ \hline 4.00E-10 & 9.398E+00 & 8.037E-04 & 8.941E-01 & 1.051E-01 \\ \hline 2.00E-10 & 9.699E+00 & 3.638E-04 & 8.094E-01 & 1.902E-01 \\ \hline 1.00E-10 & 1.000E+01 & 1.528E-04 & 6.802E-01 & 3.197E-01 \\ \hline 8.00E-11 & 1.010E+01 & 1.132E-04 & 6.298E-01 & 3.700E-01 \\ \hline 6.00E-11 & 1.022E+01 & 7.560E-05 & 5.607E-01 & 4.392E-01 \\ \hline 4.00E-11 & 1.040E+01 & 4.133E-05 & 4.598E-01 & 5.402E-01 \\ \hline 2.00E-11 & 1.070E+01 & 1.342E-05 & 2.985E-01 & 7.015E-01 \\ \hline 1.00E-11 & 1.100E+01 & 3.942E-06 & 1.754E-01 & 8.246E-01 \\ \hline \end{array}

The fourth equation

The fourth equation comes from the charge balance in the solution. The solution must be electrically neutral so the total charge on cations and anions must be equal.

$$\ce{[Na^+] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C1}$$

But in a solution of $\ce{[NaHCO3]}$ the $\ce{[Na^+] = [C_T]}$ so:

$$\ce{[C_T] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C2}$$

Which yields and exact solution as a quartic equation. But rather than solve only that particular quartic equation, let's consider the more general case. Assume we start with $\ce{H2CO3}$ in water and add amounts of $\ce{NaOH} according to the equation:

$$\ce{H2CO3 + xNaOH -> Na_xH_{2-x}CO3 + xH2O}\tag{C3}$$

where $x$ is a real number bounded by $ 0 \le x \le 2$. The special cases of $x = 0,1,2$ correspond to adding $\ce{H2CO3, NaHCO3}$ or $\ce{Na2CO3}$ to pure water. Then it is intuitively obvious to the casual observer that (a la analytical chemist Dr. Peter Carr, University of Ga.) the general equation is:

$$\begin{multline} [\ce{H+}]^4 + (K_\mathrm{a1} + xC_T)[\ce{H+}]^3 + (K_\mathrm{a1} K_\mathrm{a2} + (x - 1) C_T K_\mathrm{a1} - K_\mathrm{w})[\ce{H+}]^2\\ + K_\mathrm{a1} ((x - 2)C_T K_\mathrm{a2} - K_\mathrm{w})[\ce{H+}] - K_\mathrm{a1} K_\mathrm{a2} K_\mathrm{w} \tag{C4} \end{multline}$$

It is not hard for a computer to solve the quartic equation numerically, but to do so by hand is extremely tedious. Hence books never(?) required such a solution in problem sets.

DERIVATION OF EQUATION 1 $\ce{[H^+]} = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}$

$$K_\mathrm{a1}K_\mathrm{a2} = \dfrac{\ce{[H+][HCO3^-]}}{\ce{[H2CO3]}}\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^-]}} = \dfrac{\ce{[H+]^2[CO3^{2-}]}}{\ce{[H2CO3]}}\tag{EQ1-1}$$

rearrange (EQ1-1)

$$\ce{[H+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[H2CO3]}}{\ce{[CO3^{2-}]}}}\tag{EQ1-2}$$

assume $\ce{[H2CO3] = [CO3^{2-}]}$

Then

$$\ce{[H+]} = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}\tag{EQ1-3}$$

DERIVATION OF EQUATION 2 $\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T]}}}$

Sodium is a spectator ion. Thus it is be possible to model the system as a monoprotic Brønsted–Lowry acid which has $\ce{H^+}$ as the acid and $\ce{HCO3^-}$ as the conjugate base.

This derivation has two problems. First it ignores the autoionization of water. Second the equation models the system as a monoprotic base, but the carbonate system has three species - carbonic acid, bicarbonate, and carbonate, and all three are likely to be present in significant quantities in alkaline solutions.

Start with the reaction of the conjugate base

$$\ce{HCO3^- + H2O <=> H2CO3 + OH^-}\tag{EQ2-1}$$

and the equilibrium expression

$$\ce{K_\mathrm{b1}} = \dfrac{\ce{[H2CO3][OH^-]}}{\ce{[HCO3^-]}}\tag{EQ2-2}$$

Assume that little of $\ce{HCO3^-}$ accepts a proton, thus $\ce{[HCO3^-] = [C_T]}$ and $\ce{[H2CO3] = [OH^-]}$. Now substituting:

$$\ce{K_\mathrm{b1}} = \dfrac{\ce{[OH^-]^2}}{\ce{[C_T]}}\tag{EQ2-3}$$

but $\ce{K_\mathrm{b1}} = K_w / K_{a1}$ and $\ce{[OH^-] = K_w / [H^+]}$. Now making more substitutions

$$\dfrac{K_\mathrm{w}}{K_\mathrm{a1}} = \dfrac{(\frac{K_\mathrm{w}}{\ce{[H^+]}})^2}{[C_T]}\tag{EQ2-4}$$

rearranging and solving for $\ce{[H^+]}$ gives the desired result.

$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T]}}}\tag{EQ2-5}$$

The desired result could have been obtained more expeditiously by starting with equation (2) and letting $\ce{[H2CO3] = [OH^-] = K_w/[H^+]}$, but that route doesn't seem as clear chemically and seems more like math magic.

DERIVATION OF EQUATION 3 $\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}C_T + K_{a1}K_w}{C_T+K_{a1}}}$

Starting with equation (C2) for the charge balance of a solution of $\ce{NaHCO3}$ where $\ce{[C_T] = [Na^+]}$

$$\ce{[C_T] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C2}$$

subtract equation (S4) which is for mass balance of all the carbonate species

$$\mathrm{[C_T]} = \ce{[H2CO3] + [HCO3^-] + [CO3^{2-}]}\tag{S4}$$

yielding

$$\ce{[H^+] = [CO3^{2-}] - [H2CO3] + [OH^-]}\tag{EQ3-1}$$

rearranging

$$\ce{[OH^-] = [H2CO3] - [CO3^{2-}] + [H^+]}\tag{EQ3-2}$$

substituting $K_\mathrm{w} / \ce{[H^+] = [OH^-]}$ and rearrange

$$\dfrac{K_\mathrm{w}}{\ce{[H^+]}} \ce{+ [CO3^{2-}] = [H2CO3] + [H^+]}\tag{EQ3-3}$$

Solve equation (S1) for $\ce{[H2CO3]}$ and solve equation (S2) for $\ce{[CO3^{2-}]}$, and then substitute those results into equation (EQ3-3)

$$\dfrac{K_\mathrm{w}}{\ce{[H^+]}} + \dfrac{K_\mathrm{a2}\ce{[HCO3^-]}}{\ce{[H^+]}} = \dfrac{\ce{[H^+][HCO3^-]}}{K_\mathrm{a1}} + \ce{[H^+]} \tag{EQ3-4}$$

Multiplying both sides by $\ce{[H^+]}$, then $K_\mathrm{a1}$, and finally collecting terms

$$K_\mathrm{a1}K_\mathrm{w} + K_\mathrm{a1}K_\mathrm{a2}\ce{[HCO3^-]} = \ce{[H^+]^2([HCO3^-] + K_\mathrm{a1})} \tag{EQ3-5}$$

rearranging and solving for $\ce{[H^+]}$ yields the exact equation for $\ce{[H^+]}$ in terms of the particular species $\ce{[HCO3^-]}$ when $\ce{NaHCO3}$ is dissolved in pure water.

$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[HCO3^-]} + \mathrm{K_{a1}K_w}}{\ce{[HCO3^-] + \mathrm{K_{a1}}}}}\tag{EQ3-6}$$

As indicated previously the above equation is exact in terms of the relationship between $\ce{[H^+]}$ and the carbonate species $\ce{HCO3^-}$. Over the whole pH range $\ce{[C_T] \ne [HCO3^-]}$, however for the range $ 7.648 \lt \mathrm{pH} \lt 9.00$ the species $\ce{HCO3^-}$ is at least 95% of $\ce{[C_T]}$. So we can "reasonably" substitute $\ce{[C_T] = [NaHCO3]}$ and get the desired equation even though the equation is not exact.

$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[C_T]} + K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T] + K_\mathrm{a1}}}}\tag{EQ3-7}$$

!! CHECKS !!

Why aren't students taught to do checks any more!?!

The column $\ce{[C_T]}$ is the nominal molarity of the sodium bicarbonate solution. It is also the molarity of the spectator cation $\ce{[Na^+]}$.

The column $\Sigma \space \text{cations}$ should equal the column $\Sigma \space \text{anions}$.

The column % Diff is a measure of the inconsistency of the calculations of the concentration of anions and cations and is calculated by

$$\text{% Diff} = 200*\dfrac{\Sigma\text{ cations} - \Sigma\text{ anions}}{\Sigma\text{ cations} + \Sigma\text{ anions}}$$

Again, these calculations use concentrations instead of the more appropriate activities. Concentrations are probably fine for these solutions below 1 millimolar. Above that and activities should really be used.

EQUATION 1 $\ce{[H^+]} = \sqrt{K_{a1}K_{a2}}$

\begin{array}{|c|c|c|c|c|c|c|}\hline \ce{[C_T]} & calc \space [H^+] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff\\ \hline 1.00E-01 & 4.5733E-09 & 8.340 & 1.0000E-01& 2.1866E-06 &9.7986E-02 &2.0140E-03 &1.0000E-01 & 0.0 \\ \hline 1.00E-02 &4.5733E-09 & 8.340 &1.0000E-02 &2.1866E-06 &9.7986E-03 &2.0140E-04 &1.0002E-02 & 0.0 \\ \hline 1.00E-03 &4.5733E-09 & 8.340 &1.0000E-03 &2.1866E-06 &9.7986E-04 &2.0140E-05 &1.0022E-03 & -0.2 \\ \hline 1.00E-04 &4.5733E-09 & 8.340 &1.0000E-04 &2.1866E-06 &9.7986E-05 &2.0140E-06 &1.0219E-04 & -2.2\\ \hline 1.00E-05 &4.5733E-09 & 8.340 &1.0005E-05 &2.1866E-06 &9.7986E-06 &2.0140E-07 &1.2187E-05 & -19.7\\ \hline 1.00E-06 &4.5733E-09 & 8.340 &1.0046E-06 &2.1866E-06 &9.7986E-07 &2.0140E-08 &3.1866E-06 & -104 \\ \hline 1.00E-07 &4.5733E-09 & 8.340 &1.0457E-07 &2.1866E-06 &9.7986E-08 &2.0140E-09 &2.2866E-06 & -183 \\ \hline \end{array}

So assuming that the pH is constant only works for concentrated solutions (greater than 1 millimolar) for which activities should really be used...

EQUATION 2 $\ce{[H^+]} = \sqrt{\dfrac{K_{a1}K_w}{\ce{[C_T]}}}$

\begin{array}{|c|c|c|c|c|c|c|}\hline \ce{[C_T]} & calc \space [H^+] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff\\ \hline 1.00E-01 &2.1095E-10 & 9.676 &1.0000E-01 &4.7405E-05 &8.1748E-02 &3.6427E-02 &1.1822E-01 & -16.7 \\ \hline 1.00E-02 &6.6708E-10 & 9.176 &1.0000E-02 &1.4991E-05 &9.3287E-03 &1.3145E-03 &1.0658E-02 & -6.4\\ \hline 1.00E-03 &2.1095E-09 & 8.676 &1.0000E-03 &4.7405E-06 &9.7369E-04 &4.3388E-05 &1.0218E-03 & -2.2\\ \hline 1.00E-04 &6.6708E-09 & 8.176 &1.0001E-04 &1.4991E-06 &9.7844E-05 &1.3787E-06 &1.0072E-04 & -0.7 \\ \hline 1.00E-05 &2.1095E-08 & 7.676 &1.0021E-05 &4.7405E-07 &9.5271E-06 &4.2453E-08 &1.0044E-05 & -0.2 \\ \hline 1.00E-06 &6.6708E-08 & 7.176 &1.0667E-06 &1.4991E-07 &8.6910E-07 &1.2247E-09 &1.0202E-06 & +4.5\\ \hline 1.00E-07 &2.1095E-07 & 6.676 &3.1095E-07 &4.7405E-08 &6.7830E-08 &3.0225E-11 &1.1527E-07 & +91.8 \\ \hline \end{array}

If 1 ppt is required, then Equation 2 doesn't work. For a +/- 5% error, the equation works in the range from $1\times10^{-2}$ to $1\times10^{-6}$ molar.

EQUATION 3 $\ce{[H^+]} = \sqrt{\dfrac{K_{a1}K_{a2}C_T + K_{a1}K_w}{C_T+K_{a1}}}$

\begin{array}{|c|c|c|c|c|c|c|}\hline \ce{[C_T]} & calc \space [H] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff \\ \hline 1.00E-01 &4.5781E-09 & 8.339 &1.0000E-01 &2.1843E-06 &9.7986E-02 &2.0119E-03 &1.0000E-01 & 0.0 \\ \hline 1.00E-02 &4.6216E-09 & 8.335 &1.0000E-02 &2.1638E-06 &9.7986E-03 &1.9930E-04 &1.0000E-02 & 0.0 \\ \hline 1.00E-03 &5.0352E-09 & 8.298 &1.0000E-03 &1.9860E-06 &9.7977E-04 &1.8291E-05 &1.0000E-03 & 0.0 \\ \hline 1.00E-04 &8.0700E-09 & 8.093 &1.0001E-04 &1.2392E-06 &9.7660E-05 &1.1376E-06 &1.0004E-04 & 0.0 \\ \hline 1.00E-05 &2.1120E-08 & 7.675 &1.0021E-05 &4.7348E-07 &9.5267E-06 &4.2400E-08 &1.0043E-05 & -0.2\\ \hline 1.00E-06 &5.5624E-08 & 7.255 &1.0556E-06 &1.7978E-07 &8.8822E-07 &1.5010E-09 &1.0695E-06 & -1.3\\ \hline 1.00E-07 &9.0382E-08 & 7.044 &1.9038E-07 &1.1064E-07 &8.3082E-08 &8.6408E-11 &1.9381E-07 & -1.8 \\ \hline 1.00E-08 &9.8897E-08 & 7.005 &1.0890E-07 &1.0111E-07 &8.1785E-09 &7.7735E-12 &1.0930E-07 & -0.4\\ \hline \end{array}

If 1 ppt is required, then Equation 3 works for concentrations grater than $1\times10^{-5}$ molar. For a +/- 5% error, the equation works in the range from $0.1$ to $1\times10^{-8}$ molar.

Summary of $\ce{[H^+]}$ values

the table below is a summary of the results for the three derived equations, and the exact solution using equation (C4) by Wolfram Alpha.

\begin{array}{|c|c|c|c|c|}\hline \ce{[C_T]} & Eq(1)& Eq(2) & Eq(3) & exact\\ \hline 1.00E-01 & 4.5733E-09 & 2.1095E-10 & 4.5781E-09 & 4.5782E-9 \\ \hline 1.00E-02 & 4.5733E-09 & 6.6708E-10 & 4.6216E-09 & 4.6226E-9 \\ \hline 1.00E-03 & 4.5733E-09 & 2.1095E-09 & 5.0352E-09 & 5.0443E-9 \\ \hline 1.00E-04 & 4.5733E-09 & 6.6708E-09 & 8.0700E-09 & 8.1354E-9 \\ \hline 1.00E-05 & 4.5733E-09 & 2.1095E-08 & 2.1120E-08 & 2.1602E-8 \\ \hline 1.00E-06 & 4.5733E-09 & 6.6708E-08 & 5.5624E-08 & 5.7983E-8 \\ \hline 1.00E-07 & 4.5733E-09 & 2.1095E-07 & 9.0382E-08 & 9.1835E-8 \\ \hline 1.00E-08 & -- & -- & 9.8897E-08 & 9.9096E-8 \\ \hline \end{array}

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  • $\begingroup$ I am fascinated with your powerful answer which remediates my misconception about the assumptions and conditions at which each formula work. $\endgroup$ – Adnan AL-Amleh Jul 20 '18 at 18:29
  • 1
    $\begingroup$ @AdnanAL-Amleh - Added exact solution in last table. Bit of a struggle to Wolfram to work. $\endgroup$ – MaxW Aug 1 '18 at 4:11
  • 2
    $\begingroup$ I appreciate your effort and your work and I thank you $\endgroup$ – Adnan AL-Amleh Aug 1 '18 at 4:29
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    $\begingroup$ @AdnanAL-Amleh - You could be right. I threw all my notes on this away a while back. I'll have to check that again. $\endgroup$ – MaxW Oct 30 '18 at 12:16
  • 1
    $\begingroup$ @AdnanAL-Amleh - you're correct. equation 7 here aqion.de/site/184 $\endgroup$ – MaxW Oct 30 '18 at 18:07

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