pH of 0.05 M NaH2PO3

Question

What is the pH of a $\pu{0.050 M}$ solution of $\ce{NaH2PO3}$? I am given that $\mathrm{p}K_\mathrm{a1} = 2.0$ and $\mathrm{p}K_\mathrm{a2} = 6.58$.

I am getting two different answers, $3.94$ and $4.30$, using two different methods that I have learned:

Method 1

$\ce{H2PO3-}$ can either be protonated or deprotonated. The protonation reaction has

$$K = \frac{10^{-14}}{10^{-2}} = 10^{-12},$$

while the deprotonation reaction has

$$K = 10^{-6.58} = 2.63\times 10^{-7}.$$

Since the protonation reaction is significantly less favorable than the deprotonation ($K$ ratio is less than $1000$), we can ignore the protonation and only focus on the deprotonation. This yields:

$$10^{-6.58} = \frac{x^2}{0.05-x} \to x = [\ce{H+}] = \pu{1.15e-4 M},$$

$$\mathrm{pH} = -\log(x) = 3.94.$$

Method 2

With a diprotic acid that can either be protonated or deprotonated, the $\mathrm{pH}$ is given by the following approximation:

$$pH = 1/2(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a1}),$$

$$\mathrm{pH} = 1/2 \times (2.0 + 6.58) = 4.29.$$

This is the correct answer as presented in the text. Why is the first method invalid?

Answer 1

The problem with method 1 is that you discard the protonation process, which effectively would increase pH as it produces $\ce{OH-}$ ions.

The criteria of $K_1 > 1000K_2$ is applicable when you have a polyprotic acid in its completely (de)protonated state. The ionisation of H₂A produces only a small amount of HA^- and K₂ is small as well, so ionisation of HA^- would be negligible and we can ignore it.

However with an ampholytic species, while K₂ is smaller than K₁, the initial reactant is the same (HA^-) and it is in high concentration, so both processes occur. The formula in method 2 is the result of this consideration.

Answer 2

I think this question has some discrepancies. Asking is the pH of a $\pu{0.050 M}$ $\ce{NaH2PO3}$ solution, and given that $\mathrm{p}K_{a1}=2.0$ and $\mathrm{p}K_{a2}=6.58$ for the salt. Keep in mind that $\ce{H3PO3}$ is a diprotic acid even though it has three $\ce{H}$s, according to Wikipedia-1. If the question really meant sodium salt of $\ce{H3PO3}$, then $\mathrm{p}K_{a1}$ should have been in the range of $1.26–1.3$ at $\pu{25 ^oC}$ (reported as $1.257$ in Ref.1), and $\mathrm{p}K_{a2}=6.7$ (Wikipedia-1). See the illstration of atom arrangement of phosphorus acid:

On the other hand, if the question meant sodium salt of $\ce{H3PO4}$ instead, then $\mathrm{p}K_{a1}$ should have been $2.147$ at $\pu{25 ^oC}$ (Ref.2), and $\mathrm{p}K_{a2}=7.20$ (Wikipedia-2 & Ref.2).

Based on those literature values, $\mathrm{p}K_{a1}$ in the question is more closed to that of $\ce{H3PO4}$ ($2.0$ vs $2.15$) while given $\mathrm{p}K_{a2}$ is more closed to that of $\ce{H3PO3}$ ($6.58$ vs $6.7$). Therefore, one may determine that this question cannot resolve. Yet, I decided to use both values individually in routing calculations usually does by a biochemist to see what would happen:

If If you used $\mathrm{p}K_{a2}=6.58$, $$K_{a2} = \frac {\alpha^2}{0.05-\alpha} ~~~ \left(\text{from the equation:} ~ \ce{H2PO3- + H2O <=> HPO3^2- + H3O+} \right)$$ $$\text{when solve for} ~\alpha ~\text{gives:}~~ \alpha = \left(0.05 \cdot K_{a2} \right)^{\frac 12}~~~ \text{since}~~0.05 \gt \gt \gt K_{a2} ~~\text{and hence}~~0.05 \gt \gt \gt \alpha $$ $$\text{Thus,}~~~\mathrm{p}\ce{H} = \mathrm{p}\alpha = \frac{\mathrm{p}K_{a2}- \log (0.05)}{2}$$ $$\text{If}~\mathrm{p}K_{a2}=6.58~\text{then,}~~~\mathrm{p}\ce{H} = \frac{6.58- \log (0.05)}{2} = 3.94$$

$$\text{If}~\mathrm{p}K_{a2}=7.20~\text{then,}~~~\mathrm{p}\ce{H} = \frac{7.20- \log (0.05)}{2} = 4.25$$

Thus, I can only conclude that you have done nothing wrong on assuming neglegible protonation in first calculation. But, I assume real $\mathrm{p}K_{a2}$ is $7.20$, and calculation should be done accordingly.

References:

1. J. W. Larson, M. Pippin, “Thermodynamics of ionization of hypophosphorous and phosphorous acids. Substituent effects on second row oxy acids,” Polyhedron 1989, 8(4), 527-530 (https://doi.org/10.1016/S0277-5387(00)80751-2).
2. J. W. Larson, K. G. Zeeb, L. G. Hepler, “Heat capacities and volumes of dissociation of phosphoric acid (1st, 2nd, and 3rd), bicarbonate ion, and bisulfate ion in aqueous solution,” * Canadian Journal of Chemistry* 1982, 60(16), 2141-2150 (https://doi.org/10.1139/v82-306).

Answer 3

$\color{red}{\text{PartA.Preface:}}$

The salt ($\ce{NaH2PO3}$) is an ionic salt, and most sodium salts fully dissociate in aqueous solution; so you can consider all ($\ce{NaH2PO3}$) to be converted to ($\ce{H2PO3-}$) ions and $\ce{Na+}$ ions into the water. $\ce{Na+}$ is weak acid(a spectator ion) it won't undergo hydrolysis or take part in any reactions, so the initial amount of sodium in solution will remain constant. We can thus ignore it. ($\ce{H2PO3-}$) is amphoteric that behaves both: as an acid(releasing $\ce{H+}$ in aqueous solution forming $\ce{HPO3^{-2}}$ ions which cannot produce $\ce{PO3^{-3}}$ ions) in water and as a base(can undergo hydrolysis forming ($\ce{H3PO3}$) acid. The autoionization of water produces very few hydrogens and oxygen ions.

The main reactions which are possible to take place are : \begin{array}{ }\ce{ a)Dissociation :NaH2PO3 &~->H2PO3- + Na+&\quad\left (\mbox{Full Dissociation}\right)\\ b)Ionization:H2PO3- &~<=> HPO3^{-2} + H+ &\quad\left(K_\mathrm{2}\approx{2.63\times10^{-7}}\right) \\ c)Hydrolysis:H2PO3- + H2O &~<=> H3PO3 + OH- &\quad\left(K_\mathrm{b}=\frac{K_\mathrm{w}}{K_\mathrm{1}}=\frac{10^{-14}}{1\times10^{-2}}=1\times10^{-12}\right) \\ d)Autoionization:H2O &~<=> H+ + OH- &\quad\left(K_\mathrm{w}= 1\times 10^{-14}\right) \\ }\end{array} so :

The ($\ce{NaH2PO3}$) solution contains the following $5$ species:

the ions$\ce{ Na+ ,H+ ,OH−, H2PO3-, HPO3^{-2} }$ and undissociated acid $\ce{H3PO3}$.

$\color{red}{\text{PartB : Forming "Equilibria ,Mass balance and Proton balance " , equations :}}$

We need $6$ independent equations in order to specify the unknown concentrations of the $6$species present in an aqueous solution of ($\ce{NaH2PO4}$) . these equations are:

1- Equilibria. We have $3$ equilibrium equations:The equilibrium constants of the two stages of ionization of phosphorous acid(This acid is diprotic (readily ionizes two protons), not triprotic Wikipedia) and the dissociation equilibrium of water : \begin{align} K_\mathrm{1} & = \frac{[\ce{H+}][\ce{H2PO3-}]}{[\ce{H3PO3}]} \approx{1 \times 10^{−2}} \tag{1}\\ K_\mathrm{2} & = \frac{[\ce{H+}][\ce{HPO3^2-}]}{[\ce{H2PO3-}]} \approx{ 2.63 \times 10^{−7}} \tag{2}\\ K_\mathrm{w} & = {[\ce{H+}][\ce{OH-}]} = 1 \times 10^{−14} \tag{3} \end{align}

2- Mass balance. The mass balance equation that relates the concentrations of the various dissociation products of the substance to its initial concentration, which we designate here as $C_\ce{P}$ . For the solution of ($\ce{NaH2PO43}$), these equations would be

a- The mass balance equation for the cation ($\ce{Na+}$), but it is trivial: $$C_\ce{P} =[\ce{Na+}]= [\ce{NaH2PO3}]_0 =\pu{0.05}\tag{4}$$ b- The mass balance equation for the anion of ($\ce{H2PO3-}$) ; this concentration is the sum of all concentrations of species that are related to or derived from ($\ce{H2PO3-}$). In this solution the relevant species are: $\ce{H2PO3-, HPO3^{-2} , H3PO3}$ , so we have the following mass balance equation : $$C_\ce{P} =[\ce{H3PO3}]+[\ce{H2PO3–}] + [\ce{HPO3^2–}] \tag{5}$$ 3- Proton Balance Equation: In the proton balance equation, the sum of the concentration terms for species that form by proton consumption is equated to the sum of the concentration terms for species that are formed by the release of a proton.

When ($\ce{NaH2PO3}$ is added to water, $\ce{H2PO3-}$, does react with water and is considered to be the starting species, so:

The species $\ce{ H3PO3}$ results with the consumption of one proton(by $\ce{H2PO3-}$ acting as a base).

The species of $\ce{ H+}$ can form with the consumption of one proton (by water acting as a base).

The species of $\ce{HPO_3^{-2}}$ can form with the release of one proton (by $\ce{H2PO3-}$acting as acid).

The species of $\ce{ OH-}$ can form with the release of one proton (by water acting as acid).

Thus, we have the following proton balance equation : $$\ce{[H3PO3] + [H+] = [HPO3^{-2}] + [OH-]}$$ $$\ce{[H3PO3] + [H+] = [HPO3^{-2}] +\frac{ K_\mathrm{w}}{[\ce{H+}]}}\tag{6}$$

$\color{red}{\text{Part C}}$ : Derive quartic equation to calculate $[\ce{H+}]$ in the solution:

The sequence of derivation:

Step one : Using equations (1,2) to eliminate $\ce{ [H2PO3-], [HPO3^{-2}] }$ successively:

1- Rearrange equation($1$) to get $[\ce{ H2PO3-}]$ : $$ [\ce{H2PO3-}] =\frac{K_\mathrm{1} [\ce{H3PO3}]}{[\ce{H+}]}\tag{7}$$

,and substitute$ [\ce{H2PO3-}]$ of equation($7$) in equation($2$) to get $[\ce{ HPO3^{-2}}]$ : $$[\ce{HPO3^{-2}}] = \frac{K_\mathrm{1} K_\mathrm{2} [\ce{H3PO3}]}{[\ce{H+}]^2} \tag{8}$$

Step two:

1- Substitute the species $\ce{ [H2PO3-]}\text{of equation(7),and} [HPO_3^{-2}]\text{of equation(8) }\text {in equation}(5)$. This lead to $$C_\mathrm{P}= [\ce{H3PO3}] + \frac{K_\mathrm{1} [\ce{H3PO3}]}{[\ce{H+}]} +\frac{K_\mathrm{1} K_\mathrm{2} [\ce{H3PO3}]}{[\ce{H+}]^2 }$$ 2- Rearrange the above equation to get $[\ce{H3PO3}]$( to express as a function of $ C_\pu{P}\text{and}[\ce{H+}]$ ): $$[\ce{H3PO3}]=\frac{C_\ce{p} [ H^+]^2}{[ H^+]^2+K_\mathrm{1} [ H^+]+K_\mathrm{1} K_\mathrm{2} }\tag{9}$$ 3-Substitiute $[\ce{H3PO3}]$ of equation($9$) in the equations ($8$), in order to express $ [\ce{HPO3^{-2}}]$ as a functions of $C_\pu{p}$ and $[\ce{H+}]$:

$$[\ce{HPO3^{-2}}]=\frac{K_\mathrm{1}K_\mathrm{2}C_\pu{p}}{[ H^+]^2+K_\mathrm{1} [ H^+]+K_\mathrm{1} K_\mathrm{2} }\tag{10}$$

Step three:

1- Now, replace these expressions$(9),(10)$ in proton balance equation ($6$), to get : $\ce{\frac{C_\mathrm{p} [ H^+]^2}{[ H^+]^2 +K_\mathrm{1} [ H^+] +K_\mathrm{1} K_\mathrm{2} }+ [{H^+}]=\frac{K_\mathrm{1}K_\mathrm{2}C_\mathrm{p} }{[ H^+]^2 +K_\mathrm{1} [ H^+] +K_\mathrm{1} K_\mathrm{2}}+\frac{ K_\mathrm{w}}{[\ce{H+}]}\tag{11}}$

2- Reduce to same denominator to get : $\ce{ C_\mathrm{p} [ H^+]^3 + [ H^+]^4 +K_\mathrm{1} [ H^+]^3 + K_\mathrm{1} K_\mathrm{2} [ H^+]^2 = K_\mathrm{1}K_\mathrm{2}C_\mathrm{p} [ H+] + K_\mathrm{w}[ H^+]^2 +K_\mathrm{1}K_\mathrm{w} [ H^+] +K_\mathrm{1} K_\mathrm{2}K_\mathrm{w} \tag{12}}$ 3- Put everything on the same side:

$\ce{ [ H^+]^4 + C_\mathrm{p} [ H^+]^3 +K_\mathrm{1} [ H^+]^3 + K_\mathrm{1} K_\mathrm{2} [ H^+]^2 - K_\mathrm{w}[ H^+]^2 - K_\mathrm{1}K_\mathrm{2}C_\mathrm{p} [ H+] -K_\mathrm{1}K_\mathrm{w} [ H^+] -K_\mathrm{1} K_\mathrm{2}K_\mathrm{w} = 0 \tag{13}}$ 4- Grouping and rearranging the terms to get the final formula of quartic equation: $$\color{red}{\ce{ [ H^+]^4 + (C_\ce{p} +K_\mathrm{1}) [ H^+]^3 + (K_\mathrm{1} K_\mathrm{2} - K_\mathrm{w})[ H^+]^2 -(K_\mathrm{1}K_\mathrm{2}C_\mathrm{p} + K_\mathrm{1} K_\mathrm{w}) [ H^+]-K_\mathrm{1} K_\mathrm{2} K_\mathrm{w} = 0 \tag{14}}}$$

$\color{red}{\text{Part D:}}$ Calculations and answers :

1- Substitute the values of $k_\mathrm{1} , k_\mathrm{2}, \text{and} C_\ce{P}$ , in quartic equation ($14$) , letting $X$ denote[$\ce{H+}$] as the following :

2- Solve for (X) from the quintic equation ($18$) by usingwolframalpha

There exists two positive real solution to the quartic equation( $14$) which is$ X=\color{red}{[\ce{H+}]\approx{8.07\times{10^{-5}}}\pu{M}} ~\text{and}[\ce{H+}]\approx{3.2\times\mathrm{10^{-4}}\pu{M}}$

$$\color{red}{\pu{pH}=4.1}~\text{and}~\pu{pH}=3.5$$