# pH calculation of HK2PO4

I am struggling with a calculation I should know how to do, but I feel like I may be missing crucial information. Here's the outline:

I dissolve $\pu{6.0 g}$ of $\ce{HK2PO4}$ in $\pu{20 mL}$ of water. I then add $\pu{8.0 mL}$ of $\pu{4 M}$ $\ce{HCl}$ and dilute to $\pu{30.0 mL}$. I take $\pu{25.0 mL}$ of the resulting solution and adjust the pH to 3.5 with $\pu{1 M}$ acetic acid.

What volume of acetic acid is theoretically necessary to reach pH 3.5?

Experimentally, I attempted this and the initial pH of the $\pu{25.0 mL}$ aliquot was 5.71. Technically, I should have been able to reduce the pH to 3.5 using less than $\pu{15.0 mL}$ of $\pu{1 M}$ acetic acid. However, I observed buffering effects around pH 4.0 and the total volume required was $\sim\pu{50.0 mL}$ of $\pu{1 N}$ acetic acid.

• I don't think this calculation is as straight forward as I thought. Since this is a polyprotic acid I have to perform a speciation calculation. This results in a linear system of two mass balance equations. Please correct if I am wrong. – user58641 Feb 7 '18 at 2:51
• Thanks Vinicius Godim! That was a very thorough answer and extremely helpful. Thank you for taking the time to help me. – user58641 Feb 9 '18 at 1:37

In complex analytical chemistry problems like this, it's useful to consider a stoichiometric matrix of the reaction network.

                               | H2PO4- | H3O+ | H3PO4 | CH3COOH | HPO4-2 | PO4-3
------------------------------ | ------ | ---- | ----- | ------- | ------ | -----
1st phosporic aciddissociation | 1      | 1    | -1    | 0       | 0      | 0
2nd phosporic dissociation     | -1     | 1    | 0     | 0       | 1      | 0
3rd phosporic dissociation     | 0      | 1    | 0     | 0       | -1     | 1
Acetic acid ionization         | 0      | 1    | 0     | -1      | 0      | 0


Each score represent the stoichiometric coefficient of that species in the given reaction, with a negative sign when it's a reactant and a positive value otherwise. I am not using $\ce{KH2PO4}$ or $\ce{HCl}$ as species because they are always fully dissociated. It's better to think of them as $\ce{H2PO4-}$ and $\ce{H3O+}$ respectively in regards to their molar concentration.

We can approach the problem by consider a "mixing $\to$ reaction" two-step process. Because equilibrium is achieved regardless of the path chosen, we can do this freely, that is, consider the "initial" concentration as though we mixed all parts together with no reaction occurring.

The mixing step mass balance will be $C_{\textrm{initial}} \times V_{\textrm{pre-mixing}} / V_{\textrm{solution}}$. Here's the full table:

                               | Solvent | H2PO4-       | H3O+         | H3PO4 | CH3COOH      | HPO4-2 | PO4-3 | CH3COO- | Total
------------------------------ | ------- | ------------ | ------------ | ----- | ------------ | ------ | ----- | ------- | ------------
Initial Concentration (mol/L)  | 0       | 2,2044883383 | 4            | 0     | 1            | 0      | 0     | 0       |
Initial volume (L)             | 0,03    | 0,02         | 0,008        | 0     | 0,5614062187 | 0      | 0     | 0       | 0,6194062187
Solution concentration (mol/L) | 0       | 0,0711806976 | 0,0516623809 | 0     | 0,9063619346 | 0      | 0     | 0


Since we are basically solving a nonlinear problem, I took the liberty to assign a random value to the initial acetic acid volume, which we will later determine by minimizing an objective function.

Let's call the stoichimetric matrix $A$, and the solution concentration (post-mixing, pre-reaction) vector $C_{\textrm{post-mix}}$. Let's define the extent of reaction vector $\delta$ as the array of values representing the amount of mols (per unit volume) produced or consumed by each reaction.

Reaction                   | Extent (mol/L)
-------------------------- | --------------------
1st phosporic dissociation | 0,0034158968
2nd phosporic dissociation | 1,46105223300036E-05
3rd phosporic dissociation | -2,218252003941E-14
Acetic acid ionization     | -0,0547767774


Once again, we do not know these values a priori. They will be optimized later on. The amounts of each reactant that is produced or consumed ($C_{\textrm{reacted}}$) is then given by the matrix product

$C_{\textrm{reacted}}^T = A^T \delta$

The equilibrium concentration is then the post-mixing concentration plus the reacted concentration,

$C_{\mathrm{eq}} = C_{\text{post-mix}} + C_{\textrm{reacted}}$

Now consider the vector of potentials of the equilibrium constants fpr each reaction ($\mathrm pK = -\log(K)$),

Reaction                       | pK
------------------------------ | -------------
1st phosporic aciddissociation | 2,1611509093
2nd phosporic dissociation     | 7,2076083105
3rd phosporic dissociation     | 12,3187587626
Acetic acid ionization         | 4,7447274949


For the equilibrium step, equaling the equilibrium expression to the $K$ of each reaction is the same as imposing the following matrix product (think of why?)

$A (\mathrm pC_{\textrm{eq}})^T = \mathrm pK$

Where the vector $\mathrm pC$ is calculed by taking the negative log of the species concentrations vector $C_{\mathrm{eq}}$. One of these values is the $\mathrm pH$, which is given by the problem.

We are then left with the task of finding the initial acetic acid volume by solving:

Find $V_\ce{CH3COOH}$ and $\delta$ (remember, this is a vector) that minimize the residuals in $A (\mathrm pC_{\mathrm{eq}})^T = \mathrm pK$ and $\mathrm pC_{\mathrm{eq}} \Bigg|_{\text{species}=\ce{H3O+}} = (\mathrm pH)_{\text{given}}$.

I used libreoffice-nlpsolver to yield

$$V_{\ce{CH3COOH}} = 0.5614062187\ \mathrm L$$

But, wait, this is the volume of acetic acid we'd employ to react with the entire solution volume. But the problem asks about only $25\ \mathrm{ml}$, it is thus just a matter of scaling the proportion using the total volume ($V_\mathrm t = 0{,}6194062187$),

$$V = 0{,}5614062187 L \times \frac{0{,}025 L}{0{,}6194062187} = 0{,}022659048 L \ \mathrm L$$

Or about $22{,}7\ \mathrm{ml}$

I purposely did not take into account the self ionization of water, because of the low pH in question. But it could have been done nonetheless. The only changes would be the need to include $\ce{OH-}$ in the stoichiometric matrix and take $\mathrm pW$ into action.

• Please add the missing units. – Loong Feb 7 '18 at 16:44