How much NaOH required to change the pH of a weak acid solution by 2

Question

If I have a $\pu{20mL}$ $\pu{0.1M}$ solution of sodium dihydrogenphosphate, how much $\pu{0.1M}$ $\ce{NaOH}$ do I need to add to raise the $pH$ by 2.

Here's what I've done so far. I'm doubt it's right and I get stuck with respect to the change in volume.

Edit: I realize that this wrong as sodium dihydrogenphosphate is polyprotic.

I found using the $K_a$ that a $\pu{0.1M}$ solution will lead to a $[\ce{H+}]= 3.47\times 10^{-5}$

Since $-log(3.47\times 10^{-5}) = pH = 4.46$, the pH of the final solution is $4.46 + 2 = 6.46$.

The change in $[\ce{H+}]$ is $3.47\times10^{-5} - 10^{-6.46} = 3.44\times10^{-6}$, so there must be a concentration of $3.44\times10^{-6}M$ $\ce{OH-}$ added to the solution.

Since my solution is $0.02L$, $0.02 \times 3.44\times10^{-6} = 7\times10^{-8}$ moles of $\ce{OH-}$.

The ratio of $\ce{OH-}$ to $\ce{NaOH}$ is $1:1$, so:

$\pu{0.1M}$ $\ce{NaOH} = 7\times\dfrac{10^-8}{x}$, $x = 7\times10^{-7} mL$ of $\pu{0.1M}$ $\ce{NaOH}$ added

I know this is a VERY small amount of $\ce{NaOH}$ but is there a way to account for the fact that this change in volume will affect the $pH$ of the solution?

Answer 1

RANT - I hate problems that don't use significant figures properly. So is the answer supposed to be +/- 1 ml, +/- 0.1 ml, or +/- 0.01 ml?

First your overall approach is right. There are three parts to solving the problem:

(1) Calculate pHs
(2) Calculate how many moles NaOH
(3) Calculate how many ml of 0.1 molar solution.

So the problem is that there are four phosphate species, $\ce{H3PO4, H2PO4^-, HPO4^{2-},}$ and $\ce{PO4^{3-}}$. The precision needed for the answer is somewhat of a mystery. So what simplifying assumptions are valid?

I'm just going to guess that the volume of NaOH should be accurate to 0.01 ml.

So let's look at the problem another way. I start off with 20.00 ml of 0.1000 molar solution of phosphoric acid and titrate with 0.01 ml at a time with a solution of 0.1000 molar sodium hydroxide. As every aliquot of NaOH is added the pH is measured. So:

(1) when I have added exactly 10.00 ml I have effectively made sodium dihydrogenphosphate. I add 2 to this pH to get the target pH.
(2) I look to see volume of NaOH that gives the target pH and subtract 10.00 ml.

SIDEBAR

Just looking at the problem I'm guessing a few ml. Let's say 5.00 ml. Measuring the NaOH to 0.01 would then mean 1 part in 500 precision (2 parts per thousand). Let's say that the initial $[\ce{H+}]$ is $3.47\times 10^{-5}$. 1 part in 347 is about right. But a pH of 4.46 only has two significant figures, that is 1 part in 46 which is too little precision for my guess of about 2 parts per thousand being needed. So the pH should be noted as 4.460.

Looking at Wikipedia we can see that for phosphoric acid that pKa1 = 2.148, pKa2 = 7.198, and pKa3 = 12.319.
(1) Given these pKa's we can't do better than 2 ppt.
(2) Only a small part of the $\ce{H2PO4^-}$ will ionize to $\ce{HPO4^{2-}}$ and $\ce{H^+}$. Thus the initial pH will be lower than 7.198 but quite above 2.148.
(3) Given (2) enough $\ce{H3PO4}$ might form to lower the pH when calculating to 2 ppt. This is right on the edge for 2 ppt precision...
(4) The species $\ce{PO4^{3-}}$ can safely be ignored in all the calculations. i.e. $\ce{[PO4^{3-}] << 0.1000 molar}$

1.0 Quick and dirty method...

To avoid solving a 3rd or 4th order polynomial...

1.1 Calculate pHs

The reaction of interest is:
$\ce{H2PO4- <=> H+ + HPO4^{2-}}\quad\quad 6.339 \times 10^{-8}$

We'll let $\ce{[H+] = [HPO4^{2-}]}$ and assume that $\ce{[H2PO4-]} = 0.1000$ so

$ 6.339 \times 10^{-8} = \dfrac{\ce{[H+]^2}}{0.1000}$
$ \ce{[H+]} = 7.96 \times 10^{-5}$ pH = 4.099

Therefore end pH = 4.099 + 2.000 = 6.099

Checks

$\ce{[HPO4^{2−}]}$ of $7.96\times10^{−5}$ is just less than $1\times10^{-4}$ so the assumption that $\ce{[HPO4^{2−}]}$ is not appreciable to $\ce{[H2PO4−]} \approx 0.1000$ is just ok.

We've also assumed that an insignificant amount of $\ce{H3PO4}$ will form which isn't true...
$\dfrac{\ce{[H3PO4]}}{\ce{[H2PO4^−]}} = \dfrac{\ce{[H^+]}}{7.11\times 10^{-3}} = 0.0112$

Da fix

We have 3 species of phosphate:
0.0008 $\times \ce{[H2PO4^{−}]}$ = $\ce{[HPO4^{2−}]}$
1.0000 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H2PO4^{−}]}$
0.0112 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H3PO4]}$
$\text{--------}$
1.0120 total $\ce{H2PO4^{−}}$

Normalizing $\ce{H2PO4^{−}}$ to 1.0000 total we have for the concentrations
0.0001 = $[\ce{HPO4^{2−}}]$
0.0988 = $[\ce{H2PO4^{−}}]$
0.001107 = $[\ce{H3PO4}]$

We rearrange the equilibrium for the reaction
$\ce{H3PO4 <=> H^+ + HPO4^{2−}}$
to solve for $\ce{[H^+]}$
$\ce{[H^+]} = \dfrac{\text{K}_{a1}\ce{[H3PO4]}}{\ce{[H2PO4^-]}} = 7.97 \times 10^{-5}$

so everything now checks...

1.2 Calculate how many moles NaOH

Given:
(1) $\ce{[H^+]} = 7.96\times 10^{-7}$
(2) 0.1 molar = $\ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$

Assume:
(1) $\ce{[H3PO4]} \approx 0$
(2) $\ce{[PO4^{3-}]} \approx 0$
(3) moles change in $\ce{H^+}(\text{aq})$ is insignificant

$6.339\times10^{−8} = \dfrac{\ce{[H+][HPO4^{2-}]}}{\ce{[H2PO4^-]}}$

or

$\ce{[HPO4^{2-}] =} \dfrac{6.339\times10^{−8}\ce{[H2PO4^-]}}{\ce{[H^+]}} = 0.07962\ce{[H2PO4^-]}$

We also know that

$0.1000 = \ce{[H2PO4^-] + [HPO4^{2-}] = 1.07962[H2PO4^-]}$
$\ce{[H2PO4^-] =0.09263}$
$\ce{[HPO4^{2-}] =0.00737}$

$\ce{H2PO4^- + OH^- -> HPO4^{2-}}$ so

Moles NaOH = $(0.020 \text{ liters})[0.00737] = 1.47\times10^{-4}$

1.3 Calculate how many ml of 0.1 molar solution.

liters 0.1 molar NaOH = $\dfrac{1.47\times10^{-4}}{0.1000} = 1.47\times10^{-3}$

ml 0.1 molar NaOH = 1.47

Answer 2

A different approach: I am using the following pka values for the phosphate system: pka1 = 2.15, pka2 = 7.21 and pka3 = 12.36.

The pH of the original solution of 20,00 ml of 0,100 M sodium dihydrogenphosphate can be calculated to be 4.68. You want to add x ml of 0,100 M NaOH to increase the pH of the solution to 6.68.

The wanted pH = 6.68 is close to pka2 = 7.21. Only two protolytes of the phosphate system will be present at significant amounts at pH = 6.68 i.e. H2PO4(-) and HPO4(2-).

The charge balance can be written as: [H3O(+)] + [Na(+)] = [OH(-)] + [H2PO4(-)] + 2[HPO4(2-)] + 3[PO4(3-)]. At pH = 6.68 the [H3O(+)] will be insignificant compared to the [Na(+)]. Similarily, [OH(-)] and [PO4(3-)] will be insignificant compared to [H2PO4(-)] and [HPO4(2-)]. These statements will be obvious when plotting a logarithmic diagram of the system. Accordingly, the charge balance can be re-written as: [Na(+)] = [H2PO4(-)] + 2[HPO4(2-)].

Please note, that the addition of NaOH will not change the concentration of the sodium ions, but will decrease the concentration of the phosphate protolytes. => [H2PO4(-)] + 2[HPO4(2-)] = 0,100 or, for our convenience Ca + 2Cb = 0,100.

The equilibrium equation gives: pH = pka2 + log Cb/Ca=> Ca = 3,3884* Cb. => 3,3884*Cb +2Cb = 0,100 => Cb = 0,018558 (M) => Ca = 0,100 - 2Cb = 0,100 - 2*0,018558 = 0,062883 (M) => Ctot = Ca + Cb = 0,081441 (M) => Ctot has decreased from 0,100 (M) to 0,081441 (M) by the addition of the NaOH.

Suppose we add x ml of 0,100 M NaOH in order to reach the target pH: At pH = 6.68 we get: Ctot = 20,00*0,100 /(20,00 + x) = 0,081441 => x = (2,00 - 20,00*0,081441)/0,081441= 4,557 ≈ 4,56 ml

Answer: You need to add 4,56 ml of 0,100 M NaOH.