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?

  • $\begingroup$ Those calculations would be correct for a strong acid, not a weak one or a polyprotic one. // Quibble point -- You don't "add" a concentration of 3.44*10^-6M OH rather you change the concentration by that much. $\endgroup$
    – MaxW
    Feb 13, 2017 at 16:45
  • $\begingroup$ The problem with your answer isn't really just that phosphoric acid is polyprotic. Your calculations would work fine for sodium hydrogen sulfate. The problem is about the pKa's of the polyprotic acid. $\endgroup$
    – MaxW
    Feb 13, 2017 at 17:56

2 Answers 2


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.


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


$\ce{[HPO4^{2−}]}$ 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 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H3PO4]}$
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

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

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

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


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

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


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.


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