After following the suggestions of my RG colleagues, a sign that you are on the right track is that the pH you will find out will be above 7 because:

HCOOH & NaOH are completely soluble in water & they will undergo a neutralization reaction: HCOOH + NaOH → HCOONa + H2O .

The product is an aqueous solution of sodium methanoate "which is a basic salt".The use of a pH meter for this solution will confirm that the pH is > 7.

You have to convert mole fractions to concentrations (mol/L), but after that the "recipe" is the Hendersen-Hasselbach equation:

www4.ncsu.edu/~franzen/public_html/CH201/lecture/Lecture_15.pdf

You need to solve simultaneously: 

1. Equilibrium reaction for formic acid speciation

2. Dissocition of water

3. Mass balance of formic species

4. Charge balance.

Send me a message and I will write down the equations

After following the suggestions of my RG colleagues, a sign that you are on the right track is that the pH you will find out will be above 7 because:

HCOOH & NaOH are completely soluble in water & they will undergo a neutralization reaction: HCOOH + NaOH → HCOONa + H2O .

The product is an aqueous solution of sodium methanoate "which is a basic salt".The use of a pH meter for this solution will confirm that the pH is > 7.

Due to same concentration of HCOOH and NaOH, acid base neutralization reaction will take place  but exact pH can be measured using pH meter by making actual solution in chemistry lab.

Hi Ganesh,

I aggree with Nizar - NaOH is a stronger base than HCOOH an acid (look at the corresponding pK values). This means that the final solution will be somehow basic since NaOH will "overwhelm" HCOOH.

The concentration of water in water is equal to 55,5556 mol/L (1 L = 1000 g and 1 mol water = 18 g/mol), if you have a mole fraction of 0.0002 HCOONa (which will be produced after mixing, see Nizar's post) it should be an aqueous solution with 0,01111 mol/L HCOONa. This can be easily calculated with a formula for soft bases, you just need the pKB value of HCOONa (something like 10.25). A 0.1 mol/l solution of sodium formiate is pH = 8.375 (~ 8.4). The detailed calculation for this can be found on the attached link (it's a German forum, but the calculation is universal - see post from mexicolaLIGHT, 08 Feb 2009 - 18:41:23). Your pH should be slightly greater, I guess around 8.5-8.6.

Best regards,

Marcus

http://www.uni-protokolle.de/foren/viewt/222831,0.html

About calculating the pH of formic acid / sodium formate buffer solutions without using the Henderson-Hasselbalch equation:

I ― diluted solutions

We may consider that the (weak) formic acid dissociates accordingly: 

HCOOH + H2O ⇄ H3O+ + HCOO-

With the equilibrium constant given by: Ka = [H3O+]·[HCOO-] / [HCOOH]

Some sodium formate ― which can be considered a strong electrolyte ― would be also present in the solution, where it should dissociate nearly completely:

HCOONa → Na+ + HCOO-

We also accept that the contribution of the water autodissociation equilibrium for [H3O+] can be neglected.

From the acid and salt mass balances, with molar concentrations given by Ca and Cs, respectively, we have:

[HCOOH] = Ca - [H3O+] > 0

[HCOO-] = Cs + [H3O+]

After substitution at the equilibrium constant:

Ka = [H3O+]·{Cs + [H3O+]}/{Ca - [H3O+]}

What can be solved in terms of [H3O+], as:

[H3O+] = {- (Cs+Ka) + √[(Cs+Ka)2 + 4Ka·Ca]}/2 

Where, neglecting activity (and related ionic strength) effects:

pH = -log10[H3O+] 

where we may take Ka = 1.78·10–4 or pKa = 3.75 (25 ºC; units omitted).

Note that the acid is not to be considered utterly diluted, since it was inherently accepted that: Ca > [H3O+] ≥ [H3O+]º, where [H3O+]º refers to a pure salt solution (Ca= 0). Indeed, that should be the case for Ca >> 10-7 M. Nonetheless, the above equation is still valid for a pure acid solution (Cs = 0), as it will be further discussed at my next post. 

II ― concentrated solutions

II.a) Although our previously derived equation also holds for concentrated solutions, we could alternatively use the simpler and well-known Henderson-Hasselbalch equation, which is more coarsely approximate, generally speaking. Yet, it can be considered reliable for about Ca > 100·Ka ≈ 0.02 M.

II.b) Being Cs > 100·Ka ≈ 0.02 M, our previous equation could simplify:

[H3O+] = {- 1 + √[1 + 4Ka·Ca/Cs2]}·Cs/2

II.c) If also 4Ka·Ca > 100·Cs2 > 106·Ka2; i.e. for Ca > (1/4)106·Ka; we may simplify even further: 

[H3O+] = - Cs/2 + √(Ka·Ca)

Here, it can be noticed that we should also have also [H3O+] ≥ [H3O+]º, where [H3O+]º refers to a pure salt solution (Ca= 0), what limits the validity of last approximation to 100Ka < Cs  ≤ 2{√(Ka·Ca) - [H3O+]º} < 2√(Ka·Ca) < 1000Ca.

About calculating the pH of formic acid aqueous solutions:

I ― diluted solutions

It follows from the equation derived at my previous post (written in bold) that for aqueous solution of just the formic acid (i.e.: Cs = 0) at molar concentration Ca:

[H3O+] = Ka·[√(Ka + 4Ca)/√Ka  - 1]/2

pH = pKa - log10{[√(Ka + 4Ca)/√Ka  - 1]/2}

where we may take Ka = 1.78·10–4 or pKa = 3.75 (25 ºC; units omitted).

For instance; if Ca = 0.001 M, we obtain pH = 3.47. 

II ― concentrated solutions

For concentrated solutions simpler equations could be alternatively used, even considering that above equations still hold.

II.a) For about Ca > (100/4)·Ka = 4.4·10–3 M, we approximately have:

[H3O+] = Ka·[√(Ca/Ka)  - 1/2]

pH = pKa - log10[√(Ca/Ka)  - 1/2]

II.b) For about Ca > (104/4)·Ka = 0.44, we may further simplify:

[H3O+] = √(Ca·Ka)

pH = pKa - (1/2)log10(Ca/Ka)

For example; with Ca = 1 M, we would obtain; from either the last equation, or from the second equation of this post: pH = 1.88.

About calculating the pH of formic acid aqueous solutions:

This post completes my previous post. 

III ― Intermediate concentration for Ca = Ka

It seems interesting to note that the first couple of equations from my previous post could then rewrite itself as:

[H3O+] = Ka·(√5  - 1)/2 = Ka / Φ

pH = pKa + log10(Φ)

where Φ = 1.618 is the golden number.

IV ― Intermediate concentration for Ca = 2Ka  where pH = pKa

It can be noticed that for pH = pKa, the second equation from my previous post rewrites as √(Ka + 4Ca) = 3√Ka; showing that we would then have Ca = 2Ka. 

About calculating the pH of formic acid / sodium formate buffer solutions without using the Henderson-Hasselbalch equation:

III. On the validity of the Henderson-Hasselbalch equation

Next equation was derived at my first post to this query:

[H3O+] = {- (Cs+Ka) + √[(Cs+Ka)2 + 4Ka·Ca]}/2 

As long as Cs > 100Ka, we approximately have Cs+Ka ≈ Cs, and we may write:

2[H3O+] = {√[Cs2 + 4Ka·Ca]  - Cs}·{√[Cs2 + 4Ka·Ca]  + Cs}/{√[Cs2 + 4Ka·Ca]  + Cs}

or:

[H3O+] = 2Ka·(Ca/Cs) / {1 + √[1 + 4Ka·Ca/Cs2]}

If we have also Ca > Ka, then we shall have (2Ka/Cs)2 < 4Ka·Ca/Cs2 < 0.01, i.e. 2Ka/Cs 20Ka, what is ensured by our previous assumption. Then:  

[H3O+] = Ka·(Ca/Cs)

and:

pH = pKa + log10(Cs/Ca)

This is the Henderson-Hasselbalch equation, thus shown to constitute a reasonable approach for both Ca > Ka and Cs > 100Ka. For Ca ~ Cs, we may alternatively express this as: Ca > 100Ka.

For formic acid, Ka = 1.78·10–4; showing that we should approximately have Cs > 0.02 M to use the Henderson-Hasselbalch equation with satisfactory accuracy.

Similar pH calculation can use for mixture of formic acid/sodium hydroxide.

Ganesh Yadhav

Dear Ganesh Yadhav:

I happened to find a similar case during my research.

I solved it by applying first principle models.

I will explain you how in a concise way:

The interaction you will have in your system will be:

HCOOH dissociation (ka, weak acid)

NaOH complete dissolution (starting from solid poured into water) (full disolution, strong base)

NaOH / HCOOH neutralization (water and salt formation)

Salt SL equilibrium (ksp)

Water dissociation (kw)

Interaction with the gas phase? Open/closed system? (inert load, water vapor pressure...).

+ Residual species

These waste water residual species are tricky, because they seem to be vaguely defined, but do affect a precise calculation.

In case you DID NOT HAVE these residual species, you will now need to determine the displacement of the dissociation, involving the whole aqueous system.

What I did was to find the displacements that satisfied Aspen's molar fraction equilibrium constants.

I could read them from Aspen: by using e-NRTL one can see and expression of kx as ln(kx)=A+B/T+C*ln(T) T[K]

Things to keep in mind are:

• As the other researchers have stated, pKa depends on initial loading. This is why pKa is given for an initial weak acid loading. Thus, the final pH also depends on the initial loading.
• The relationship between molar concentration and molar fraction depend on the total concentration. This is why it would be more precise to determine the residual species.
• Aspen offers these data, but it is licensed. Maybe it is available on papers, but I used that for academic purposes by then.
• I am working on a project to model this, so if you are still seeking for more concrete advice, I will send you the link to my work once it is finished.
• The mathematical system is highly interrelated and requires implicit solvers. I suggest you use Gekko or APMonitor (free), because there are loads of tutorials and other material. https://apmonitor.com/

Hope this helps.

Best regards,

Enrique Garcia-Franco