Sciencemadness Discussion Board - Sodium Bisulfate?

Sodium Bisulfate?

MttLsp - 17-7-2010 at 18:09

So here's what I did:
I added 25 mL of concentrated (I assumed 18 Molar when figuring out the stoichiometric quantities) H2SO4 into a jar filled halfway with water, which was in an ice bath. Then I slowly added 225 mL of 0.5 Molar NaOH.

The amounts I used are based on the equation:
H2SO4 + NaOH --> NaHSO4 + H2O

When the reaction was complete I tested the pH of the solution, which was acidic. Then I took a small sample of the solution and put in an evaporation dish and heated it. Steam came off of the top and I assumed it was water, which was correct because the vapor had no affect on blue litmus. After all of the water was cooked out there was a thick bubbling mass. Here's the part I do not understand:

I dipped a piece of newspaper in the thick shit, and it turned black. The fact that whatever substance was in the dish dehydrated the paper made me think it was H2SO4 and the experiment failed. But as the blob (molten NaHSO4?) cooled, it solidified. Could this be crystalline H2SO4?

I don't understand what I did wrong...

per.y.ohlin - 17-7-2010 at 18:48

The reaction is 1mol H2SO4 to 1mol NaOH.

H2SO4:
25mL=0.025L
0.025L*18M=0.45mol

NaOH:
225mL=0.225L
0.225L*0.5M=0.1125mol

0.45M/0.1125M=4

You have four moles of H2SO4 for one mol of NaOH. At the end of the reaction, you would have a mole ratio of 3:1:1 H2SO4:NaHSO4:H2O. This might solidify in an amorphous blob, but I doubt it would form a crystal structure.

Given the concentration of sulfuric acid, I would expect this to be a very violent reaction. The dilution by water is vigorous enough, when water is added to the acid.

It is usually impossible to get an exact stoichiometric mixture, given the limits of measuring equipment. I don't know what your intended use is, but I would expect an impurity of Na2SO4 would have little effect, whereas excess H2SO4 could be hazardous. For this reason, I suggest adding a slight excess of NaOH.

Melgar - 17-7-2010 at 18:57

If the goal is preparing sodium bisulfate, mix sodium bicarbonate or sodium carbonate with the sulfuric acid instead. It's a lot cheaper and won't react as exothermically (although there will be a lot of bubbling). Just a thought.

MttLsp - 17-7-2010 at 20:34

Oh OK. I'm new to chemistry so my stoichiometry is below average. So is this right?

M= m/L

H2SO4
18M X .001L = .018 m/mL
Then I need to set the # of moles of H2SO4 equal to the NaOH...
.018m/mL = .5M X (xL)
.018/.5 = .036 L
36 mL

For every mL of Sulfuric Acid I would need 36 mL of Lye, and therefore for 25mL of H2SO4 I would need 900 mL of NaOH.

Is that correct?

[Edited on 18-7-2010 by MttLsp]

[Edited on 18-7-2010 by MttLsp]

[Edited on 18-7-2010 by MttLsp]

MttLsp - 17-7-2010 at 20:52

And yes the reaction is EXTREMELY violent. I learned that the hard way. But if I add small amounts very slowly and carefully and keep the temperature down with and ice bath theres no problems.

blogfast25 - 18-7-2010 at 09:01

I haven't checked your reviewed stoichiometry, so no comment on that.

Mixing strong acids and strong alkali always generates much heat, so much so the liquid may end up boiling violently. Cooling and very gradual mixing are order of the day.

The heat basically comes from making water: H3O+ + OH- ---> H2O, aka the 'heat of neutralisation', about 57 kJ/mol of water formed.

From it you can estimate what will be the temperature rise of a given mixture of dissolved alkali and dissolved acid.

Calculate from the stoichoi how many moles of water will be formed, say n. The heat liberated on neutralisation is thus Q = n x 57 kJ.

Assume the mixture was a total of m g(ram) of water. Ignoring the heat soaked up by the receptacle and the formed salt, then Q = 4.2 x m x ΔT, with 4.2 J / (C.g) the heat capacity of water and ΔT the increase in temperature of the mix.

Thus Q = n x 57 kJ = 4.2 J/ (C.g) x m x ΔT

and ΔT ≈13,600 x n/m

For instance, neutralising 5 mol of HCl with 5 mol of NaOH in a total volume of 1,000 mL of water causes a temperature rise of about 68 C.

entropy51 - 18-7-2010 at 09:26

Quote: Originally posted by Melgar

If the goal is preparing sodium bisulfate, mix sodium bicarbonate or sodium carbonate with the sulfuric acid instead. It's a lot cheaper and won't react as exothermically (although there will be a lot of bubbling). Just a thought.

If I'm not mistaken, the "heat of neutralization" per mole is the same, no matter which acid and base is used. I use Na2CO3 to neutralize waste acid, and it can easily foam out of the container, so best to set the flask in a bucket. And I usually need to add ice cubes or the stuff will boil.

MttLsp - 18-7-2010 at 09:50

Thanks alot I think I understand what I did wrong (First of all I made a 2 M solution of NaOH a few weeks ago and labeled it 0.5 M. So I needed 225 mL NaOH after all.

If I dilute the H2SO4 with water, will this affect the amount of base needed to neutralize it. I don't see why it would...those moles of acid don't just jump out of the flask and walk away right?

blogfast25 - 18-7-2010 at 11:30

Quote: Originally posted by entropy51

Yes, that's roughly correct.

In the case of the neutralisation of a carbonate, the equilibria:

CO3(2-) + H2O <--->HCO3- + OH-

and HCO3- + H2O <---> H2CO3 + OH-

H2CO3 <---> H2O + CO2 ↑

are driven 100 % to the right because of the neutralisation reaction: H3O+ + OH- ---> 2 H2O

Each mol of Na2CO3 + 2 HCl ---> 2 NaCl + H2O + CO2 ↑ produces one mol of water and thus generates about the same amount of heat as NaOH + HCl ---> NaCl + H2O`

[Edited on 18-7-2010 by blogfast25]

bbartlog - 18-7-2010 at 12:21

Na2CO3 + HCl is endothermic, presumably because of the entropy change that comes from one of the products being a gas. So practically speaking, no, the (observed, net) heat of neutralization is not the same - sodium carbonate or bicarbonate will leave you with a much cooler solution. However in my experience the foaming is obnoxious and I would usually sooner deal with the heat.

blogfast25 - 18-7-2010 at 12:38

Quote: Originally posted by bbartlog

Na2CO3 + HCl is endothermic

I'm skeptical about that. Do you have proof? That some of the heat is carried off by the escaping CO2 will run an actual reaction slightly cooler is true.

Gas escaping would affect ΔG, not ΔH, off the top of my head.

I'm inclined to put it to the test...

bbartlog - 18-7-2010 at 13:04

I know from personal experience that NaHCO3 + HCl is endothermic (ice cold beakers...), and it's a popular demonstration of such a reaction (see for example: http://www.chem.umn.edu/services/lecturedemo/info/Endothermi...).
I am less sure of Na2CO3. Some claim it is (see: http://www.physicsforums.com/showthread.php?t=165216) but the overall substantiation-by-google seems thin on the ground. Obviously since the second *part* of the HCl reaction with Na2CO3 is the NaHCO3 reaction, that part is endothermic, but I am uncertain of the net. Still it would be quite a bit less exothermic than NaOH+HCl even if it's not an endotherm.

blogfast25 - 18-7-2010 at 13:09

I will see if I can figure this out from NIST date of HoF's and entropies of the reaction species and reaction.

Bar that I might run a simple test.

entropy51 - 18-7-2010 at 15:18

Quote: Originally posted by bbartlog

Na2CO3 + HCl is endothermic, presumably because of the entropy change that comes from one of the products being a gas.

If you say so, but H2SO4 + Na2CO3 is quite exothermic as I related above. And the gaseous product is CO2 in either case, so I am less than convinced.

bbartlog - 18-7-2010 at 15:52

I don't doubt that. But I do doubt, practically speaking, that '...the "heat of neutralization" per mole is the same' regardless of the acid and base involved (in fact the case of NaHCO3 and HCl makes it clear that that just can't be right). The heat from H+ and OH- combining to make water may be the same, but the heat of formation from the other side (Na+ and Cl- versus Na+ and SO4--, or whatever it is) is going to be different for different compounds, even before the entropy for gaseous products comes into play. Looking at the heat of formation for sodium chloride (-411kJ/mole) versus sodium sulfate (-1387kJ/mole, or cut that in half to get the figure per mole of sodium atoms...) I see a pretty substantial difference.

entropy51 - 18-7-2010 at 16:34

Quote: Originally posted by bbartlog

I don't doubt that. But I do doubt, practically speaking, that '...the "heat of neutralization" per mole is the same' regardless of the acid and base involved (in fact the case of NaHCO3 and HCl makes it clear that that just can't be right). The heat from H+ and OH- combining to make water may be the same, but the heat of formation from the other side (Na+ and Cl- versus Na+ and SO4--, or whatever it is) is going to be different for different compounds, even before the entropy for gaseous products comes into play. Looking at the heat of formation for sodium chloride (-411kJ/mole) versus sodium sulfate (-1387kJ/mole, or cut that in half to get the figure per mole of sodium atoms...) I see a pretty substantial difference.

Strictly speaking, I think it just applies to strong acids and bases, and NaHCO3 is neither. But I thought we were speaking of H2SO4 and Na2CO3 vs. NaOH, in which case, I think it is 13700 cal/mol in both cases. But what do I know?

[Edited on 19-7-2010 by entropy51]

Formatik - 18-7-2010 at 22:21

Quote: Originally posted by MttLsp

I dipped a piece of newspaper in the thick shit, and it turned black. The fact that whatever substance was in the dish dehydrated the paper made me think it was H2SO4 and the experiment failed. But as the blob (molten NaHSO4?) cooled, it solidified. Could this be crystalline H2SO4?

I know a mixture of saturated NaHSO4 in sulfuric acid can be liquid around room temperature at least for some time, but as it's cooled, the NaHSO4 crystallizes.

blogfast25 - 19-7-2010 at 04:57

I didn't know the reaction of NaHCO3 with HCl is endothermic but in the conditions described in bbartblog's interesting link, the HCl is added onto DRY bicar.

But I'm running a test tomorrow: 1 mol NaOH (+ 1 mol HCl) back to back against 1 mol Na2CO3 (+ 2 mol HCl), using a small excess of HCl. I'm drying some Na2CO3 as I'm writing this. May the highest heat of neutralisation win!

blogfast25 - 20-7-2010 at 06:42

Talk about an experiment that kind of went wrong and succeeded anyway!

So I had prepared the day before:

1) 0.5 mol Na2CO3 (washing soda) in 250 mL of water and 1.2 mol HCl in 140 mL. The carbonate was a bit mucky, this washing soda being suitable for washing but not much else! Next time will use decarbonated bicar.

2) 0.5 mol technical NaOH in 250 mL of water and 0.6 mol of HCl in 140 mL.

The NaOH was expected to raise about 17.4 C upon addition of the HCl according to the equation in the post above.

All were allowed to acclimatise overnight next to each other and well stoppered.

First up, the soda. I added the acid over a period of about 1'30". Starting temp. was 22.3 C, end temperature 24.9 C, so a considerable deficit in temperature rise.

Next the NaOH. Adding the HCl over about 1'30" the solution went from 22.5 C to 31.7, or ΔT = 9.2 C, well below expectations! I checked the pH of the solution... 13.4! So there's something up with my HCl and I don't know what. Both solutions were made for the same technical 32 w% HCl! My calcs were correct.

Not to worry, I then added small amounts of the "32 %" HCl until pH < 7. End temperature was 37.5 C and measured end volume 470 mL. Expected ΔT = 13600 x (0.5/470) = 14.5 C, actual ΔT = 15.0 C, so good accordance with theory there.

Checking also the pH of the carbonate solution it was about 9.5, so barely half neutralised! Adding strong HCl the bubbling continued briskly but temperature hardly budged: end temperature about 25.5 C and ΔT = 3.2 C!

So despite the HCl mishap it's safe to conclude that Na2CO3 barely heats up on neutralising, presumably because any heat of neutralisation is offset by the endotherm of the bicarbonate neutralisation...

blogfast25 - 20-7-2010 at 06:55

Quote: Originally posted by bbartlog

Unfortunately this is completely incorrect, bbartblog: the heats of formation of for example NaCl are obtained by reacting sodium metal with chlorine gas, stoichiometrically and measuring the amount of enthalpy with a calorimeter (most of that enthalpy is the so-called 'lattice energy': the huge amount of energy freed when the positively charged and negatively charged ions settle in a lattice: it's basically Coulombic energy).

In solution, no NaCl is formed: the Na is already there as Na+ ions and the Cl as Cl- ions. No lattice is formed and no lattice energy released.

Assuming the NaOH and HCl were fully dissociated at the time of mixing (this is 99.9 % the case), the only actual reaction that occurs is H3O+ + OH- ---> 2 H2O and the only heat generated the neutralisation heat.

This is certainly true for strong acids/bases but also almost completely true for weak acids/bases, except those where significant entropy changes take place.

Continued below.

bbartlog - 20-7-2010 at 07:39

If you're right, then neutralizing H2SO4 (or any aqueous acid) with NaHCO3 should be endothermic? I get the sense that the heats of solution come in to play here somehow, but my grasp of the overall thermodynamics is clearly weak. I'll have to read up on it some more. Anyway thanks for running the NaOH/Na2CO3 experiment.

blogfast25 - 20-7-2010 at 07:41

Continued from above.

Take a hypothetical weak acid, HA, with a pKa = 4. Assume a solution of Ca = 1 M nominal concentration.

HA + H2O <---> H3O+ + A-

And Ka = [H3O+]x[A-]/[HA]

A little theory shows that the pH of the solution is approx. (for dilute solutions):

pH = 1/2 (pKa + pCa) = 2.0

Theory also shows that [A-] ≈ [H3O+], or 10^-2 mol/l. The acid is thus only dissociated only by about 2 %!

Now assume we neutralise the acid with an equivalent amount of solid NaOH (thereby keeping the volume more or less constant and I'll neglect here the heat of solvation of NaOH for the sake of simplicity). Now we have a solution of sodium acetate, nominal concentration of Ca = 1 M.

Theory shows that for a weak base (the conjugated base of a weak acid) the pOH is:

pOH = 1/2 ( pKb + pCa)

With pKa + pKb = pKw and Kw = 10^-14, thus pKb = 10 and

pOH = 5 or [H3O+] = 10^-9 mol/l

It can also be shown that [HA] ≈ [OH-], so basically after neutralisation the residual concentration of acid in the solution is 10^-5 and the amount neutralised is 1 mol/l - 10^-5 mol/l ≈1 mol/l. No significant difference with a strong acid there.

The neutralisation proceeds of course by the equilibrium H3O+ + OH- <---> 2H2O with a Kw = 10^-14 pushing the equilibrium HA + H2O ---> H3O+ + A- to the right by constantly reacting away the H3O+ (and generating heat in the process).

There is a small snake in the grass:

The reaction HA + H2O ---> H3O+ + A- is probably slightly endothermic because the Gibbs Free energy for chemical equilibria is given by:

ΔG = - RT lnK and with Ka so small ΔG becomes positive (which is why it's a weak acid in the first place).

Adjust Ka to the true constant by dividing it by [H2O] ≈ 55.5 mol/l, the equilibrium constant becomes 1.8 10^-6 and ΔG = - 8.3 x 298 x ln(1.8 10^-6) = + 32.7 kJ/mol.

Assume no entropic effects, then ΔG ≈ ΔH and that would be a significant amount of heat (compared to the neutralisation heat of - 57 kJ/mol of water).

Rummaging around in my lab dungeon, I found an old but intact Dewar flask and in the coming days I'll precisely determine the heat released by neutralising a known amount of acetic acid (well, distilled vinegar!) with NaOH (in solution)... To be continued!

blogfast25 - 20-7-2010 at 07:48

Quote: Originally posted by bbartlog

If you're right, then neutralizing H2SO4 (or any aqueous acid) with NaHCO3 should be endothermic? I get the sense that the heats of solution come in to play here somehow, but my grasp of the overall thermodynamics is clearly weak. I'll have to read up on it some more. Anyway thanks for running the NaOH/Na2CO3 experiment.

'Yes' to the first question.

As regards the heats of dissolution or 'solvation heats', they've already been accounted for at the point of dissolution: that heat has then been drained away by cooling after preparation of the stocks. No more heats of solvation play any part if relatively diluted solutions are used.

Concentrated H2SO4 + NaHCO3 is a different matter: the water released will cause solvation energy to be released by the H2SO4.

[Edited on 20-7-2010 by blogfast25]

blogfast25 - 21-7-2010 at 07:09

Heat of Neutralisation of Acetic Acid

For the heat of reaction of the neutralisation of vinegar (HAc solution) the caloric value k of the 1 l Dewar flask was first independently determined by 4 experiments, to be about 86 J/K. This is essentially the heat the calorimeter soaks up when its temperature is raised by 1 Kelvin.

For acetic acid solution was used distilled vinegar (Tesco value!) at nominal acidity of 4.6 %, that's volume percent. I didn't verify the actual strength but previous experiments with this grade showed that it does what it says on the bottle, in terms of acid content. 4.6 V% of HAc is about Ca = 0.80 mol/l, reworked from density and MW of HAc.

800 ml of this solution at RT were loaded into the calorimeter, stoppered and allowed to stabilise by gentle swirling for 2 mins and the temperature found to be 21.4 C (it's really hot here for the moment).

Then the required equivalent of NaOH, dissolved in 150 ml iced water were added: the solution was still quite warm at 38.5 C.

The neutralised and stoppered solution was then allowed to achieve thermal equilibrium for 2 mins, gently swirling and the final temperature was 32.4 C.

Additional observations:

pH of the vinegar was 2.9 where theory predicts 2.4 based on pKa = 4.76 (Wiki).

pH of the neutralised solution was 12.8, significantly higher than the theoretical 9.3. Remember that we're in the equivalent zone were even small amounts of acid or alkali tip the pH balance strongly. Obviously there was a small excess of alkali but this should not appreciably affect the results and at least we know for sure all acid has been neutralised.

The calculation of the heat of the reaction ΔHr is as follows.

Initial enthalpy of the flask + acid:

H1 = (4.1813 x 800 + 86) x 21.4 = 73.4 kJ

Enthalpy added when adding NaOH:

H2 = ΔH + 4.1813 x 150 x 38.5 = ΔH + 24.1 kJ

Measured enthalpy content of the flask + contents:

H3 = [4.1813 x (800 + 150) + 86] x 32.5 = 132 kJ

Since as H3 = H1 + H2

73.4 + 21.4 + ΔH = 132

Then ΔH = 34.5 kJ but this needs a sign inversion because it's an exotherm, so ΔH = - 34.5 kJ. During reaction 0.8 mol/l x 0.8 l = 0.64 mol were neutralised so that ΔHr = - 34.5 / 0.64 = -54.5 kJ/mol which is rather in line with the standard enthalpy of neutralisation ΔHn = - 57.3 kJ.

One thing can be inferred with reasonable certainty: the neutralisation of HAc seems to generate about the same heat than the neutralisation of an equivalent strong acid.

I think I'll repeat this experiment with another monoprotonic weak acid: NH4Cl...