# Last edited on 2019-10-14 09:36:52 by jstolfi

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Goal

In this experiment we will crystallize the water-soluble salt sodium hydogen sulfate Na+ HSO4-, also called "monosodium sulfate".

Introduction

When only one of the hydrogen ions from sulfuric acid is removed, what is left behind is the ion HSO4-, with a single negative charge, called "hydrogensulfate" or "bisulfate".  Sodium bisulfate is a crystalline solid that gives strongly acidic solutions in water.

Sodium hydrogen sulfate can occur either as anhydrous (melting point 315 C; molar mass 120.06 g; CAS 7681-38-1), or monohydrate  (melting point 58.5 C; molar mass 138.08 g; CAS 10034-88-5).   Thus crystallizing it at 60 C or higher should give the anhydrous salt, and below that temperature we should get the monohydrate.

Solubility in water

The solubility of the salt in water increases sharply with temperature.  The solubility Sm of the monohydrate is infinite at 58.5 C.  The solubility Sa of the anhydrous salt remains finite.  In g per 100 g (100 mL) of water:

  temp (C) ! Sa   ! source !  Sm ! source   
  0        |      |        |  50 | [1]      
  25       | 28.5 | [2]    |  67 | [3]  
  58.5     |      |        |     |
  100      | 100  | [2]    | 100 | [4]         
  
  [1] https://www.ams.usda.gov/sites/default/files/media/Sodium%20Bi%20report%202015.pdf
  
  [2] https://en.wikipedia.org/wiki/Sodium_bisulfate

  [3] https://www.fishersci.com/shop/products/sodium-bisulfate-monohydrate-99-analysis-acros-organics-3/AC125005000
  
  [4] https://www.kimyaborsasi.com.tr/en/s/sodium-bisulfate-126.html

Refs [2] and [4] says 100 g / 100 mL at 100 C, but it is not clear which version is that.

To convert solubility from anhydrous to monohydrate:

  Suppose Sa grams of NHSO4 dissolve in 100 g of H2O.
  
  Sa grams of NaHSO4 is N = Sa/Ma moles of NaHSO4 where Ma = 120.06.
  
  That plus N moles of H2O will form N moles of the monohydrate.
  
  N moles of H2O is G = N x Mw grams of H2O where Mw = 18.02.
  
  That will be dissolved in H = 100 - G grams of water.
  
  N moles of the monohydrate will be M = N x Mm grams of monohydrate wher Mm = 138.08.
  
  Thus Sm = 100 x (M / H) grams of the monohydrate will dissolve in 100 g of water.
  
Putting that together:

  Sm = 100 x Sa x Mm /(100 x Ma - Sa x Mw)
  
  Sm x (100 x Ma - Sa x Mw) = 100 x Sa x Mm
  
  100 x Sm x Ma - Sm x Sa x Mw = 100 x Sa x Mm
  
  100 x Sm x Ma = Sa x ( 100 x Mm + Sm x Mw ) 
  
  Sa = 100 x Sm x Ma / ( 100 x Mm + Sm x Mw )
  
At the melting point of the monohydrate (58.5 C), if the result was just one liquid phase, the solubility Sm would be infinite, and Sa would be 100 x Ma / Mw = 666 g in 100 mL.  However, the sources say that either Sa or Sm is 100 at 100 C.  Thus we must conclude that the melting is incongruous.  That is, at 58.5 C the solidNaHSO4.H2O breaks down into two phases: a saturated solution, and some solid anhydrous NaHSO5.

Other properties

It is decomposed by alcohol into sodium sulfate and H2SO4. [3]

Goals and methods

The reagent

Reagent amount

Water amount

Preparing the solution

Crystallization