Preparation of a saturated (6M) HCl in IPA solution.

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# Constants
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d(85% H3PO4) = 1.69 g/ml
M(H3PO4) = 97.994 g/mol

d(IPA) = 0.786 g/ml
M(IPA) = 60.1 g/mol

M(HCl) = 36.46 g/mol
M(NaCl) = 58.44 g/mol
M(NaOH) = 39.997 g/mol

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# Writeup
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n = 3 mol
6M HCl-IPA 500ml
175g NaCl
345g (205ml) 85% H3PO4

A slight excess of NaCl and H3PO4 is recommended even though the list above and
the following writeup uses stoichiometric amounts.

 1. Load 175g NaCl and 345g (205ml) 85% H3PO4 into a 1L RBF.
 2. Load 500mL of IPA into a 600mL beaker.
 3. Connect the RBF with a gas outlet adapter and a PVC hose (or some other
    HCl(g)-resistant hose) through an inverted funnel trap to the IPA beaker.
 4. Heat the RBF gently until we see a steady stream of bubbles. For heating,
    a heating mantle *without* stirring is recommended.
 5. Continue heating until all the acid dries up in the RBF or the concentration
    of HCl no longer increases in the receiving HCl-IPA beaker.
    CRITICAL: Readjust the height of the inverted funnel trap as necessary.
 6. Verify that the concentration of the resulting HCl-IPA is 6M (saturated).

NOTE: 500mL IPA expands to 550mL 6M HCl-IPA after gassing with HCl(g)!
      I should fix this somehow in the text above, but I'm too lazy.

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# Verification
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m(NaOH) = 0.24 g
n(NaOH) = 0.24 g / 39.997 g/mol = 0.006000450033752531 mol
V(6M IPA-HCl) = n(NaOH) / 6M = 1.0000750056254217 mL = ~1 mL

The basic idea is that we can use 1 mL of (saturated) 6M IPA-HCl to neutralize
0.24 grams of NaOH.

 1. Dissolve 0.24 g of NaOH into a small amount of distilled water.
 2. Neutralize by adding 1 mL of IPA-HCl.
 3. Measure pH (near to 7 if the IPA-HCl is 6M, i.e., saturated).