Hach Micro Dist Manuel D'utilisation page 25

Note: Some distillates can contain compounds that have a chlorine demand. One minute after the addition of
chloramine T, measure for residual chlorine with KI-starch paper. If the test is negative, add an additional 0.060 mL
of chloramine T. After one minute, measure the sample again.
Note: More than 0.060 mL of chloramine T will prevent the color from developing with pyridine-pyrazolone.
Equivalency of releasing agents
Method Concentrations of releasing agents
Micro Dist 0.75 mL of 7.11 M H
Method H SO
2
Cyanide–1
7.11 M H
= 0.79 M H
This same solution is also 0.79 M in MgCl
0.79 M MgCl
= (0.088 M MgCl
USEPA Method 50 mL of 9 M (18 N) H
335.2 in H
2
0.9 M H
= 0.79 M H
20 mL of a 510 g MgCl
gives a concentration of:
2.5 M MgCl
= (0.088 M MgCl
USEPA Method 5 mL of 9 M (18 N) H
335.4 H SO
2
9 M H
= 0.79 M H
2 mL of a 510 g MgCl
a concentration of:
2.5 M MgCl
= (0.088 M MgCl
Equivalency of trapping solutions
Method
Micro Dist Method
Cyanide-1
USEPA Method 335.2
USEPA Method 335.4
Sulfide-2 Method
Comparison of the Micro Dist Sulfide-2 Method and the SW–846 Method 9030B,
Revision 2
Both methods use a trapping solution which is 0.043 M in zinc acetate and contains 1.60%
formaldehyde. The stock 37% formaldehyde solution contains 10–15% methanol. 2.0 mL of this
trapping solution is in each Sulfide–1 collector tube.
SO is added to 6.0 mL of sample. This gives a sample concentration in
2 4
:
4
SO × 0.75 mL ⁄(6.0 mL + 0.75 mL)
2 4
SO
2 4
⋅ 6 H
O × 0.75 mL ⁄(6.75 mL total)
2 2
⋅ 6 H
O)
2 2
SO is added to 500 mL of sample. This gives a sample concentration
2 4
SO :
4
SO × 50 mL ⁄(500 mL+50 mL+20 mL)
2 4
SO
2 4
· 6 H
2 2
⋅ 6 H
O × 20 mL ⁄(570 mL total)
2 2
⋅ 6 H
O)
2 2
SO is added to 50 mL of sample. This gives a sample concentration in
2 4
:
4
SO × 5 mL ⁄(50 mL+5 mL+2 mL)
2 4
SO
2 4
· 6 H O/L (2.5 M) solution is also added to 50 mL of sample. This gives
2 2
⋅ 6 H
O × 2 mL ⁄(57 mL total)
2 2
⋅ 6 H
O)
2 2
Concentrations of trapping solutions
In the Micro Dist Method, 1.5 mL of 1.0 M NaOH becomes 6.0 mL solution after
distillation and dilution to the mark:
1.0 M NaOH x (1.5 mL / 6.0 mL) = 0.25 M NaOH
In the macro distillation, 50 mL of 1.25 M NaOH becomes 250 mL solution after
distillation and dilution to the mark:
1.25 M NaOH x (50 mL / 250 mL) = 0.25 M NaOH
In the macro distillation, 50 mL of 0.25 M NaOH stays undiluted.
· 6 H O. This gives a concentration of:
2 2
O/L (2.5 M) solution is also added to 500 mL of sample. This
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