34
Cyanide–1
8.7.1 Pyridine-barbituric acid method: Add 0.24 mL of chloramine
T (7.12) and mix. After one to two minutes, add 0.60 mL of
pyridine-barbituric acid solution (7.13.1) and mix. [Now add 3.36 mL
of deionized water] and mix again. Allow eight minutes for color
development, then read absorbance at 578 nm in a 1 cm cell within
15 minutes.
8.7.2 Pyridine-pyrazolone method: Add 0.060 mL of chloramine T
(7.12) and mix. After one to two minutes add 0.60 mL of
pyridine-pyrazolone solution (7.13.1) and mix. [Now add 3.54 mL
deionized water] and mix again. After 40 minutes read absorbance
at 620 nm in a 1 cm cell.
Note: Some distillates may contain compounds that have a chlorine
demand. One minute after the addition of chloramine T, test for residual
chlorine with KI-starch paper. If the test is negative, add an additional 0.060
mL of chloramine T. After one minute, recheck the sample.
Note: More than 0.060 mL of chloramine T will prevent the color from
developing with pyridine-pyrazolone.
8.5 Calculations Showing Equivalency of Concentrations of Releasing
Solutions
Equivalency of releasing solutions in Micro Dist Method Cyanide–1
and in USEPA Method 335.2 (see section 8.3 on page 33).
8.5.1 Micro Dist Method Cyanide–1
To 6.0 mL of sample is added 0.75 mL of 7.11 M H
2
SO
4
. This gives
a sample concentration in H
2
SO
4
:
This same solution is also 0.79 M in MgCl
2
· 6 H
2
O. This gives a
concentration of:
8.5.2 USEPA Method 335.2
To 500 mL of sample is added 50 mL of 9 M (18 N) H
2
SO
4
. This
gives a sample concentration in H
2
SO
4
:
To 500 mL of sample is also added 20 mL of a 510 g MgCl
2
· 6
H
2
O/L (2.5 M) solution. This gives a concentration of:
8.5.3 USEPA Method 335.4
To 50 mL of sample is added 5 mL of 9 M (18 N) H
2
SO
4
. This gives
a sample concentration in H
2
SO
4
:
7.11 M H
2
SO
4
0.75 mL× 6.0 mL 0.75 mL+()⁄ 0.79 M H
2
SO
4
=
0.79 M MgCl
2
6 H
2
O 0.75 mL 6.75 mL total()⁄×⋅
0.9 M H
2
SO
4
50 mL× 500 mL 50 mL 20 mL++()⁄ 0.79 M H
2
SO
4
=
2.5 M MgCl
2
6 H
2
O⋅ 20 mL× 570 mL total()⁄
9 M H
2
SO
4
5 mL× 50 mL 5 mL 2 mL++()⁄ 0.79 M H
2
SO
4
=
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