Transcript Topic 9.1 Volumetric Analysis and the Winker method

Happy Days video
(2:30)
in both titration labs so far this year…
 recall if you know 2 out of 3 in in one
solution:

M = # moles /volume(L)
›

›
›
›
›

𝑴=
𝒏
𝑽
you can always solve for the unknown
then look at the balanced equation
determine the number of moles reacted with in
the other solution
then solve for its unknown
this is actually called volumetric analysis

from this, one can derived the following
formula to solve for molarity in one step
›
›
𝟏
(nA)
𝒗𝑨
𝟏
(VA x
𝒗𝑨
=
𝟏
(nB)
𝒗𝑩
MA) =
and hence…
𝟏
(VB
𝒗𝑩
x MB)
 VA = volume of reactant A
 MA = molarity of reactant A
 VB = volume of reactant B
 MB = molarity of reactant B
 va and vb are the coefficients in the
balanced equation

Consider the balanced redox reaction of
potassium manganate(VII) with ammonium
iron(II) sulfate.
5Fe2+ + MnO4- + 8H+  5Fe3+ + Mn2+ + 4H2O
notice spectator ions are left out of the equation and only
focusing what was reduce and oxidized

In a titration to determine the concentration
of a potassium manganate(VII) solution, 28.0
cm3 of the potassium manganate(VII)
reacted completely with 25.0 cm3 of a
0.0100 mol/dm-3 solution of iron(II) sulfate.
Determine the concentration, in g dm-3, of
the potassium manganate(VII) solution.
𝒏
𝒏
𝑴𝑨 = … look at balanced equation…. 𝑴𝑩 =
𝑽
𝑽
𝒏
𝒏
𝟎. 𝟎𝟏𝟎𝟎𝑴 =
…mole(n) ratio… 𝑴𝑩 =
.𝟎𝟐𝟓𝟎𝑳
.𝟎𝟐𝟖𝟎𝑳
n = .000250
mole(n) ratio is 5 to 1
n = .0000500
𝑴𝑩 =
.𝟎𝟎𝟎𝟎𝟓𝟎𝟎
.𝟎𝟐𝟖𝑳
𝑴𝑩 = 𝟎. 𝟎𝟎𝟏𝟕𝟗
need answer in g dm-3, not Molarity
𝟏𝟓𝟖. 𝟎𝟒 𝒈 𝐊𝐌𝐧𝐎𝟒
𝟎. 𝟎𝟎𝟏𝟕𝟗 𝐦𝐨𝐥 𝐊𝐌𝐧𝐎𝟒 𝒙
𝟏 𝒎𝒐𝒍 𝐊𝐌𝐧𝐎𝟒
0.283 g dm -3
from the
periodic
table
 aquatic
life depends on CO2 an O2
dissolved in water
 O2 is non-polar, while H20 is polar
› therefore, solubility of oxygen in water is
very low
 it decreases with increase in temperature
 0°C is 14.6 ppm (parts per million)
 20°C is 7.6 ppm (parts per million)

ppm is used for very dilute solutions

=



𝒎𝒂𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒊𝒏 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒈𝒓𝒂𝒎𝒔
𝒕𝒐𝒕𝒂𝒍 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒈𝒓𝒂𝒎𝒔
x 106
.00419 g in 1 liter of solution (1 liter H2O = 1000 ml = 1000 g)
.00419 g / 1000 g x 106
= 4.19 ppm
or

=



𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝒎𝒈
𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒅𝒎𝟑
.00419g would be 4.19 mg / 1 dm3
= 4.19 ppm
to sustain a healthy aquatic environment,
dissolved oxygen should be >6 ppm

can be used to measure the degree of
pollution in a water sample
› saturate a sample of water with oxygen
› incubate for 5 days to allow microorganisms to
oxidize the organic matter
› use an iodine/thiosulfate redox titration to
measure amount of remaining dissolved oxygen
› the measurement is biochemical oxygen
demand (BOD)
 defined as the amount of O2 required by
aerobic biological organisms to break down
organic material present in a given water
sample for 5 days
 measured in ppm or mg/L (same thing)

when organic matter is discharged into
water, it provides a source of food for
bacteria
› when bacteria multiply and their uptake
of oxygen is greater than what is
replaced by photosynthesis, the body of
water becomes depleted of oxygen
 in these anaerobic conditions
 methane (CH4) is produced instead of CO2
 hydrogen sulfide (H2S) and ammonia (NH3)
instead of water, sulfates, and nitrates

An iodine/thiosulfate redox titration is carried out to
measure the dissolved oxygen present in a water
sample:
A 50 cm3 water sample is taken and saturated with
oxygen for five days
› The following reactions (don’t need to know) are
related to the Winkler method
 2 Mn2+ (aq) + O2 (aq) + 4OH– → 2 MnO(OH)2(s)
 MnO(OH)2(s) + 2 I–(aq) + 4H+ → Mn2+(aq) + I2(aq) + 3H2O
 2 S2O32-(aq) + I2 (aq) → S4O62-(aq) + 2 I– (aq)
› It was found that 5.25 cm3 of a 0.00500 mol dm-3
solution of sodium thiosulfate (Na2S2O3) was required
to react with the iodine produced.

Determine the concentration of
dissolved oxygen in ppm in the sample
of water
› molarity of O2 is 1.31 x 10-4 mol dm-3
› ppm is 4.19

Deduce the BOD, in ppm, of the water
sample assuming the maximum solubility
of oxygen in water is 9.00 ppm at 293K.
› BOD is 4.81 ppm

Comment on the BOD value obtained.
› Results show reasonable water quality
 untreated sewage has a BOD range of 100-400
ppm.