

 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 𝒗
𝑨
𝟏
(V 𝒗
𝑨
 V
A
 M
A
 V
B
 M
B v a
A
A
x M
𝟏 𝒗
𝑩
A
B
) = 𝟏 𝒗
𝑩
(V
B x M )
= molarity of reactant A
= volume of reactant B
= molarity of reactant B and v b are the coefficients in the balanced equation
B
= volume of reactant A


Consider the balanced redox reaction of potassium manganate(VII) with ammonium iron(II) sulfate.
5Fe 2+ + MnO
4
+ 8H +  5Fe 3+ + Mn 2+ + 4H focusing what was reduce and oxidized
2
O notice spectator ions are left out of the equation and only

In a titration to determine the concentration of a potassium manganate(VII) solution, 28.0 cm 3 of the potassium manganate(VII) reacted completely with 25.0 cm 3 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.

𝑴
𝑨
= 𝒏
𝑽
… look at balanced equation….
𝑴
𝑩
= 𝒏
𝑽
𝟎. 𝟎𝟏𝟎𝟎𝑴 = 𝒏
.𝟎𝟐𝟓𝟎𝑳
…mole(n) ratio… 𝑴
𝑩
= 𝒏
.𝟎𝟐𝟖𝟎𝑳 n = .000250 mole(n) ratio is 5 to 1 n = .0000500
𝑴
𝑩
𝑩
.𝟎𝟎𝟎𝟎𝟓𝟎𝟎
=
.𝟎𝟐𝟖𝑳
need answer in g dm -3 , not Molarity
𝟎. 𝟎𝟎𝟏𝟕𝟗 𝐦𝐨𝐥 𝐊𝐌𝐧𝐎𝟒 𝒙
0.283 g dm -3
𝟏𝟓𝟖. 𝟎𝟒 𝒈 𝐊𝐌𝐧𝐎
𝟒
𝟏 𝒎𝒐𝒍 𝐊𝐌𝐧𝐎
𝟒 from the periodic table

 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 O
2 required to oxidize organic matter in a sample of water at a given temp. for 5 days
 measured in ppm = mg/L (or mg dm -3 )


2
›
2
therefore, solubility of oxygen in water is very low (max of 9 ppm)
 it decreases with increase in temperature
› to sustain a healthy aquatic environment, dissolved oxygen should be >6 ppm
 ppm = mg/L (or mg dm -3 )

 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 (CH
4
) is produced instead of CO
2
 hydrogen sulfide (H
2
S) and ammonia (NH instead of water, sulfates, and nitrates
3
)