Winkler titration for dissolved ${\mathit{O}}_{\mathbf{2}}\mathbf{.}$Dissolved ${\mathit{O}}_{\mathbf{2}}$ is a prime indicator of the ability of a body of natural water to support aquatic life. If excessive nutrients run into a lake from fertilizer or sewage, algae and phytoplankton thrive. When algae die and sink to the bottom of the lake, their organic matter is decomposed by bacteria that consume ${\mathit{O}}_{\mathbf{2}}$from the water. Eventually, the water can be sufficiently depleted of${\mathit{O}}_{\mathbf{2}}$so that fish cannot live. The process by which a body of water becomes enriched in nutrients, some forms of life thrive, and the water eventually becomes depleted of${\mathit{O}}_{\mathbf{2}}$is called eutrophication. One way to measure dissolved${\mathit{O}}_{\mathbf{2}}$ is by the Winkler method that involves an iodometric titration: 35

Dissolved oxygen or biochemical oxygen demand

1. Collect water in a $\mathbf{~}\mathbf{300}\mathbf{}\mathit{m}\mathit{L}$bottle with a tightly fitting, individually matched ground glass stopper. The manufacturer indicates the volume of the bottle $\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{mL}\mathbf{)}$with the stopper inserted on the bottle. Submerge the stoppered bottle at the desired depth in the water to be sampled. Remove the stopper and fill the bottle with water. Dislodge any air bubbles before inserting the stopper while the bottle is still submerged.

2. Immediately pipet 2.0 mL of $\mathbf{2}\mathbf{.}\mathbf{15}\mathbf{M}\mathbf{}\mathbf{}{\mathbf{MnSO}}_{\mathbf{4}}$and 2.0 mL of alkali solution containing$\mathbf{500}\mathbf{}\mathbf{g}\mathbf{}\mathbf{NaOH}\mathbf{/}\mathbf{L}\mathbf{,}\mathbf{}\mathbf{135}\mathbf{}\mathbf{g}\mathbf{}\mathbf{NaL}\mathbf{/}\mathbf{L}$ and $\mathbf{10}\mathit{g}\mathbf{}\mathit{N}\mathit{a}{\mathit{N}}_{\mathbf{3}}\mathbf{/}\mathit{L}$ (sodium azide). The pipet should be below the liquid surface during addition to avoid introducing air bubbles. The dense solutions sink and displace close to 4.0 Ml of natural water from the bottle. 3. Stopper the bottle tightly, remove displaced liquid from the cup around the stopper, and mix by inversion. ${\mathit{O}}_{\mathbf{2}}$is consumed and$\mathit{M}\mathit{n}\mathbf{(}\mathit{O}\mathit{H}{\mathbf{)}}_{\mathbf{3}}$ precipitates:

$\mathbf{4}\mathit{M}{\mathit{n}}^{\mathbf{2}\mathbf{+}}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{+}\mathbf{8}\mathit{O}{\mathit{H}}^{\mathbf{-}}\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\to }\mathbf{4}\mathit{M}\mathit{n}\mathbf{\left(}\mathit{O}\mathit{H}{\mathbf{)}}_{\mathbf{3}}\mathbf{\right(}\mathit{s}\mathbf{)}$

Azide consumes any nitrite$\mathbf{\left(}\mathit{N}{\mathit{O}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\right)}$in the water so that nitrite cannot subsequently interfere in the iodometric titration:

$\mathbf{2}\mathit{N}{\mathit{O}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{+}\mathbf{6}{\mathit{N}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}\mathbf{4}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\to }\mathbf{10}{\mathit{N}}_{\mathbf{2}}\mathbf{+}\mathbf{8}\mathit{O}{\mathit{H}}^{\mathbf{-}}$

4. Back at the lab, slowly add 2.0 mL of $\mathbf{18}\mathit{M}{\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$below the liquid surface, stopper the bottle tightly, remove the displaced liquid from the cup, and mix by inversion. Acid dissolves which reacts quantitatively with

$\mathbf{2}\mathbf{Mn}\mathbf{\left(}\mathbf{OH}{\mathbf{)}}_{\mathbf{3}}\mathbf{\right(}\mathbf{s}\mathbf{)}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{+}\mathbf{3}{\mathbf{I}}^{\mathbf{-}}\mathbf{\to }\mathbf{2}{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}}\mathbf{+}{\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}\mathbf{3}{\mathbf{SO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

5. Measure 200.0 mLof the liquid into an Erlenmeyer flask and titrate with standard thiosulfate. Add 3mL of starch solution just before the end point and complete the titration.

A bottle of 297.6 mLof water from a creek at$\mathbf{0}\mathbf{°}\mathit{C}$in the winter was collected and required 14.05 mL 10.22 mM thiosulfate.

(a) What fraction of the 297.6 mL sample remains after treatment with and alkali solution?

(b) What fraction remains after treatment with ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$? Assume that ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$sinks into the bottle and displaces 2.0 mL of solution prior to mixing.

(c) How many mL of the original sample are contained in the 200.0 mLthat are titrated?

(d) How many moles of${\mathit{l}}_{\mathbf{3}}^{\mathbf{-}}$ are produced by each mole of${\mathit{O}}_{\mathbf{2}}$ in the water?

(e) Express the dissolved ${\mathit{O}}_{\mathbf{2}}$content in (f) Pure water that is saturated with ${\mathit{O}}_{\mathbf{2}}$ contains $\mathbf{14}\mathbf{.}\mathbf{6}\mathit{m}\mathit{g}{\mathit{O}}_{\mathbf{2}}\mathbf{/}\mathit{L}\mathit{a}\mathit{t}\mathbf{}\mathbf{0}\mathbf{°}\mathbf{}\mathit{C}\mathbf{.}$ What is the fraction of saturation of the creek water with ${\mathit{O}}_{\mathbf{2}}$?

(g) Write a reaction of $\mathit{N}{\mathit{O}}_{{}_{\mathbf{2}}}^{\mathbf{-}}$with${\mathit{l}}^{\mathbf{-}}$ that would interfere with the titration if${\mathit{N}}_{\mathbf{3}}^{\mathbf{-}}$were not introduced. See Table 16-5.

Step by step solution

01

Concept used

Bacteria in the water devour their organic materials and degrade it.

02

Step 2:

They displace 4.0 mL of natural water after treatment with $MnS{O}_4$and alkali solution, leaving the following volume of water sample:

$$\begin{gathered} V\left(water\right)=V\left(watersample\right)-4.0mLV\left(water\right) \\ =297.6mL-4.0mLV\left(water\right) \\ =293.6mLgathered \end{gathered}$$

The remaining water percentage is then:

$$\begin{gathered} V\left(water\right)=\frac{V\left(water\right)}{V\left(originalwatersample\right)} \\ V\left(water\right)=\frac{293.6.nL}{297.6mE} \\ V\left(water\right)=98.66\% \end{gathered}$$

03

Step 3:

(b)

It displaces another 2.0 mL of natural water after treatment with$H_{2}S{O}_4,$leaving the following volume of water sample:

$$\begin{gathered} V\left(water\right)=V\left(watersampleafteralkalization\right)-4.0mL \\ V\left(water\right)=293.6mL-2.0mL \\ V\left(water\right)=291.6mL \end{gathered}$$

The remaining water percentage is then:

role="math" localid="1664882104663" $$\begin{gathered} V\left(water\right)=\frac{V\left(water\right)}{V\left(originalwatersample\right)} \\ V\left(water\right)=\frac{291.6.nL}{297.6mE} \\ V\left(water\right)=97.98\% \end{gathered}$$

04

Step 4:

(c)

The initial sample that is contained in the 200 mL sample that is titrated is:

$$\begin{gathered} V\left(water\right)=V\left(watersampleafteralkalization\right)-4.0mL \\ V\left(water\right)=200mL-4.0mL \\ V\left(water\right)=196mL \end{gathered}$$

05

Step 5:

(d)

Let can calculate how many moles of$l_3-$are created by each mole of$O_2:$using these two reactions:

1. $O_2:4M{n}^{2+}+O_2+8O{H}^{-}+2H_{2}O\rightleftharpoons 4Mn{(OH)}_3$
   
2. $2Mn{\left(OH\right)}_3+3H_{2}S{O}_4+3I^{-}\rightleftharpoons 2M{n}^{2+}+I_3^{-}+3S{O}_4^{2-}+6H_{2}O$
   

The ratio of oxygen and triiodide may be calculated using the ratio of moles of triiodide and$Mn{\left(OH\right)}_3$and$Mn{\left(OH\right)}_3$and oxygen:

localid="1664883390783" $n(I_3^{-})=\frac{1}{2}n(Mn{(OH)}_3)n\left(Mn{\left(OH\right)}_3\right)=4n\left(O_2\right)$

Equation (2) is obtained by plugging into equation (1).

localid="1664882962222" $$\begin{gathered} n\left(I_3^{-}\right)=\frac{1}{2}·4n\left(O_2\right) \\ n\left(I_3^{-}\right)=2n\left(O_2\right) \end{gathered}$$

06

Step 6:

e) To describe the dissolved$O_2$content, we must additionally write the thiosulfate and triiodide titration reaction:

$I_3^{-}+2S_2{O}_3^{2-}\rightleftharpoons 3I^{-}+S_4{O}_6^{2-}$

These two ratios can be used to calculate the thiosulfate and oxygen ratios:

$$\begin{gathered} n\left(l_3^{-}\right)=\frac{1}{2}n\left(S_2{O}_3^{2-}\right) \\ n\left(l_3^{-}\right)=2n\left(O_2\right) \end{gathered}$$

When these two equations are combined, we get:

role="math" localid="1664884077876" $$\begin{gathered} \frac{1}{2}n\left(S_2{O}_3^{2-}\right)=2n\left(O_2\right)/:2 \\ n\left(O_2\right)=\frac{1}{4}n\left(S_2{O}_3^{2-}\right)n\left(O_2\right)=\frac{1}{4}.c\left(S_2{O}_3^{2-}\right).N.V\left(S_2{O}_3^{2-}\right) \\ n\left(O_2\right)=\frac{1}{4}.10.22\times {10}^{-3}M.2.14.05\times {10}^{-3}L \\ n\left(O_2\right)=7.1796\times {10}^{-5}mol \end{gathered}$$

The normality of sodium thiosulfate is N

Now we can figure out how much oxygen there is in the air:

$$\begin{gathered} m(O_2)=n\left(O_2\right)\cdot M(O_2) \\ m(O_2)=7.1796\times {10}^{-5}mol\cdot 31.998g/mol \\ m(O_2)=2.2973\times {10}^{-3}g=2.297mg \end{gathered}$$

By dividing the mass of$O_2$with the volume of the original sample contained in 200.0 mL from task c), the content of $O_2$in mg/L is calculated:

$$\begin{gathered} \gamma \left(O_2\right)=\frac{m\left(O_2\right)}{V\left(sample\right)} \\ \gamma \left(O_2\right)=\frac{2.297mg}{196\times {10}^{-3}L} \\ \gamma \left(O_2\right)=11.72mg/L \end{gathered}$$

07

Step 7:

f) The saturation fraction is:

$$\begin{gathered} saturation=\frac{\gamma \left(sample\right)}{\gamma (purewater)}·100\% \\ saturation=\frac{11.72mg/L}{14.6mg/L}·100\% \\ saturation=80.27\% \end{gathered}$$

08

Step 8:

g) If$N_3$- was not added,$HN{O}_2$would react with iodide and cause the titration to fail:

$HN2HN{O}_2+2H^{+}+3I^{-}\rightleftharpoons 2NO+I_3^{-}-+2H_2{O}_2$

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