Question

The volume of 0.0125 mol/L sodium thiosulfate used for titration is 98.7 ml. What is the amount of dissolved O2(mg/L) in the sample?

Answer 1

Q150586

The volume of 0.0125 mol/L sodium thiosulfate used for titration is 98.7 ml. What is the amount of dissolved O₂ (mg/L) in the sample?

Solution:

Winkler method is used to determine the dissolved oxygen in aqueous solution. The steps

involved in the Winkler method are as follows:

1) Mn2+(aq) + 2OH- (aq) ----> Mn(OH)₂ (s)

2) 4Mn(OH)₂ (s) + 1 O₂ (aq) + 2H₂O(l) ---> 4 Mn(OH)₃ (s)

3) 2 Mn(OH)₃ (s) + 3H₂SO₄ (aq) ---> 2 Mn³⁺ (aq) + 3 SO₄^2- (aq) + 6 H₂O(l)

4) 2 Mn³⁺ (aq) + 2 I^- (aq) ----> 2 Mn²⁺ (aq) + 1 I₂ (aq)

5) 1 I₂ (aq) + I^- (aq) ------> 1 I₃ ^- (aq)

6) 1 I₃ ^- (aq) + 2 S₂O₃^2- (aq) ----> 3 I^- (aq) + S₄O₆^2- (aq)

Using these equation we will find the mole to mole ratio of O₂ and 2S₂O₃^2- (aq)

$\frac{1mol O_2}{4mol Mn(OH)_3}* \frac{2mol Mn(OH)_3}{2mol Mn^{+3}}*\frac{2mol Mn^{+3}}{1mol I_2}*\frac{1mol I_2 }{1mol I_3^{-1}}$ $*\frac{1mol I_3^{-1}}{2 mol S_2O_3^{2-}}$

= $\frac{1mol O_2}{4 mol S_2O_3^{2-}}$

The mole to mole ratio of O₂ and S₂O₃^2- is 1 : 4. Using this we can find the moles of O₂ present in the given sample of water.

Step 1 : To find the moles of thiosulfate (S₂O₃^2- ) present in 98.7 mL of 0.0125 mol/L sodium thiosulfate solution.  

$Molarity = \frac{moles \space of \space solute}{volume \space of \space solution \space in \space 'L'}$

98.7 mL in ‘L’ = 98.7 mL * 1 L/1000mL = 0.0987 L;

plug molarity = 0.0125 mol /L and volume = 0.0987 L in the formula, we have

0.0125 mol /L = moles of S₂O₃^2- / 0.0987 L

multiply both the side by 0.0987 L, we have

0.0125 mol /L * 0.0987L = moles of S₂O₃^2- / 0.0987 L * 0.0987 L

moles of S₂O₃^2- = 0.001234 moles .

Step 2 : To find the moles of O₂ from moles of S₂O₃^2-

The mole to mole ratio of O₂ and S₂O₃^2- is 1 : 4.

$So, moles \space of\space O_2 = 0.01234 mol S_2O_3^{2-}* \frac{1 mol O_2}{4 mol S_2O_3^{2-} }$

0.0003084 mol O₂ .

Step 3 : Convert 0.0003084 mol O₂ to grams using molar mass of O_{2 .}

molar mass of O₂ = 2 * atomic mass of O = 2 * 15.999 g/mol = 31.998 g/mol

grams of O₂ = 0.0003084 mol O₂ * $\frac{31.998 \space g \space O_2}{1\space mol \space O_2} = 0.009869 grams$

in 'mg' the mass of O₂ is = $0.009869 g * \frac{1000mg}{1g} = 9.87 mg;$

We are not provided the volume of sample of water. If we assume that the volume in 1L. Then the solubility of O₂ in (mg/L) will be 9.87 mg/L