contestada

The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq) (thiosulfate ion). The determination is based on the net ionic equation 2S2O2−3(aq)+I3(aq)⟶S4O2−6(aq)+3I−(aq) Given that it requires 29.6 mL of 0.260 M Na2S2O3(aq) to titrate a 30.0 mL sample of I−3(aq), calculate the molarity of I−3(aq) in the solution.

Respuesta :

Answer : The molarity of [tex]I_3^-[/tex] in the solution is, 0.128 M

Explanation :

The given balanced chemical reaction is,

[tex]2S_2O_2^{-3}(aq)+I_3(aq)\rightarrow S_4O_2^{-6}(aq)+3I^-(aq)[/tex]

First we have to calculate the moles of [tex]Na_2S_2O_3[/tex].

[tex]\text{Moles of }Na_2S_2O_3=\text{Molarity of }Na_2S_2O_3\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }Na_2S_2O_3=0.260mole/L\times 0.0296L=0.007696mole[/tex]

Conversion used : (1 L = 1000 ml)

Now we have to calculate the moles of [tex]I_3^-[/tex].

From the balanced chemical reaction, we conclude that

As, 2 moles of [tex]S_2O_2^{-3}[/tex] react with 1 mole of [tex]I_3^-[/tex]

So, 0.007696 moles of [tex]S_2O_2^{-3}[/tex] react with [tex]\frac{0.007696}{2}=0.003848[/tex] mole of [tex]I_3^-[/tex]

The moles of [tex]I_3^-[/tex] = 0.003848 mole

Now we have to calculate the molarity of [tex]I_3^-[/tex].

[tex]\text{Molarity of }I_3^-=\frac{\text{Moles of }I_3^-}{\text{Volume of solution}}[/tex]

Now put all the given values in this formula, we get:

[tex]\text{Molarity of }I_3^-=\frac{0.003848mole}{0.03L}=0.128mole/L=0.128M[/tex]

Therefore, the molarity of [tex]I_3^-[/tex] in the solution is, 0.128 M

pstnonsonjoku

0.128 M of iodide solution was reacted with thiosulphate.

The equation of the reaction is;

[tex]2S2O2^-3(aq) + I3^-(aq)------->S4O2^-6(aq) + 3I^- (aq)[/tex]

Number of moles of  S2O2^-3- = 29.6/1000 × 0.260 M

= 0.0077 moles

Since 2 moles of thiosulphate reacts with 1 mole of iodide

0.0077 moles of thiosulphate reacts with 0.0077 moles × 1 mole/ 2 moles

= 0.00385 moles of iodide.

Since;

Number of moles = concentration × volume

concentration of iodide = Number of moles/volume

Volume of iodide = 30/1000 = 0.03 L

Concentration of iodide =  0.00385 moles/0.03 L

= 0.128 M

Learn more: https://brainly.com/question/9847214

Otras preguntas