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Crazy boy [7]
3 years ago
12

Given that it requires 27.9 ml of 0.270 m na2s2o3(aq) to titrate a 15.0-ml sample of i3–(aq), calculate the molarity of i3–(aq)

in the solution.
Chemistry
1 answer:
Digiron [165]3 years ago
6 0

The molar concentration of the KI_3 solution is 0.251 mol/L.

<em>Step 1</em>. Write the <em>balanced chemical equation</em>

I_3^(-) + 2S_2O_3^(2-) → 3I^(-) + S_4O_6^(2-)

<em>Step 2</em>. Calculate the <em>moles of S_2O_3^(2-)</em>

Moles of S_2O_3^(2-)

= 27.9 mL S_2O_3^(2-) ×[0.270 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)]

= 7.533 mmol S_2O_3^(2-)

<em>Step 3</em>. Calculate the <em>moles of I_3^(-) </em>

Moles of I_3^(-) = 7.533 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 3.766 mmol I_3^(-)

<em>Step 4</em>. Calculate the <em>molar concentration of the I_3^(-) </em>

<em>c</em> = "moles"/"litres" = 3.766 mmol/15.0 mL = 0.251 mol/L

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