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
Answer:
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Explanation:
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Explanation:
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