Suppose a student diluted and titrated a bleach unknown exactly as described in the experimental procedure, except only a single titration was performed which required 16.22 mL of 0.100 M Na2S2O3.
Calculate the number of moles of ClO- in the sample titrated.
To calculate the number of moles of ClO- in the sample titrated, we need to use the balanced chemical equation for the reaction between ClO- and Na2S2O3.
The balanced equation is:
2ClO- + 2Na2S2O3 → 2NaCl + S4O6-2 + H2O
From the balanced equation, we can see that 2 moles of ClO- react with 2 moles of Na2S2O3.
Given that the volume of Na2S2O3 required for titration is 16.22 mL and the concentration of Na2S2O3 is 0.100 M, we can calculate the number of moles of Na2S2O3 used in the reaction.
moles of Na2S2O3 = volume × concentration
= 16.22 mL × 0.100 M
= 1.622 mmol
Since the mole ratio between Na2S2O3 and ClO- is 1:1, the number of moles of ClO- present in the sample is also 1.622 mmol.
Therefore, the number of moles of ClO- in the sample titrated is 1.622 mmol.
To calculate the number of moles of ClO- in the sample titrated, we can use the stoichiometry of the reaction between ClO- and Na2S2O3.
The balanced equation for the reaction is:
2 Na2S2O3(aq) + ClO-(aq) -> Na2SO4(aq) + NaCl(aq) + H2O(l) + SO2(g)
From the balanced equation, we can see that 2 moles of Na2S2O3 are required to react with 1 mole of ClO-.
Given that 16.22 mL of 0.100 M Na2S2O3 is required for the titration, we can calculate the moles of Na2S2O3 used.
Moles of Na2S2O3 = Volume of Na2S2O3 solution (in L) x Molarity of Na2S2O3
To convert the volume of Na2S2O3 from milliliters (mL) to liters (L), we divide by 1000:
Volume of Na2S2O3 solution (in L) = 16.22 mL / 1000 = 0.01622 L
Substituting this value into the equation:
Moles of Na2S2O3 = 0.01622 L x 0.100 mol/L = 0.001622 mol
Since 2 moles of Na2S2O3 react with 1 mole of ClO-, the number of moles of ClO- in the sample titrated is half the moles of Na2S2O3 used:
Moles of ClO- = 0.001622 mol / 2 = 0.000811 mol
Therefore, the number of moles of ClO- in the sample titrated is 0.000811 mol.