Regular-strength chlorine bleach is a 5.25% solution of NaClO. how many milliliters of 0.02005 M sodium thiosulfate will be required to reach the equivalence point in the titration of a 1.000ml sample of Bleach by reacting the ClO- ions with I, then titrating the resulting I2 with S2O3-2?
How much of a 10 g sample of cesium 124 remains after 93 samples?
How does this work
To find the number of milliliters of 0.02005 M sodium thiosulfate required to reach the equivalence point in the titration, we need to know the stoichiometry of the reaction between sodium thiosulfate (Na2S2O3) and chlorine bleach (NaClO).
Let's write the balanced chemical equation for the reaction between ClO- ions and I- ions:
2NaClO + 2KI → I2 + 2NaCl + K2O
From the balanced equation, we can see that it takes 2 moles of sodium thiosulfate (Na2S2O3) to react with 1 mole of iodine (I2):
2Na2S2O3 + I2 → Na2S4O6 + 2NaI
Now, let's calculate the number of moles of ClO- in the given 1.000 ml sample of bleach. The bleach is a 5.25% solution of NaClO, which means it contains 5.25 grams of NaClO per 100 ml:
5.25 grams NaClO * (1 mole NaClO / molar mass NaClO) * (1/1000 L) = moles of NaClO
Now, knowing the moles of NaClO, we can determine the moles of I2 produced in the reaction:
moles of NaClO * (1 mole I2 / 2 moles NaClO) = moles of I2
At the equivalence point, all the I2 formed would be reacted with the sodium thiosulfate. So, the moles of sodium thiosulfate required to reach the equivalence point can be calculated by:
moles of I2 * (2 moles Na2S2O3 / 1 mole I2) = moles of Na2S2O3
Finally, we can calculate the volume (in ml) of 0.02005 M sodium thiosulfate required using the molarity and the moles of Na2S2O3:
moles of Na2S2O3 * (1/0.02005 M) * 1000 = volume of sodium thiosulfate in ml
By plugging in the values, you can now calculate the volume of 0.02005 M sodium thiosulfate required to reach the equivalence point in the titration.