1. A solution was prepared by adding 12.00 mL of 0.130 M KI, 3.00 mL of 0.130 M
ID: 712188 • Letter: 1
Question
1. A solution was prepared by adding 12.00 mL of 0.130 M KI, 3.00 mL of 0.130 M KNO3, 5 mL of 0.0100 M Na2S2O3, 1.00 mL of starch solution, and 15.00 mL of 0.130 M (NHa)2S20s. a. Assuming the volumes of the components are additive, calculate the concentration (M) ofI in the final mixture before the chemical reaction. Show the calculation. b. Calculate the concentration (M) of S20s in the mixture prior to the reaction. Show the calculation. c. Calculate the concentration (M) of S203- in the mixture prior to the reaction. Show the calculation.Explanation / Answer
a.)
number of mol of KI in 12.00 ml of 0.130 M solution = 0.130 mol /L * 0.012 L
= 0.00156 mol
after addition of all solution, final volume of the mixture is = (12 + 3+5+1+15) ml
= 36 ml
= 0.036 lit
final concentration of iodide ion is = 0.00156 mol / 0.036 lit
= 0.0433 M
b)
mol of S2O82- = 0.130 mol /L * 0.015 L
= 0.00195 mol
after addition of all solution, final volume of the mixture is = (12 + 3+5+1+15) ml = 36 ml = 0.036 lit
final concentration of S2O82- =0.00195 mol / 0.036 lit = 0.054 M
c.)
mol of S2O32- = 0.0100 mol /L * 0.005 L
= 0.00005 mol
after addition of all solution, final volume of the mixture is = (12 + 3+5+1+15) ml = 36 ml = 0.036 lit
final concentration of S2O32- =0.00005 mol / 0.036 lit
= 0.00138 M
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