Suppose that the KMnO4 (aq) were standardized by reaction with As2O3.
As2O3 + MnO4- + H+ +H2O yields H3AsO4 + Mn2+ (not balanced).
If a 0.1322-g sample that is 99.16 % As2O3 by mass had been used in the titration, how many milliliters of the 0.02140 M KMnO4 (aq) would have been required?
First, balance the titration equation.
g As2O3 = 0.1322 x 0.9916 = ??
moles As2O3 = g/molarmass
Using the coefficients in the balanced equation to find moles KMnO4.
mols As2O3 x (xxmoles KMnO4/yy moles As2O3).
M KMnO4 = moles KMnO4/L KMnO4
To find the number of milliliters of the 0.02140 M KMnO4 (aq) required in the titration, we need to calculate the moles of As2O3 used and then use stoichiometry to determine the corresponding amount of KMnO4.
First, we will determine the number of moles of As2O3 used in the titration. We are given that the sample has a mass of 0.1322 g and is 99.16% As2O3 by mass. So, we can calculate:
Mass of As2O3 = 0.9916 * 0.1322 g = 0.1310 g
Next, we need to convert the mass of As2O3 to moles using its molar mass. The molar mass of As2O3 is:
As2O3: As (2 * 74.92 g/mol) + O (3 * 16.00 g/mol)
= 74.92 g/mol + 48.00 g/mol
= 122.92 g/mol
Moles of As2O3 = Mass of As2O3 / Molar mass of As2O3
= 0.1310 g / 122.92 g/mol
= 0.00106 mol
Now, using the balanced equation provided, we can determine the stoichiometric ratio between As2O3 and KMnO4. According to the equation:
As2O3 + MnO4- + H+ + H2O → H3AsO4 + Mn2+
From the balanced equation, we know that the stoichiometric ratio between As2O3 and MnO4- is 1:1. This means that 1 mole of As2O3 reacts with 1 mole of MnO4-.
Since we have determined that there are 0.00106 moles of As2O3 used in the titration, it means that we would require the same number of moles of KMnO4. Therefore, we need to find the volume (in liters) of 0.02140 M KMnO4 (aq) that contains 0.00106 moles using the formula:
Moles of solute = Molarity × Volume (in liters)
0.00106 mol = 0.02140 mol/L × Volume (in liters)
Solving for the volume:
Volume (in liters) = 0.00106 mol / 0.02140 mol/L
= 0.0495 L
Finally, we will convert the volume from liters to milliliters:
Volume (in milliliters) = 0.0495 L × 1000 mL/L
= 49.5 mL
Therefore, approximately 49.5 milliliters of the 0.02140 M KMnO4 (aq) would have been required in the titration.
To find the number of milliliters of the 0.02140 M KMnO4 (aq) required for the titration, we need to set up a balanced equation and use stoichiometry.
First, let's balance the equation:
As2O3 + 8MnO4- + 13H+ + 5H2O → 2H3AsO4 + 8Mn2+
From the equation, we can see that 1 mole of As2O3 reacts with 8 moles of KMnO4.
Next, we can calculate the number of moles of As2O3 in the sample:
Mass of As2O3 = 0.1322 g
Percentage purity of As2O3 = 99.16%
Mass of pure As2O3 = (Percentage purity / 100) x Mass of sample
Mass of pure As2O3 = (99.16 / 100) x 0.1322 g
Mass of pure As2O3 = 0.1309 g
Now, we can calculate the number of moles of As2O3:
Molar mass of As2O3 = 197.84 g/mol
Number of moles of As2O3 = Mass of As2O3 / Molar mass of As2O3
Number of moles of As2O3 = 0.1309 g / 197.84 g/mol
Number of moles of As2O3 = 0.000661 mol
From the balanced equation, we know that 1 mole of As2O3 reacts with 8 moles of KMnO4. Therefore, 0.000661 mol of As2O3 will react with X moles of KMnO4.
Using the molar ratio:
Number of moles of KMnO4 = 0.000661 mol As2O3 x (8 mol KMnO4 / 1 mol As2O3)
Number of moles of KMnO4 = 0.005288 mol
Finally, we can calculate the volume of the 0.02140 M KMnO4 solution that contains 0.005288 mol of KMnO4:
Volume of KMnO4 solution = Number of moles of KMnO4 / Molarity of KMnO4 solution
Volume of KMnO4 solution = 0.005288 mol / 0.02140 mol/L
Volume of KMnO4 solution = 0.247 L = 247 mL
Therefore, approximately 247 milliliters of the 0.02140 M KMnO4 (aq) would have been required for the titration.