#5: Titration of 0.1212 g of pure Na2C2O4 (134 g/mol) required 19.14 mL of KMnO4. What is the molarity of the KMnO4 solution? The net ionic equation for this reaction is: 2MnO4- + 5C2O42- + 16H+ ---> 2Mn2+ + 8H2O + 10CO2 Choices: 0.0189 M 0.0473 M 0.0722 M 0.1181 M
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#5: Titration of 0.1212 g of pure Na2C2O4 (134 g/mol) required 19.14 mL of KMnO4. What is the molarity of the KMnO4 solution?
The net ionic equation for this reaction is:
2MnO4- + 5C2O42- + 16H+ ---> 2Mn2+ + 8H2O + 10CO2
Choices:
0.0189 M
0.0473 M
0.0722 M
0.1181 M
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- A sample of impure NaOH, which has been partially converted to Na2CO3 by exposure to CO2, is analyzed by titrating a 188.5 mg sample with 0.1065 M HCl. The volume required to reach phenolphtalein end point is 39.19 ml while the volume required to reach bromocresol green end point is 40.67 ml. Calculate the percentages of NaOH (MWt= 40 g/mole) and Na2CO3 (MWt=83 g/mole) in the sample.A student weighs out 1.118g of impure HPT, dissolves the sample in deionized water and tirates it with 0.1001M of NaOH solution. If the titration requires 2710mL of the NaOH solution, and none of the impurities react with NaOH, what is the percent KHP in the sample? I'm confused why none of the impurities would react. I multipled 0.1001M by 2.71L and got 0.2172M of NaOH. As it's 1:1 ratio, shouldn't the point of equivalence be 0.2712 mol of KHP? But I do not have the volume of KHP so how can I calculate the molarity with only grams (which can be turned into mol) and no volume?The SO2 present in air is mainly responsible for the phenomenon of acid rain. The concentration of SO2 can be determined by titrating against a standard permanganate solution as follows: 5SO2 + 2MnO4- + 2H2O ---> 5SO42- + 2Ms2 + 4H+ Calculate the number of grams of SO2 in a sample of air if 4.90mL of 0.00700M KMnO4 solution are required for the titration. Be sure your answer has the correct number of significant digits.