Molarity of KMnO4:0.2M
Volume of H2O2: 10mL
Initial volume 50mL
Final Volume: 31.90mL
Volume of KMnO4 (new H2O2)
Would it be the 31.90mL or do I subtract 50mL-31.90mL= 18.10mL
Your question is confusing because you have a new H2O2 and we don't know if that is the 10 mL or not. In addition, you don't label the 50 mL and 31.90 mL. I suspect your question is that you started with a full buret of KMnO4 at 50 mL and you used down to 31.90 mL KMnO4. Therefore, you subtract. The amount of KMnO4 used to titrate the H2O2 was 50.00 mL - 31.90 mL = the 18 or so you have in the post. (I assume the 50 mL you wrote actually means 50.00 mL.)
To determine the volume of KMnO4 that reacts with the given volume of H2O2, you need to consider the stoichiometry of the reaction between KMnO4 and H2O2. Without the balanced equation or any other information regarding the reaction, it is not possible to calculate the specific volume of KMnO4 required.
However, if you have additional information, such as the balanced chemical equation and the stoichiometry of the reaction, it is possible to calculate the volume of KMnO4 needed using the provided information. The volume of H2O2 (10 mL) and the molarity of KMnO4 (0.2 M) alone do not provide enough information to determine the volume of KMnO4 required for the reaction.