Help with percent concentration of hydrogen peroxide if diluted?

  1. :frown: I did an experiment on the effect of substrate concentration on the rate of a catalase.

    I was given 6% substrate concentration of hydrogen peroxide. We had to do 3 trials with different concentrations of the hydrogen peroxide but we had to dilute the peroxide ourselves in the second trial I added 10 mL of water and in the third trial I added 20 mL of water.

    What I dont know is if I started with 6% and added 10 mL of water how did the adding of the water change the percent concentration of the peroxide? What is the concentration of the peroxide after 10 and 20 mL of water are added?
     
  2. jcsd
  3. iansmith

    iansmith 1,430
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    First, what was the volume of the 6% hydrygen peroxide that you added to the 10 mL and the 20 mL of water?

    For percetage concentration, assume that 6% is volume of hydrogen peroxide per volume of solution. Basicly, in 100 mL of solution you have 6 mL of hydrogen peroxide.

    To find the concentration after you added 10 mL or 20 mL, you have to find the volume of hydrogen peroxide that you added, which is calculated from the volume of 6% hydrogen peroxide you added to the 10 mL or 20 mL. When you have that number divid it by the total volume of the diluted hydrogen peroxide of second or third trial.
     
  4. Moonbear

    Moonbear 12,266
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    What volume of peroxide did you add?

    Oops, Iansmith beat me to it.
     
  5. well in the first trial I have only a 6% concentration of hydrogen peroxide-the amount of the peroxide is 75 mL.

    In the second trial I have 6% concentration of hydrogen peroxide but i have diluted it with 10 mL of water. There is 65 mL of hydrogen peroxide in the test tube

    In the third trial I have used 6% concentration of hydrogen peroxide but i have diluted it with 20 mL of water. There is 55 mL of hydrogen peroxide in the test tube.

    Each time am i increasing the concentration of the substrate or decreasing the concentration??? I think I am decreasing but im not sure and if I am then why would this make the rate of the reaction increase?
     
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