Okay so i have this titration question and I don't really know the steps because everyone explains it differently. IF someone could explain me the steps of how i could solve this questions and other like this. Thanks

In three runs of titration 22.8mL, 221.mL, 22.2mL of 0.200 Ba(OH)2 were required to neutralize 10mL of HCl. Calculate the acid concentration

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asked by Sherry
  1. Step 1. Write and balance the equation.
    Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O

    Step 2.
    mols Ba(OH)2 = M x L = ? That's approx 0.0046 but you need an accurate answer.

    Step 3.
    Using the coefficients in the balanced equation, convert mols Ba(OH)2 to mols HCl. That's
    approx 0.0046 mols Ba(OH)2 x (2 mols HCl/1 mol Ba(OH)2) = approx 0.0092.

    Step 4.
    M HCl = mols HCl/L HCl = approx 0.0092/0.01 = approx 0.9 M for HCl.

    1. You can make it easier by working in millimoles. In order of above, 22.8 x 0.2 = approx 4.6.
    4.6 x 2 = 9.2. Then M = mollimoles/mL = 9.2/10 = about 0.92M

    2. In practice, you can work this three times (by the way I think that 221 mL must be 22.1 mL), then average the three values you get for the final molarity OR, and this easier, average the three mL [(22.8+22.1+22.2/3)] and use that avg mL.

    Post your work if you have questions on this. These four steps will work any titration problem like this.

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  2. Thank you so much! this helped a lot!

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    posted by Sherry

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