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Chem 101 problem

  1. Oct 13, 2004 #1
    Help, I need to know how to do these problems!

    1. How many grams of Mg(OH)2 (subscript 2, just in case) will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M HCl?

    2 How many mL of a 0.10 NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4 (subscripts) solution?

    Thanks in advance!
  2. jcsd
  3. Oct 13, 2004 #2


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    in order to get a neutral solution, the amount of H+ ion needs to be equal to the amount of OH- ions.

    in both cases, it gives you the amount and the substance of the acidic part, you just need to find out how many moles of hydroxide ions to add to make it even.
    to check, you can use these, neutral = 7
    to find pH, pH = -log(H+)
    to find pOH, pOH = -log(OH-)

    so in number 1 you would start off with 25 mL of .1 M HCl, which means you have this:
    .1HCl --> .1H+ + .1Cl-
    so to find how many moles of H+ ions you have, multiply .1 * .025 L = .0025,
    that means you have a pH of -log(.0025) = 2.6
    you need to add the same number of moles of hydroxide to get a pH of 7

    Mg(OH)2 --> Mg+2 + 2OH-, that is 2 moles of hydroxide for ever one mole of magnessium hydroxide.
    and you need .0025 moles of hyroxide, so,
    .0025 moles of hydroxide / 2 moles per moles of magnesium hydroxide =
    .0025/2 = moles needed of magnessium hydroxide = .00125 moles of magnessium hydroxide.
    and magnessium hydroxide has a molar weight of (24.31+16+1.01) =
    41.32 g/mol, so you need (41.32*.00125) = .05165 grams of magnessium hydroxide to neutralize the solution.

    I sure hope I didnt make any careless errors in doing that, that really wouldne help much when trying to explain it.
  4. Oct 13, 2004 #3
    Thank you for the answer and your time. I need more detail ( I'm very
    chemistry illiterate) preferably deriving the equations without the use of-log(OH-) since I havent used that terminology yet... thanks again for your help
  5. Oct 14, 2004 #4


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    well, I cant always explain things very well, but I will try.

    when you have water (pure H2O) it will naturaly break down:
    H2O --> H+1 + OH-1
    into hydroxide ions (OH-1) and a positve hydrogen atom (H+1), I forgot what that is called.
    , but normally, this break down happens and there are equal numbers of H+ and OH-, so it evens itself out, and the water stays neutral at a pH of 7.
    but when you introduce other stuff into the water, you will set off balance the natural equalibrium, and throw it one way or the other, so for example if you add hydrochloric acid, that will become part of the solution as:
    HCl --> H+ + Cl-
    so that adds excess H+ ions, and lowers the pH down, likewise, if you add sodium hydoxide to the solution it will raise the pH by adding hydoxide ions:
    NaOH --> Na+ +OH-
    the Na+ doesnt matter, it wont effect the pH, but the OH- will, it will rais it up.
    if the concentration of H+ and OH- is equal, than you have a neutral solution with pH at 7.
    to find the pH you can use the formula -log(concentration of H+), but in your question, it isnt really needed.

    in your question, all you realy need to do, is calculate the amount of the H+ ions that are allready in solution, then figure out to yourself what needs to be added to bring that number eqaul to the number of OH- ions.
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