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how many mL of 0.15 M NaOH solution are required to neutralize 35.00 mL of 0.22 M HCl

  • chemistry -

    mL NaOH x M NaOH = mL HCl x M HCl

  • chemistry -

    The easy way is simple substitution.
    mL NaOH x M NaOH = mL HCl x M HCl
    mL NaOH x 0.15M = 35.00mL x 0.22M
    Solve for mL NaOH. I get (35.00*0.22/0.15) = approximately 51 mL NaOH.
    The only problem with working all titration problems that way is that not all titrations are 1:1 as these are; for example,
    NaOH + HCl ==> NaCl + H2O

    The more fundamental way, but a little long, and it works with all types of titrations is to convert to moles.
    1. Write the equation and balance it.
    NaOH + HCl => NaCl + H2P

    2. How many moles HCl do we have? That is M x L = 0.22M x 35.00 mL = 7.70 moles.

    3. Usint the coefficients in the balanced equation, convert moles HCl to moles NaOH.
    7.70moles HCl x (1 mole NaOH/1 mole HCl) = 7.70 x (1/1) = 7.70 moles NaOH.

    4. Then the definition of molarity is M = moles/L. 0.15M = 7.70moles/L and
    L = 7.70/0.15 = approximately 51.

  • chemistry -

    Thanks alot! So Ma x Va= Mb x Vb or Ma=MbVb/Va :)

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