Chemistry

A 0.2299 N acid is prepared by mixing 100 ml of 0.1234 N H2SO4 and 0.3200 N HCl. How much HCl is needed in the preparation of the solution?

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asked by Summer
  1. Let x = amount of 0.3200 N HCl needed.
    m.e. = # milliequivelents
    Then m.e. H2SO4 + m.e. HCl = m.e. 0.2299 solution.
    (100mL*0.1234N) + (XmL*0.3200N) = (100mL + xmL)*0.2299
    Solve for X.

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  2. 118.2

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    posted by lala

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