What is the concentration of all ionic species present when 100.mL of 0.200 M NH4OH and 100.mL of 0.200 M H2SO4 are mixed?

I got 0.100 M NH4+, 0.100 M SO4(2-), 0 M OH-, and 0.100 M H+

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  • 4 weeks ago

    Supposing additive volumes, and supposing complete ionization:

    (100. mL) x (0.200 M NH4OH) x (1 mol NH4{+} / 1 mol NH4OH) /

    (100. mL + 100. mL) = 0.100 M NH4{+}

    (100. mL) x (0.200 M H2SO4) x (1 mol SO4{2-} / 1 mol H2SO4) /

    (100. mL + 100. mL) = 0.100 M SO4{2-}

    H2SO4 + 2 NH4OH → 2 H2O + (NH4)2SO4

    H{+} + OH{-} → H2O

    (100. mL) x (0.200 M NH4OH) x (1 mol OH{-} / 1 mol NH4OH) = 20.0 mmol OH{-}

    (100. mL) x (0.200 M H2SO4) x (2 mol H{+} / 1 mol H2SO4) = 40.0 mmol H{+}

    20.0 millimoles of OH{-} ions would react completely with 20.0 x (1/1) = 20.0 millimoles of H{+} ions, but there are more H{+} ions present than that, so H{+} is in excess.

    ((40.0 mmol H{+} initially) - (20.0 mmol H{+} reacted)) / (100. mL + 100. mL) =

    0.100 mmol/mL = 0.100 M H{+}

    There are no appreciable OH{-} ions left -- they were all reacted by the excess H{+}.

    So you are correct four times.

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