# 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+

Relevance
• 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.