Question

In: Chemistry

A 0.4000 M solution of nitric acid is used to titrate 50.00 mL of 0.237 M...

A 0.4000 M solution of nitric acid is used to titrate 50.00 mL of 0.237 M barium hydroxide. Using this information, complete the following:

A.) Write a net ionic equation for the reaction that takes place during titration

B.) What species are present at the equivalence point?

C.) What volume is nitric acid is required to reach the equivalence point?

Expert Solution

it's titration between strong acid and strong base.

A. balanced chemical equation is

$Ba(OH)_{2}_{(aq)} + 2HNO_{3} _{(aq)} = Ba(NO_{3})_{2}_{(aq)} + 2H_{2}O_{(l)}$

now, break it into ionic form...

$Ba^{2+}_{(aq)}+2OH^{-}_{aq} + 2H^{+}_{(aq)} + 2NO_{3}^{-}_{(aq)} = Ba^{2+}_{(aq)} + 2NO_{3}^{-}_{(aq)} + 2H_{2}O_{(l)}$

cancel the spectator ions....
and net ionic equation is...

or,

B. It is clear from balanced chemical equation that at equivalence point a salt of Ba(NO₃)₂ and H₂O will be there.

C. From standard balanced chemical equation it is also clear that 2 mole of nitric acid required to titrate 1 mole of barium hydroxide to reach the equivalance point.

here, mole of barium hydroxide = molarity * vol ( L) = 0.237 * 0.05 = 0.01185

therefore, required moles of nitric acid = 2*0.01185 = 0.0237

it is assumed that V lit of nitric acid of 0.400 M is required to titrate barium hydroxide.

so, 0.0237 = 0.400 * V

or, V = 0.05925 lit =59.25 ml (answer)

queen_honey_blossom answered 2 years ago

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