A 75*mL volume of 0.100*mol*L^-1 HNO_3(aq) and a 50*mL volume of 0.150*mol*L^-1 Ba(OH)_2(aq) are mixed...will the reaction proceed quantitatively...?

Feb 11, 2017

We need (i) a stoichiometric equation:

$B a {\left(O H\right)}_{2} \left(s\right) + 2 H N {O}_{3} \left(a q\right) \rightarrow B a {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

Explanation:

And need (ii) equivalent quantities of the reagents we use:

$\text{Moles of nitric acid:}$ $75 \times {10}^{-} 3 \cdot L \times 0.100 \cdot m o l \cdot {L}^{-} 1$ $= 7.50 \times {10}^{-} 3 \cdot m o l$.

$\text{Moles of barium hydroxide:}$ $50 \times {10}^{-} 3 \cdot L \times 0.150 \cdot m o l \cdot {L}^{-} 1$ $= 7.50 \times {10}^{-} 3 \cdot m o l$.

And thus there is an equivalent quantity of $1.50 \times {10}^{-} 2 \cdot m o l$ hydroxide ion. Why?

And thus this question SHOULD NOT have been proposed. The given volume, the given quantity, of nitric acid will not neutralize the given quantity of barium hydroxide. A $150 \cdot m L$ volume of nitric acid of the given concentration is required. Note also that barium hydroxide is quite insoluble, and the question should have proposed an equivalent quantity of sodium or potassium hydroxide.