A $75 \cdot m L$ $75mL$ volume of $0.100 \cdot m o l \cdot L^{- 1}$ $0.100molL^-1$ $H N O_{3} (a q)$ $HNO_3(aq)$ and a $50 \cdot m L$ $50mL$ volume of $0.150 \cdot m o l \cdot L^{- 1}$ $0.150molL^-1$ $B a {(O H)}_{2} (a q)$ $Ba(OH)_2(aq)$ are mixed...will the reaction proceed quantitatively...?

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A $75 \cdot m L$ $75mL$ volume of $0.100 \cdot m o l \cdot L^{- 1}$ $0.100molL^-1$ $H N O_{3} (a q)$ $HNO_3(aq)$ and a $50 \cdot m L$ $50mL$ volume of $0.150 \cdot m o l \cdot L^{- 1}$ $0.150molL^-1$ $B a {(O H)}_{2} (a q)$ $Ba(OH)_2(aq)$ are mixed...will the reaction proceed quantitatively...?

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Answer 1 · 2017-02-11T05:59:00

We need (i) a stoichiometric equation:

$B a {(O H)}_{2} (s) + 2 H N O_{3} (a q) \to B a {(N O_{3})}_{2} (a q) + 2 H_{2} O (l)$ $Ba(OH)_2(s) + 2HNO_3(aq) rarr Ba(NO_3)_2(aq) + 2H_2O(l)$

Explanation:

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

$Moles of nitric acid:$ $"Moles of nitric acid:"$ $75 \times 10^{- 3} \cdot L \times 0.100 \cdot m o l \cdot L^{- 1}$ $75xx10^-3*Lxx0.100*mol*L^-1$ $= 7.50 \times 10^{- 3} \cdot m o l$ $= 7.50xx10^-3*mol$ .

$Moles of barium hydroxide:$ $"Moles of barium hydroxide:"$ $50 \times 10^{- 3} \cdot L \times 0.150 \cdot m o l \cdot L^{- 1}$ $50xx10^-3*Lxx0.150*mol*L^-1$ $= 7.50 \times 10^{- 3} \cdot m o l$ $= 7.50xx10^-3*mol$ .

And thus there is an equivalent quantity of $1.50 \times 10^{- 2} \cdot m o l$ $1.50xx10^-2*mol$ 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 $150 \cdot m L$ $150*mL$ 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.

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A $75 \cdot m L$ $75mL$ volume of $0.100 \cdot m o l \cdot L^{- 1}$ $0.100molL^-1$ $H N O_{3} (a q)$ $HNO_3(aq)$ and a $50 \cdot m L$ $50mL$ volume of $0.150 \cdot m o l \cdot L^{- 1}$ $0.150molL^-1$ $B a {(O H)}_{2} (a q)$ $Ba(OH)_2(aq)$ are mixed...will the reaction proceed quantitatively...?

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Explanation:

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A 75⋅mL75*mL volume of 0.100⋅mol⋅L−10.100*mol*L^-1 HNO3(aq)HNO_3(aq) and a 50⋅mL50*mL volume of 0.150⋅mol⋅L−10.150*mol*L^-1 Ba(OH)2(aq)Ba(OH)_2(aq) are mixed...will the reaction proceed quantitatively...?

1 Answer

Explanation:

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A $75 \cdot m L$ $75mL$ volume of $0.100 \cdot m o l \cdot L^{- 1}$ $0.100molL^-1$ $H N O_{3} (a q)$ $HNO_3(aq)$ and a $50 \cdot m L$ $50mL$ volume of $0.150 \cdot m o l \cdot L^{- 1}$ $0.150molL^-1$ $B a {(O H)}_{2} (a q)$ $Ba(OH)_2(aq)$ are mixed...will the reaction proceed quantitatively...?