Question #c87b5

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Question #c87b5

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Answer 1 · 2017-12-27T14:14:23

In some cases.

More often, when a strong acid results with a strong base, they effectively "neutralize" each other,

$H C l (a q) + N a O H (a q) \to N a C l (a q) + H_{2} O (l)$ $HCl(aq) + NaOH(aq) to NaCl(aq) + H_2O(l)$

A scenario where your statement may be correct is using a hydrated strong base salt (e.g. $S r {(O H)}_{2} \cdot 8 H_{2} O$ $Sr(OH)_2*8H_2O$ ) and dissolving it in a strong acid.

Answer 2 · 2017-12-27T20:33:56

Yes, this is one of the general types of a double displacement reactions.

Explanation:

A double displacement reaction where;

A salt with an acid that forms a salt of the acid and a second acid that is volatile; i.e.,
$2 K N O_{3} (a q) + H_{2} S O_{4} (a q) \to K_{2} S O_{4} (a q) + 2 H N O_{3} (g)$ $2KNO_3(aq)+H_2SO_4(aq)->K_2SO_4(aq)+color(red)(2HNO_3(g)$ (balanced)
Note:
$H N O_{3} is one of the a highly volatile acids$ $color(red)(HNO_3) " is one of the a highly volatile acids"$
Same reaction of a salt with an acid that may yield a compound that can be decomposed into a $gas$ $color(red)("gas")$ and a $liquid$ $color(blue)("liquid")$ ; that is,
$C a C O_{3} (a q) + 2 H C l (a q) \to C a C l_{2} (a q) + H_{2} C O_{3} (a q)$ $CaCO_3(aq)+2HCl(aq)->CaCl_2(aq)+color(orange)(H_2CO_3)(aq)$ (balanced)
where:
$H_{2} C O_{3} (a q) \to C O_{2} (g) + H_{2} O (l)$ $color(orange)(H_2CO_3)(aq)->color(red)(CO_2(g))+color(blue)(H_2O(l))$

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Question #c87b5

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