Question c87b5

Dec 27, 2017

In some cases.

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

$H C l \left(a q\right) + N a O H \left(a q\right) \to N a C l \left(a q\right) + {H}_{2} O \left(l\right)$

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

Dec 27, 2017

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

Explanation:

A double displacement reaction where;

1. A salt with an acid that forms a salt of the acid and a second acid that is volatile; i.e.,
2KNO_3(aq)+H_2SO_4(aq)->K_2SO_4(aq)+color(red)(2HNO_3(g)#(balanced)
Note:
$\textcolor{red}{H N {O}_{3}} \text{ is one of the a highly volatile acids}$
2. Same reaction of a salt with an acid that may yield a compound that can be decomposed into a $\textcolor{red}{\text{gas}}$ and a $\textcolor{b l u e}{\text{liquid}}$; that is,
$C a C {O}_{3} \left(a q\right) + 2 H C l \left(a q\right) \to C a C {l}_{2} \left(a q\right) + \textcolor{\mathmr{and} a n \ge}{{H}_{2} C {O}_{3}} \left(a q\right)$(balanced)
where:
$\textcolor{\mathmr{and} a n \ge}{{H}_{2} C {O}_{3}} \left(a q\right) \to \textcolor{red}{C {O}_{2} \left(g\right)} + \textcolor{b l u e}{{H}_{2} O \left(l\right)}$