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Inessa [10]
3 years ago
6

Net ionic equation of HF(aq)+RbOH(aq)=H2O(l)+RbF(aq)

Chemistry
1 answer:
Tamiku [17]3 years ago
4 0

net ionic equation

H⁺(aq)+OH⁻(aq)⇒H₂O(l)

<h3>Further explanation  </h3>

The electrolyte in the solution produces ions.  

The equation of a chemical reaction can be expressed in the equation of the ions  

For strong electrolytes (the ionization rate = 1) is written in the form of separate ions, while the weak electrolyte (degree of ionization <1) is still written as an un-ionized molecule  

In the ion equation, there is a spectator ion that is the ion which does not react because it is present before and after the reaction  

When these ions are removed, the ionic equation is called the net ionic equation  

For gases and solids including water (H₂O) can be written as an ionized molecule  

So only the dissolved compound is ionized ((expressed in symbol aq)  

Reaction

HF(aq)+RbOH(aq)⇒H₂O(l)+RbF(aq)

ionic equation

H⁺(aq)+F⁻(aq)+Rb⁺(aq)+OH⁻(aq)⇒H₂O(l)+Rb⁺(aq)+F⁻(aq)

net ionic equation

H⁺(aq)+OH⁻(aq)⇒H₂O(l)

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What is the ph of a solution that has a [h ] = 0. 0039 m?
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What is the solubility of m(oh)2 in a 0.202 m solution of m(no3)2? ksp 4.45*10^-12?
umka2103 [35]

<span>M(NO</span>₃<span>)</span>₂<span> fully separates into M</span>²⁺<span> and NO</span>₃<span> </span>²⁻<span> </span><span>and M(OH)</span>₂<span> partially separates as <span>M</span></span>²⁺<span><span> and 2OH</span></span>⁻

<span>M(NO</span>₃<span>)</span>₂<span><span>     </span>→        M</span>²⁺<span>      +          2NO</span>₃²⁻

<span>0.202 M                 0.202 M</span>

 

<span>     M(OH)</span>₂<span>(s)        ↔        <span>M</span></span>²⁺<span><span> (aq) +          2OH</span></span>⁻<span><span>(aq)</span></span>

<span>I                                         -                            -</span>

<span>C   -X                                +X                        +2X</span>

<span>E                                        X                          2X</span>

 

<span>Ksp                  =          [M</span>²⁺<span> (aq)] [OH</span>⁻<span>(aq)]</span>²

4.45 * 10∧-12        =          (0.202 + X ) (2X)²

Since X is very small, (0.202 + X ) = 0.202

<span>4.45 * 10<span>-12        </span>=          0.202 * 4X</span>²

<span>            X          =          2.347 </span>× 10∧-6 M

Hence the solubility of <span>M(OH)2 is 2.347 </span>× 10∧-6 M

6 0
3 years ago
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