Respuesta :

form the molar concentration of the Ca(OH)2 which is a base we can deduct the pOH

pOH=-lg[OH-]

[OH-]=8.8 x 10-4 M

pOH=4

pH=14-pOH

pH=10

-lg[H+]=10

[H+]=10^-10

The concentration of Hydrogen ion [H⁺] in 8.8×10¯⁴ M Ca(OH)₂ is 5.68×10¯¹² M

 

The concentration of Hydrogen ion [H⁺] in a solution talks about the acidicity  the solution. Thus, we can obtain the concentration of Hydrogen ion [H⁺] in 8.8×10¯⁴ M Ca(OH)₂ solution as illustrated below:

Step 1: 

Data obtained from the question

Concentration of Ca(OH)₂ = 8.8×10¯⁴ M  

Concentration of Hydrogen ion [H⁺] =..?  

Step 2:

Determination of the concentration of Hydroxide ion [OH¯]

 Concentration of Ca(OH)₂ = 8.8×10¯⁴ M

Concentration of Hydroxide ion [OH¯] =?  

Ca(OH)₂ (aq)⇄Ca²⁺ (aq) + 2OH¯(aq)  

From the balanced equation above,  

1 mole Ca(OH)₂ produced 2 moles of OH¯.

Therefore, 8.8×10¯⁴ M Ca(OH)₂ will produce = 8.8×10¯⁴ × 2 = 1.76×10¯³ M

Thus, the Concentration of Hydroxide ion [OH¯] is 1.76×10¯³ M  

Step 3:

Determination of concentration of Hydrogen ion [H⁺].  Concentration of Hydroxide ion [OH¯] = 1.76×10¯³ M

Concentration of Hydrogen ion [H⁺] =..?    

[tex]H^{+} * OH^{-} = 1*10^{-14}\\H^{+} * 1.76^{-3} = 1*10^{-14}[/tex]

Divide both side by 1.76×10¯³  

[tex]H^{+} = \frac{1*10^{-14} }{1.76*10^{-3}}\\[/tex]

[H⁺] = 5.68×10¯¹² M  

Therefore, the concentration of Hydrogen ion [H⁺] in the solution is 5.68×10¯¹² M  

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