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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