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The pH of the solution at 25 degree celsius of 1.3 × 10⁻⁶ moles of a sample of Sr(OH)₂ is 10.02.
The pH of any solution gives an idea about the acidic and basic nature of the solution and the equation of pH will be represented as:
pH + pOH = 14
Given that,
Moles of Sr(OH)₂ = 1.3 × 10⁻⁶ mol
Volume of solution = 25mL = 0.025L
The concentration of Sr(OH)₂ in terms of molarity = 1.3×10⁻⁶/0.025
= 5.2×10¯⁵M
Dissociation of Sr(OH)₂ takes place as:
Sr(OH)₂ → Sr²⁺ + 2OH⁻
From the stoichiometry of the reaction 1 mole of Sr(OH)₂ produces 2 moles of OH⁻.
Given that the base is a strong base and that it entirely dissociates into its ions, the hydroxide ion concentration is 5.2×10¯⁵×2 = 1.04×10¯⁴ M.
pOH = -log[OH⁻]
pOH = -log(1.04×10¯⁴)
pOH = 3.98
Now we put this value on the first equation we get,
pH = 14 - 3.98 = 10.02
Therefore, the value of pOH is 10.02.
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